Maharashtra Board 10th Class Maths Part 2 Practice Set 5.3 Solutions Chapter 5 Co-ordinate Geometry

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Practice Set 5.3 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x1, y1) = A (2, 3) and B (x2, y2) = B (4, 7)
Here, x1 = 2, x2 = 4, y1 = 3, y2 = 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 1
∴ The slope of line AB is 2.

ii. P (x1, y1) = P (-3, 1) and Q (x2, y2) = Q (5, -2)
Here, x1 = -3, x2 = 5, y1 = 1, y2 = -2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 2
∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x1, y1) = C (5, -2) and D (x2, y2) = D (7, 3)
Here, x1 = 5, x2 = 7, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 3
∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x1, y1) = L (-2, -3) and M (x2,y2) = M (-6, -8)
Here, x1 = -2, x2 = – 6, y1 = – 3, y2 = – 8
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 4
∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x1, y1) = E (-4, -2) and F (x2, y2) = F (6, 3)
Here,x1 = -4, x2 = 6, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 5
∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x1, y1) = T (0, -3) and S (x2, y2) = S (0, 4)
Here, x1 = 0, x2 = 0, y1 = -3, y2 = 4
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 6
∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 8
∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 9
∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 10
∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 11a
∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 12
∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 13
∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 14
∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 15
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 16
∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x1, y1) = R (1, -1), S (x2, y2) = S (-2, k)
Here, x1 = 1, x2 = -2, y1 = -1, y2 = k
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 17
But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x1, y1) = B (k, -5), C (x2, y2) = C (1, 2)
Here, x1 = k, x2 = 1, y1 = -5, y2 = 2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 18
But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 19
But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Maharashtra Board 10th Class Maths Part 2 Practice Set 5.2 Solutions Chapter 5 Co-ordinate Geometry

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Practice Set 5.2 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x1, y1) B (x2, y2) be the given points.
Here, x1 = -1, y1 = 7, x2 = 4, y2 = -3, m = 2, n = 3
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 1
∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x1, y1), Q (x2, y2) be the given points.
Here, x1 = -3, y1 = 7, x2 = 1, y2 = -4, a = 2, b = 1
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 2
∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x1,y1), Q (x2, y2) be the given points.
Here, x1 = -2, y1 = -5, x2 = 4, y2 = 3, a = 3, b = 4
By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 3
∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x1, y1), Q (x2, y2) be the given points.
Here,x1 = 2,y1 = 6, x2 = -4, y2 = 1, a = 1,b = 2
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 4
∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x1, y1), Q (x2, y2) and T (x, y) be the given points.
Here, x1 = -3, y1 = 10, x2 = 6, y2 = -8, x = -1, y = 6
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 5
∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x1, y1), B (x2, y2) and P (x, y) be the given points.
Here, x1 = 2, y1 =-3,
x = -2, y = 0
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 6
Point P is the midpoint of seg AB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 7
∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x1, y1), B (x2, y2) and P (x, y) be the given points.
Here, x1 = 8, y1 = 9, x2 = 1, y2 = 2, x = k, y = 7
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 8
∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x1, y1) = A (22, 20),
B (x2,y2) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 9
The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x1, y1) = A (-7, 6),
B (x2, y2) = B (2, -2),
C (x3, y3) = C(8, 5)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 10
∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x1 y1) = A (3, -5),
B (x2, y2) = B (4, 3),
C(x3, y3) = C(11,-4)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 11
∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x1, y1) = A (4, 7),
B (x2, y2) = B (8,4),
C (x3, y3) = C(7,11)
∴ By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 12
∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x1, y1) = A (-14, -19),
B(x2, y2) = B(3,5)
Let the co-ordinates of point C be (x3, y3).
G is the centroid.
By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 13
∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x1,y1) = A(h, -6),
B (x2, y2) = B(2, 3),
C (x3, y3) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 14
∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 15
∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 16
∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 17
Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 18
Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 19
Point D is the midpoint of seg AB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 20
∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 21
∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 22
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 23
∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 24
∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 25
∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 26
∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 27
∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.2 28
Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Maharashtra Board 10th Class Maths Part 2 Practice Set 6.2 Solutions Chapter 6 Trigonometry

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

Practice Set 6.2 Geometry 10th Std Maths Part 2 Answers Chapter 6 Trigonometry

Question 1.
A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Solution:
Let AB represent the height of the church and point C represent the position of the person.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 1
BC = 80 m
Angle of elevation = ∠ACB = 45°
In right angled ∆ABC,
tan 45° = \(\frac { AB }{ BC } \) … [By definition]
∴ 1 = \(\frac { AB }{ 80 } \)
∴ AB = 80m
∴ The height of the church is 80 m.

Question 2.
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( \(\sqrt { 3 }\) = 1.73)
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.
AB = 90 m
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 2
Angle of depression = ∠PAC = 60°
Now, ray AP || seg BC
∴ ∠ACB = ∠PAC … [Alternate angles]
∴ ∠ACB = 60°
In right angled ∆ABC,
tan 60° = \(\frac { AB }{ BC } \) … [By definition]
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 3
∴ The ship is 51.90 m away from the lighthouse.

Question 3.
Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 4
Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = 60°
AB = 10 m
BD= 12 m
In ꠸ABDM,
∠B = ∠D = 90°
∠M = 90° … [seg AM ⊥ seg CD]
∴ ∠A = 90° … [Remaining angle of ꠸ABDM]
∴ ꠸ABDM is a rectangle …. [Each angle is 90°]
∴ AM = BD = 12 m opposite sides
DM = AB = 10 m of a rectangle
In right angled ∆AMC,
tan 60° = \(\frac { CM }{ AM } \) …[By definition]
∴ \(\sqrt { 3 }\) = \(\frac { CM }{ 12 } \)
∴ CM = 12\(\sqrt { 3 }\) m
Now, CD = DM + CM … [C – M – D]
∴ CD = (10 + 12\(\sqrt { 3 }\))m
= 10 + 12 × 1.73
= 10 + 20.76 = 30.76
∴ The height of the second building is 30.76 m.

Question 4.
Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal.
Solution:
Let AB and CD represent the heights of two poles, and AC represent the length of the wire.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 5
Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = θ
AB = 7 m
CD = 18 m
AC = 22 m
In ꠸ABDM,
∠B = ∠D = 90°
∠M = 90° …[seg AM ⊥ seg CD]
∴ ∠A = 90° … [Remaining angle of ꠸ABDM]
∴ □ABDM is a rectangle. … [Each angle is 90°]
∴ DM = AB = 7 m … [Opposite sides of a rectangle]
Now, CD = CM + DM … [C – M – D]
∴ 18 = CM + 7
∴ CM = 18 – 7 = 11 m
In right angled ∆AMC,
sin θ = \(\frac { CM }{ AC } \) …..[By definition]
∴ sin θ = \(\frac { 11 }{ 22 } \) = \(\frac { 1 }{ 2 } \)
But, sin 30° = \(\frac { 1 }{ 2 } \)
∴ θ = 30°
∴ The angle made by the wire with the horizontal is 30°.

Question 5.
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Solution:
Let AB represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
AC is the broken part of the tree which takes position CD such that ∠CDB = 60°
∴ AC = CD …(i)
BD = 20 m
In right angled ∆CBD,
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 6
tan 60° = \(\frac { BC }{ BD } \) … [By definition]
∴ \(\sqrt { 3 }\) = \(\frac { BC }{ 20 } \)
∴ BC = 20\(\sqrt { 3 }\) m
Also, cos 60° = \(\frac { BC }{ CD } \) … [By definition]
∴ \(\frac { 1 }{ 2 } \) = \(\frac { 20 }{ CD } \)
∴ CD = 20 × 2 = 40 m
∴ AC = 40 m …[From(i)]
Now, AB = AC + BC ….[A – C – B]
= 40 + 20\(\sqrt { 3 }\)
= 40 + 20 × 1.73
= 40 + 34.6
= 74.6
∴ The height of the tree is 74.6 m.

Question 6.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (\(\sqrt { 3 }\) = 1.73)
Solution:
Let AB represent the height at which kite is flying and point C represent the point where the string is tied at the ground.
∠ACB is the angle made by the string with the ground.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 7
∠ACB = 60°
AB = 60 m
In right angled ∆ABC,
sin 60° = \(\frac { AB }{ AC } \) … [By definition]
Maharashtra Board Class 10 Maths Solutions Chapter 6 Trigonometry Practice Set 6.2 8
∴ AC = 40 \(\sqrt { 3 }\) = 40 × 1.73 = 69.20 m
∴ The length of the string is 69.20 m.

Maharashtra Board 10th Class Maths Part 2 Practice Set 5.1 Solutions Chapter 5 Co-ordinate Geometry

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Practice Set 5.1 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x1, y1) and B (x2, y2) be the given points.
∴ x1 = 2, y1 = 3, x2 = 4, y2 = 1
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 3
∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x1, y1 ) and Q (x2, y2) be the given points.
∴ x1 = -5, y1 = 7, x2 = -1, y2 = 3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 1
∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x1, y1) and S (x2, y2) be the given points.
∴ x1 = 0, y1 = -3, x2 = 0, y2 = \(\frac { 5 }{ 2 } \)
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 2
∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x1, y1) and M (x2, y2) be the given points.
∴ x1 = 5, y1 = -8, x2 = -7, y2 = -3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 4
∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x1,y1) and R (x2, y2) be the given points.
∴ x1 = -3, y1 = 6,x2 = 9,y2 = -10
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 5
∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x1, y1) and X (x2, y2) be the given points.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 6
∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 8
∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 9
On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 10
On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 11
On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 12
∴ (x + 3)2 + (-4)2 = (x- 1)2 + 42
∴ x2 + 6x + 9 + 16 = x2 – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 13
Consider, PQ2 + QR2 = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR2 = PQ2 + QR2 … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 14
PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 15

∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 17
∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 18
∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 19
Solution:
In ∆ABC, ∠B = 900
∴ (AB)2 + (BC)2 = [(Ac)2 …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x1,y1)
co-ordinate of A is (2, 3) = (x2, Y2)
Since, AB || to Y-axis,
d(A, B) = Y2 – Y1
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x1,y1)
co-ordinate of B is (2, 2) = (x2, y2)
Since, BC || to X-axis,
d(B, C) = x2 – x1
d(B,C) = 2 – -2 = 4
∴ AC2 = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Maharashtra Board 10th Class Maths Part 1 Practice Set 6.3 Solutions Chapter 6 Statistics

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

Practice Set 6.3 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics

Question 1.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 2
Here, the maximum frequency is 80.
∴ The modal class is 4 – 5.
L = lower class limit of the modal class = 4
h = class interval of the modal class = 1
f1 = frequency of the modal class = 80
f0 = frequency of the class preceding the modal class = 70
f2 = frequency of the class succeeding the modal class = 60
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 3
∴ The mode of the fat content is 4.33%.

Question 2.
Electricity used by some families is shown in the following table. Find the mode of use of electricity.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 20
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 5
Here, the maximum frequency is 100.
∴ The modal class is 60 – 80.
L = lower class limit of the modal class = 60
h = class interval of the modal class = 20
f1 = frequency of the modal class = 100
f0 = frequency of the class preceding the modal class = 70
f2 = frequency of the class succeeding the modal class = 80
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 6
∴ The mode of use of electricity is 72 units.

Question 3.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 7
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 8
Here, the maximum frequency is 35.
∴ The modal class is 9 – 11.
L = lower class limit of the modal class = 9
h = class interval of the modal class = 2
f1 = frequency of the modal class = 35
f0 = frequency of the class preceding the modal class = 20
f2 = frequency of the class succeeding the modal class = 18
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 9
∴ The mode of the supply of milk is 9.94 litres (approx.).

Question 4.
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 10
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 11
Here, the maximum frequency is 50.
The modal class is 9.5 – 14.5.
L = lower class limit of the modal class = 9.5
h = class interval of the modal class = 5
f1 = frequency of the modal class = 50
f0 = frequency of the class preceding the modal class = 32
f2 = frequency of the class succeeding the modal class = 36
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 12
∴ The mode of the ages of the patients is 12.31 years (approx.).

Maharashtra Board 10th Class Maths Part 1 Practice Set 6.2 Solutions Chapter 6 Statistics

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

Practice Set 6.2 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics

Statistics Practice Set 6.2 Question 1.
The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 2
Cumulative frequency which is just greater than (or equal) to 500 is 650.
∴ The median class is 10 – 12.
Now, L = 10, f = 500, cf = 150, h = 2
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 3
∴ The median of the number of hours the workers work is 11.4 hours.

10th Class Algebra Practice Set 6.2 Question 2.
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 4
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 5
Here, total frequency = ∑fi = N = 250
∴ \(\frac { N }{ 2 } \) = \(\frac { 250 }{ 2 } \) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 153.
∴ The median class is 150 – 200.
Now, L = 150, f = 90, cf = 63, h = 50
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 6
∴ The median of the given data is 184 mangoes (approx).

Statistics Class 10 Practice Set 6.2 Question 3.
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 7
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 8
Here, total frequency = ∑fi = N = 200
∴ \(\frac { N }{ 2 } \) = \(\frac { 200 }{ 2 } \) = 100
Cumulative frequency which is just greater than (or equal) to 100 is 184.
∴ The median class is 74.5 – 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 9
∴ The median of the given data is 75 km/hr (approx.).

Practice Set 6.2 Geometry Class 10 Question 4.
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 10
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 11
Cumulative frequency which is just greater than (or equal) to 52.5 is 67.
∴ The median class is 50 – 60.
Now, L = 50, f = 20, cf = 47, h = 10
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 12
∴ The median of the productions is 52750 bulbs (approx.).

Practice Set 6.2 Question 1.
If the number of scores is odd, then the (\(\frac { n+1 }{ 2 } \))th score is the median of the data. That is, the number of scores below as well as above \({ K }_{ \frac { n+1 }{ 2 } }\) is \(\frac { n-1 }{ 2 } \) Verify the fact by taking n = 2m + I. (Textbk pg. no. 139)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 13
The sequence of the terms of scores is 1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1
Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1
i.e. to prove
number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1,d = 1, tn1 = m
tn1 = a + (n1 – 1) d
∴ m = 1 + (n1 – 1)1
∴ m = 1 + n1 – 1
∴ m = n1
Consider the R.H.S. of equation (ii)
The sequence is an A.P. with a = m + 2, d = 1, tn2 = 2m + 1
tn2 = a + (n2 – 1)d
∴ 2m + 1 = m + 2 + (n2 – 1)1
∴ 2m + 1 = m + n2 + 1
∴ m = n2
∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = \(\frac { n-1 }{ 2 } \)
∴ m + 1 is the middle term if the number of scores is 2m + 1.

Question 2.
If the number of the scores is even, then the mean of the middle two terms is the median. This is because the number of terms below \({ K }_{ \frac { n }{ 2 } }\) and above \({ K }_{ \frac { n+2 }{ 2 } }\) is equal, which is \(\frac { n-2 }{ 2 } \). Verify this by taking n = 2m. (Textbook pg. no. 139)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.2 14
The sequence of the terms of scores is 1, 2, 3 … m – 1, m, m + 1, m + 2,…., 2m
Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.
i.e. to prove
number of terms from 1 to m – 1 = number of terms from m + 2 to 2m …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, tn1 = m – 1
tn1 = a + (n1 – 1)d
∴ m – 1 = 1 + (n1 – 1)1
∴m – 1 = 1 + n1 – 1
∴ n1 = m – 1
Consider the R.H.S. of equation (i)
The sequence is an A.P. with a = m + 2, d = 1, tn2= 2m
tn2= a + (n2 – 1) d
∴ 2m = m + 2 + (n2 – 1)1
∴ 2m = m + 2 + n2 – 1
∴ n2 = m – 1
∴ number of terms from 1 to m – 1 = number of terms from m + 2 to 2m = m – 1 = \(\frac { n-2 }{ 2 } \)
∴ m and m + 1 are the middlemost terms if the number of scores is 2m.

Maharashtra Board 10th Class Maths Part 2 Problem Set 3 Solutions Chapter 3 Circle

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Problem Set 3 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Problem Set 3 Geometry Class 10 Question 1.
Four alternative answers for each of the following questions are given. Choose the correct alternative.

i. Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centres?
(A) 4.4 cm
(B) 8.8 cm
(C) 2.2 cm
(D) 8.8 or 2.2 cm
Answer: (D)
Two circles can touch each other internally or externally.
∴ Distance between centres = 5.5 + 3.3 or 5.5 – 3.3 = 8.8 or 2.2

ii. Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle?
(A) 6 cm
(B) 12 cm
(C) 24 cm
(D) can’t say
Answer: (B)
PQ is the radius = 12 cm
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 1

iii. A circle touches all sides of a parallelogram. So the parallelogram must be a __________.
(A) rectangle
(B) rhombus
(C) square
(D) trapezium
Answer: (B)
꠸ABCD is a rhombus.
Note: It cannot be square as the angles are not mentioned as 90°.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 2

iv. Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.
(A) 25 cm
(B) 24 cm
(C) 7 cm
(D) 14 cm
Answer: (C)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 3
In ∆OAB, ∠B = 90° [Tangent theorem]
∴ OA2 = OB2 + AB2 [Pythagoras theorem]
∴ 12.52 = OB2 + 122
∴ OB2 = 156.25- 144
∴ OB = \(\sqrt { 12.25 }\) = 3.5 cm
∴ Diameter = 2 × OB = 2 × 3.5 = 7 cm

v. If two circles are touching externally, how many common tangents of them can be drawn?
(A) One
(B) Two
(C) Three
(D) Four
Answer: (C)
line l, line m and line n are the tangents.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 4

vi. ∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find m(arc ACB).
(A) 65°
(B) 130°
(C) 295°
(D) 230°
Answer: (D)
m∠ACB = \(\frac { 1 }{ 2 } \) m(arc AB) [Inscribed angle theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 5
∴ m(arc AB) = 2 m∠ACB =
= 2 × 65
= 130°
m(arc ACB) = 360° – m(arc AB) [Measure of a circle is 360°]
= 360° – 130°
= 230°

vii. Chords AB and CD of a circle intersect inside the circle at point E. If AE = 5.6, EB = 10, CE = 8, find ED.
(A) 7
(B) 8
(C) 11.2
(D) 9
Answer: (A)
Chords AB and CD intersect
internally at E.
AE × EB = CE × ED [Theorem of internal division of chords]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 6
∴ 5.6 × 10 = 8 × ED
∴ ED = 7 units

viii. In a cyclic ꠸ABCD, twice the measure of ∠A is thrice the measure of ∠C. Find the measure of ∠C?
(A) 36°
(B) 72°
(C) 90°
(D) 108°
Answer: (B)
∠A + ∠C = 180° [Theorem of cyclic quadrilateral]
∴ 2∠A + 2∠C = 2 × 180° [Multiplying both sides by 2]
∴ 3∠C + 2∠C = 360° [∵ 2∠A = 3∠C]
∴ 5∠C = 360°
∴ ∠C = 72°

ix. Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = 120°. No point, except point B, is common to the arcs. Which is the type of ∆ABC?
(A) Equilateral triangle
(B) Scalene triangle
(C) Right angled triangle
(D) Isosceles triangle
Answer: (A)
m(arc AB) + m(arc BC) + m(arc AC) = 360° [Measure of a circle is 360°]
∴ 120° + 120° + m (arc AC) = 360°
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 7
∴ m(arc AC) = 120°
∴ arc AB = arc BC = arc AC
∴ seg AB ≅ seg BC ≅ seg AC [Corresponding chords of congruents arcs of a circle are congruent]
∴ ∆ABC is an equilateral triangle.

x. Seg XZ is a diameter of a circle. Point Y lies in its interior. How many of the following statements are true?
(i) It is not possible that ∠XYZ is an acute angle.
(ii) ∠XYZ can’t be a right angle.
(iii) ∠XYZ is an obtuse angle.
(iv) Can’t make a definite statement for measure of ∠XYZ.
(A) Only one
(B) Only two
(C) Only three
(D) All
Answer: (C)
x. seg XZ is the diameter.
∴ ∠XWZ is a right angle. [Angle inscribed in a semicircle]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 8
Since, Y lies in the interior of ∆XWZ,
∴ ∠XYZ > 90°
i.e., ∠XYZ is an obtuse angle.

Problem Set 3 Question 2.
Line l touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following.
i. What is d(O, P) = ? Why?
ii. If d(O, Q) = 8 cm, where does the point Q lie?
iii. If d(O, R) = 15 cm, how many locations of point R are on line l ?
At what distance will each of them be from point P?
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 9
Solution:
i. seg OP is the radius of the circle.
∴ d(0, P) = 9 cm
ii. Here, 8 cm < 9 cm
∴ d(0, Q) < d(0, P)
∴ d(0, Q) < radius
Point Q lies in the interior of the circle.
iii. There can be two locations of point R on line l.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 10
d(0, R) = 15 cm
Now, in ∆OPR, ∠OPR = 90° [Tangent theorem]
∴ OR2 = OP2 + PR2 [Pythagoras theorem]
∴ 152 = 92 + PR2
∴ 225 = 81 + PR2
∴ PR2 = 225 – 81 = 144 [Taking square root of both sides]
∴ PR = \(\sqrt { 144 }\)
= 12 cm

Standard 10th Geometry Problem Set 3 Question 3.
In the adjoining figure, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = 6\(\sqrt { 3 }\) ,then find
i. Radius of the circle.
ii. Measures of ∠K and ∠M.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 11
Solution:
i. Line KL is the tangent to the circle at point L and seg ML is the radius. [Given]
∴ ∠MLK = 90°…………. (i) [Tangent theorem]
In ∆MLK, ∠MLK = 90°
∴ MK2 = ML2 + KL2 [Pythagoras theorem]
∴ 122 = ML2 + (6\(\sqrt { 3 }\))2
∴ 144 = ML2 + 108
∴ ML2 = 144 – 108
∴ ML2 = 36
∴ ML = \(\sqrt { 36 }\) = 6 units. [Taking square root of both sides]
∴ Radius of the circle is 6 units.

ii. We know that,
ML = \(\frac { 1 }{ 2 } \) MK
∴ ∠K = 30° …………… (ii) [Converse of 30° – 60° – 90° theorem]
In ∆MLK ,
∠L = 90° [From (i)]
∠K = 30° [From (ii)]
∴ ∠M = 60° [Remaining angle of ∆MLK]

10th Class Geometry Problem Set 3 Question 4.
In the adjoining figure, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and l(AB) = r. Prove that, □ABOC is a square.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 12
Given: O is the centre of circle.
seg AB and seg AC are the tangents, radius = r, /(AB) = r.
To prove: □ABOC is a square.
Construction: Draw seg OB and seg OC.
Proof:
seg AB and seg AC are the tangents to the circle. [Given]
∴ AB = AC [Tangent segment theorem]
But, AB = r [Given]
∴ AB = AC = r ……….. (i)
Also, OB = OC = r ……….. (ii) [Radii of the same circle]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 13
∴ AB = AC = OB = OC [From (i) and (ii)]
∴ □ABOC is a rhombus.
∠OBA = 90° [Tangent theorem]
∴ □ABOC is a square [A rhombus is a square, if one of its angles is a right angle]

Question 5.
In the adjoining figure, ꠸ABCD is a parallelogram. It circumscribes the circle with centre T. Points E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 14
Solution:
Let the values of DH and CF be x and y respectively.
[ AE = AH = 4.5
BE = BF = 5.5
DH = DG = x
CF = CG = y ] [Tangent segment theorem]
□ABCD is a parallelogram. [Given]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 15
∴ AB = CD [Opposite sides of a parallelogram]
∴ AE + BE = DG + CG [A – E – B, D – G – C]
∴ 4.5 + 5.5 = x + y
∴ x + y = 10 ……….. (i)
Also, AD = BC [Opposite sides of a parallelogram]
∴ AH + DH = BF + CF [A – H – D, B – F – C]
∴ 4.5 + x = 5.5 + y
∴ x – y = 1 ………… (ii)
Adding equations (i) and (ii), we get
2x = 11
∴ x = \(\frac { 11 }{ 2 } \) = 5.5
∴ AD = AH + DH [A – H – D]
= 4.5 + 5.5
∴ AD = 10 units

Question 6.
In the adjoining figure, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions, hence find the ratio MS : SR.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 16
i. Find the length of segment MT.
ii. Find the length of seg MN.
iii. Find the measure of ∠NSM.
Solution:
i. MT = 9 cm [Radius of the bigger circle]
ii. MT = MN + NT [M – N – T]
∴ 9 = MN + 2.5
∴ MN = 9 – 2.5
∴ MN = 6.5 cm
iii. seg MR is a tangent to the smaller circle and NS is its radius.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 17
∴ ∠NSM = 90° [Tangent theorem]
iv. In ∆NSM, ∠NSM = 90°
∴ MN2 = NS2 + MS2 [Pythagoras theorem]
∴ 6.52 = 2.52 + MS2
∴ MS2 = 6.52 – 2.52
= (6.5 + 2.5) (6.5 – 2.5) [∵ a2 – b2 = (a + b)(a – b)]
= 9 × 4=36
∴ MS = \(\sqrt { 36 }\) [Taking square root of both sides]
= 6 cm
But, MR = MS + SR [M – S – R]
∴ 9 = 6 + SR
∴ SR = 9 – 6
∴ SR = 3cm
Now, \(\frac { MS }{ SR } \) = \(\frac { 6 }{ 3 } \) = \(\frac { 2 }{ 1 } \)
∴ \(\frac { MS }{ SR } \) = 2 : 1

10th Std Geometry Circle Problem Set Question 7.
In the adjoining figure, circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA || radius YB. Fill in the blanks and complete the proof.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 18
Given: X and Y are the centres of circle.
To prove: radius XA || radius YB
Construction: Draw segments XZ and YZ.
Proof:
By theorem of touching circles, points X, Z,
Y are collinear.
∴ ∠XZA ≅ ∠BZY [Vertically opposite angles]
Let ∠XZA = ∠BZY = a …………….. (i)
Now, seg XA seg XZ [Radii of the same circIe]
∴ ∠XAZ ≅∠XZA = a …………….. (ii) [Isosceles triangle theorem]
Similarly, seg YB ≅ seg YZ [Radii of the same circie]
∴ ∠BZY = ∠ZBY = a …………….. (iii) [Isosceles triangle theorem]
∴ ∠XAZ = ∠ZBY [From (i), (ii) and (iii)]
∴ radius XA || radius YB [Alternate angles test]

Circle Problem Set 3 Question 8.
In the adjoining figure, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it intersects smaller circle at point A. Prove that, seg AX || seg BY.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 19
Given: X and Y are the centres of the circle.
To prove: seg AX || seg BY
Proof:
In ∆XAZ,
seg XA ≅ seg XZ [Radii of the same circle]
∴ ∠XZA ≅ ∠XAZ ………… (i) [Isosceles triangle theorem]
Also, in ∆YBZ,
seg YB ≅ seg YZ [Radii of the sanie circle]
∴ ∠YZB ≅∠YBZ [Isosceles triangle theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 20
∴ ∠XZA ≅ ∠YBZ ………….. (ii) [Y – X – Z,B – A – Z]
∴ ∠XAZ ≅ ∠YBZ [From (i) and (ii)]
∴ seg AX || seg BY [Corresponding angles test]

Question 9.
In the adjoining figure, line l touches the circle with centre O at point P, Q is the midpoint of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12, find the radius of the circle.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 21
Solution:
Let the radius of the circle be r.
line l is the tangent to the circle and [Given]
seg OP is the radius.
∴ seg OP ⊥ line l [Tangent theorem]
chord RS || line l [Given]
∴ seg OP ⊥ chord RS
∴ QS = \(\frac { 1 }{ 2 } \) RS [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= \(\frac { 1 }{ 2 } \) × 12 = 6 cm
Also, OQ = \(\frac { 1 }{ 2 } \) OP [Q is the midpoint of OP]
= \(\frac { 1 }{ 2 } \) r
In ∆OQS, ∠OQS = 90° [seg OP ⊥ chord RS ]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 22
∴ OS2 = OQ2 + QS2 [Pythagoras theorem]
∴ r2 = (\(\frac { 1 }{ 2 } \)r)2 + 62
∴ r2 = \(\frac { 1 }{ 4 } \) r2 + 36
∴ r2 – \(\frac { 1 }{ 4 } \) r2 = 36
∴ \(\frac { 3 }{ 4 } \) r2 = 36
∴ r2 = \(\frac{36 \times 4}{3}\)
∴ r2 = 48
∴ r = \(\sqrt { 48 }\) [Taking square root of both sides]
= 4 \(\sqrt { 3 }\)
∴ The radius of the given circle is 4\(\sqrt { 3 }\) cm.

Question 10.
In the adjoining figure, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. Seg AP ⊥ line PQ and seg BQ ⊥ line PQ. Prove that seg CP ≅ seg CQ.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 23
Given: C is the centre of circle.
seg AB is the diameter of circle.
line PQ is a tangent, seg AP ⊥ line PQ and seg BQ ⊥ line PQ.
To prove: seg CP ≅ seg CQ
Construction: Draw seg CT, seg CP and seg CQ.
Proof:
Line PQ is the tangent to the circle at point T. [Given]
∴ seg CT ⊥ line PQ (i) [Tangent theorem]
Also, seg AP ⊥ line PQ,
seg BQ ⊥ line PQ [Given]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 24
∴ seg AP || seg CT || seg BQ [Lines perpendicular to the same line are parallel to each other]
∴ \(\frac { AC }{ CB } \) = \(\frac { PT }{ TQ } \) [Property of intercepts made by three parallel lines and their transversals]
But, AC = CB [Radii of the same circle]
∴ \(\frac { AC }{ AC } \) = \(\frac { PT }{ TQ } \)
∴ \(\frac { PT }{ TQ } \) = 1
∴ PT = TQ ………… (ii)
∴ In ∆CTP and ∆CTQ,
seg PT ≅ seg QT [From (ii)]
∠CTP ≅ ∠CTQ [From (i), each angle is of measure 90° ]
seg CT ≅ seg CT [Common side]
∴ ∆CTP ≅ ∆CTQ [SAS test of congruence]
∴ seg CP ≅ seg CQ [c.s.c.t]

Question 11.
Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.
Analysis:
Let the circles with centres A, B, C touch each other at points P, Q, R.
[∴ A – P – B
A – Q – C
B – R – C ] [Theorem of touching circles]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 25
∴ AB = AP + BP [A – P – B]
∴ AB = 3 + 3 = 6 cm
Similarly, BC = 6 cm, AC = 6 cm
So, if we construct triangle ∆ABC of side 6 cm each, then with A, B, C as the centres and radius 3 cm, the touching circles can be drawn.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 26

Question 12.
Prove that any three points on a circle cannot be collinear
Given: A circle with centre O.
Points A, B and C lie on the circle.
To prove: Points A, B and C are not collinear.
Proof:
OA = OB [Radii of the same circle]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 27
∴ Point O is equidistant from the endpoints A and B of seg AB.
∴ Point O lies on the perpendicular bisector of AB. [Perpendicular bisector theorem]
Similarly, we can prove that,
Point O lies on the perpendicular bisector of BC.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 28
∴ Point O is the point of intersection of perpendicular bisectors of AB and BC (i.e., circumcentre of ∆ABC) ……… (i)
Now, suppose that the points A, B, C are collinear.
Then, the perpendicular bisector of AB and BC will be parallel. [Perpendiculars to the same line are parallel]
∴ The perpendicular bisector do not intersect at O.
This contradicts statement (i) that the perpendicular bisectors intersect each other at O.
∴ Our supposition that A, B, C are collinear is false.
∴ Points A, B and C are non collinear points.

Question 13.
In the adjoining figure, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
i. What is the sum of ∠TAQ and ∠TSQ?
ii. Find the angles which are congruent to ∠AQP.
iii. Which angles are congruent to ∠QTS?
iv. ∠TAS = 65°, find the measures of ∠TQS and arc TS.
v. If ∠AQP = 42° and ∠SQR = 58° find measure of ∠ATS.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 29
Solution:
i. ꠸AQST is a cyclic quadrilateral. [Given]
∴ ∠TAQ + ∠TSQ = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
ii. line PR is the tangent and seg AQ is the secant. [Given]
∴∠AQP = \(\frac { 1 }{ 2 } \) m(arc AQ) [Theorem of angle between tangent and secant]
But,∠ASQ = \(\frac { 1 }{ 2 } \) m(arc AQ) [Inscribed angle theorem]
∴∠AQP ≅ ∠ZASQ
Similarly, we can prove that,
∠AQP ≅ ∠ATQ
iii. ∠QTS = \(\frac { 1 }{ 2 } \) m(arc QS) [Inscribed angle theorem]
But, ∠SQR = \(\frac { 1 }{ 2 } \) m(arc QS) [Theorem of angle between tangent and secant]
∴ ∠QTS ≅ ∠SQR
Also, ∠QTS = ∠QAS [Angles inscribed in the same arc]
iv. ∠TQS = ∠TAS [Angles inscribed in the same arc]
∴ ∠TQS = 65°
Now, ∠TQS = \(\frac { 1 }{ 2 } \) m(arc TS) [Inscribed angle theorem]
∴ 65°= \(\frac { 1 }{ 2 } \) m(arcTS)
∴ m(arc TS) = 65° × 2
∴ m(arc TS) = 130°
v. ∠AQP + ∠AQS + ∠SQR = 180° [Angles in a linear pair]
∴ 42° + ∠AQS + 58° = 180°
∴∠AQS + 100° = 180° ………… (i)
But, ꠸AQST is a cyclic quadrilateral.
∴ ∠AQS + ∠ATS = 180° ………… (ii)  [Theorem of cyclic quadrilateral]
∴ ∠ATS = 100° [From (i) and (ii)]

Question 14.
In the adjoining figure, O is the centre of a circle, chord PQ ≅ chord RS.
If ∠POR = 70° and (arc RS) = 80°, find
i. m (arc PR)
ii. m (arc QS)
iii. m (arc QSR).
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 30Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 31
Solution:
i. m(arc PR) = m∠POR [Definition of measure of arc]
∴ m(arc PR) = 70°
ii. chord PQ chord RS [Given]
∴ m(arc PQ) = m(arc RS) = 80° [Corresponding arcs of congruents chords of a circle are congruent]
Now, m(arc QS) + m(arc PQ) + m(arc PR) + m(arcRS) = 360°
∴ m(arc QS) + 80° + 70° + 80° = 360° [Measure of a circle is 360°]
∴ m(arc QS) + 230° = 360°
∴ m(arc QS) = 130°
iii. m(arc QSR) = m(arc QS) + m(arc SR) [Arc addition property]
= 130° + 80°
∴ m(arc QSR) = 210°

Question 15.
In the adjoining figure, m(arc WY) = 44°, m(arc ZX) = 68°, then
i. Find the measure of ∠ZTX.
ii. If WT = 4.8, TX = 8.0, YT = 6.4, find TZ.
iii. If WX = 25, YT = 8, YZ = 26, find WT.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 31
Solution:
i. Chords WX and YZ intersect internally at point T.
∴ ∠ZTX = \(\frac { 1 }{ 2 } \) m(arc WY) + m(arc ZX)]
= \(\frac { 1 }{ 2 } \) (44° + 68°)
= \(\frac { 1 }{ 2 } \) × 112°
∴ m ∠ZTX = 56°
ii. WT × TX = YT × TZ [Theorem of internal division of chords]
∴ 4.8 × 8.0 = 6.4 × TZ
∴ TZ = \(\frac{4.8 \times 8.0}{6.4}\)
∴ l(TZ) = 6.0 units
iii. Let the value of WT be x. [W – T – X]
WT + TX = WX
∴ x + TX = 25
∴ TX = 25 – x
Also, YT + TZ = YZ [Y – T – Z]
∴ 8 + TZ = 26
∴ TZ = 26 – 8
= 18 units
But, WT × TX = YT × TZ [Theorem of internal division of chords]
∴ x × (25 – x) = 8 × 18
∴ 25x – x2 = 144
∴ x2 – 25x + 144 = 0
∴ (x – 16)(x – 9) = 0
∴ x = 16 or x = 9
∴ WT = 16 units or WT = 9units

Question 16.
In the adjoining figure,
i. Mn (arc CE) = 54°, m (arc BD) = 23°, find measure of ∠CAE.
ii. If AB = 4.2,BC = 5.4, AE = 12.0, find AD.
iii. If AB = 3.6, AC = 9.0, AD = 5.4, find AE.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 32
Solution:
i. Chords BC and ED intersect each other externally at point A.
∴ ∠CAE = \(\frac { 1 }{ 2 } \) [m(arc CE) – m(arc BD)]
= \(\frac { 1 }{ 2 } \) (54° – 23°)
= \(\frac { 1 }{ 2 } \) × 31°
∴ m∠CAE = 15.5°
ii. AC = AB + BC [A – B – C]
= 4.2 + 5.4
= 9.6 units
Now, AB × AC = AD × AE [Theorem of external division of chords]
∴ 4.2 × 9.6 = AD × 12.0
∴ AD = \(\frac{4.2 \times 9.6}{12.0}\)
∴ AD = 3.36 units
iii. AB × AC = AD × AE [Theorem of external division of chords]
∴ 3.6 × 9.0 = 5.4 × AE
∴ AE = \(\frac{3.6 \times 9.0}{5.4}\)
∴ AE = 6 units

Geometry Problem Set 3 Question 17.
In the adjoining figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH.
Fill in the blanks and write the proof.
Given: chord EF || chord GH
To prove: chord EG = chord FH
Construction: Draw seg GF.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 33
Proof:
∠EFG = ∠FGH (i) Alternate angles
∠EFG = \(\frac { 1 }{ 2 } \) m (arcEG)] (ii) [Inscribed angle theorem]
∠FGH = \(\frac { 1 }{ 2 } \)m(arcFH) (iii) [Inscribed angle theorem]
∴ m(arcEG) = m (are FH) [From (i), (ii) and (iii)]
∴ chord EG ≅ chord FH The chords corresponding to congruent arcs of a circle are congruent

SSC Geometry Circle Chapter Solutions Pdf Question 18.
In the adjoining figure, P is the point of contact.
i. If m (arc PR) = 140°, L POR = 36°, find m (arc PQ)
ii. If OP = 7.2, OQ = 3.2, find OR and QR
iii. If OP = 7.2, OR = 16.2, find QR.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 34
Solution:
i. ∠PQR m(arc PR) [Inscribed angle theorem]
= \(\frac { 1 }{ 2 } \) × 140° = 70°
∠PQR is the exterior angle of ∆POQ. [Remote interior angle theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 35
∴ ∠PQR = ∠POQ + ∠QPO [R – Q – O]
∴ 70° = ∠POR + ∠QPO
∴ 70 = 36° + ∠QPO
∴ ∠QPO = 70° – 36° = 340
Now, ray OP is tangent at point P and segment PQ is a secant.
∴ ∠QPO = \(\frac { 1 }{ 2 } \) m(arcPQ) [Theorem of angle between tangent and secant]
∴ 34° = \(\frac { 1 }{ 2 } \) m(arc PQ)
∴ m(arc PQ) = 68°
ii. Here, OP = 7.2, OQ = 3.2
Line OP is the tangent at point P [Given]
and seg OR is the secant.
∴ OP2 = OQ × OR [Tangent secant segments theorem]
∴ 7.22 = 3.2 × OR
∴ 51.84 = 3.2 × OR
∴ OR \(\frac { 51.84 }{ 3.2 } \)
∴ OR = 16.2 units
Now, OR = OQ + QR [O – Q – R]
∴ 16.2,= 3.2 + QR
∴ QR = 16.2 – 3.2
∴ QR = 13 units
iii. Here, OP = 7.2, OR = 16.2
OP2 = OQ × OR [Tangent secant segments theorem]
∴ 7.22 = OQ × 16.2
∴ OQ = \(\frac { 51.84 }{ 16.2 } \)
∴ OQ = 3.2 units
Now, OR = OQ + QR [O – Q – R]
∴ 16.2 = 3.2 + QR
∴ QR = 16.2 – 3.2
∴ QR = 13 units

Question 19.
In the adjoining figure, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅ seg AB.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 36
Given: Circles with centres C and D touch each other internally.
To prove: seg EA ≅ seg AB
Construction: Join seg ED and seg DA.
Proof:
E – C – D [Theorem of touching circles]
seg ED is the diameter of smaller circle.
∴∠EAD = 90° [Angle inscribed in a semicircle]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 37
∴ seg AD ⊥ chord EB
∴ seg EA ≅ seg AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord]

Question 20.
In the adjoining figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling the blanks.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 38
Given: seg AB is a diameter, seg CD bisects ∠ACB.
To prove: seg AD ≅ seg BD
Construction: Draw seg OD.
Proof:
∠ACB = 90° [Angle inscribed in a semicircle]
∠DCB = ∠DCA = 45° [CD is the bisector of ∠C]
m(arcDB) = 2∠DCA = 90° [Inscribed angle theorem]
∠DOB = m(arc DB) = 90° ………… (i) [Definition of measure of arc]
segOA ≅ segOB …………. (ii) [[Radii of the same circle]
∴ line OD is the perpendicular biscctor of [From (i) and (ii)]
seg AB.
∴ seg AD ≅ seg BD

10th Geometry Circle Question 21.
In the adjoining figure, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and d(0, L) = 5. Find the radius of the circle.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 39
Construction: Draw seg OK ⊥ chord MN. Join OM.
Solution:
seg OK ⊥chord MN [Construction]
∴ MK = \(\frac { 1 }{ 2 } \) MN [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= \(\frac { 1 }{ 2 } \) × 25
= 12.5 units
MK = ML + LK [M – L – K]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 40
∴ 12.5 = 9 + LK
∴ LK= 12.5 – 9 = 3.5 units
In ∆OKL, ∠OKL = 90°
∴ OL2 = KL2 + OK2 [Pythagoras theorem]
∴ 52 = 3.52 + OK2
∴ OK2 = 25 – 12.25 = 12.75
Now, in ∆OKM, ∠OKM = 90°
∴ OM2 = OK2 + MK2
= 12.75 + 12.52
= 12.75 + 156.25
= 169
∴ OM = \(\sqrt { 169 }\)
= 13 units [Taking square root of both sides]
∴ The radius of the given circle is 13 units.

Question 22.
In the adjoining figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q.
Prove that, ∠PRQ + ∠PSQ = 180°.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 41
Given: Two circles intersect each other at points S and R.
line PQ is a common tangent.
To prove: ∠PRQ + ∠PSQ = 180°
Proof:
Line PQ is the tangent at point P and seg PR is a secant.
∴ [∠RPQ = ∠PSR …………. (i)
and ∠PQR = ∠QSR] ………… (ii) [Tangent secant theorem]
In ∆ PQR,
∠PQR + ∠PRQ + ∠RPQ = 180° [Sum of the measures of angles of a triangle is 180°]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 42
∴ ∠QSR + ∠PRQ + ∠PSR = 180° [From (i) and (ii)]
∴ ∠PRQ + ∠QSR + ∠PSR = 180°
∴ ∠PRQ + ∠PSQ = 180° [Angle addition property]

Question 23.
In the adjoining figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that: seg SQ || seg RP.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 43
Given: Two circles intersect each other at points M and N.
To prove: seg SQ || seg RP
Construction: Join seg MN.
Proof:
□RMNP is a cyclic quadrilateral.
∴ ∠MRP = ∠MNQ …………. (i) [Corollary of cyclic quadrilateral theorem]
Also, □MNQS is a cyclic quadrilateral.
∴ ∠MNQ+ ∠MSQ = 180° [Theorem of cyclic quadrilateral]
∴ ∠MRP + ∠MSQ = 180° [From (i)]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 44
But, they are a pair of interior angles on the sarpe side of transversal RS on lines SQ and RP.
∴ seg SQ || seg RP [Interior angles test]

Question 24.
In the adjoining figure, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that □ABCD is cyclic.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 45
Given: Two circles intersect each other at A and E. seg BC and seg CD are the tangents to the circles.
To prove: □ABCD is cyclic.
Construction: Draw AB, AE and AD.
Proof:
[∠EBC = ∠BAE (i)
∠EDC = ∠DAE ] (ii) [Tangent secant theorem]
In ∆BCD,
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 46
∠DBC + ∠BDC + ∠BCD = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠EBC + ∠EDC + ∠BCD = 180° (iii) [B – E – D]
∴ ∠BAE + ∠DAE + ∠BCD = 180° [From (i), (ii) and (iii)]
∴ ∠BAD + ∠BCD = 180° [Angle addition property]
∴ □ABCD is cyclic. [Converse of cyclic quadrilateral theorem]

Question 25.
In the adjoining figure, seg AD ⊥ side BC, seg BE ⊥ side AC, seg CF ⊥ side AB. Point O is the orthocentre. Prove that, point O is the incentre of ∆DEF.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 47
Given: seg AD ⊥ side BC,
seg BE ⊥ side AC,
seg CF ⊥ side AB.
To prove: Point O is the incentre of ∆DEF.
Construction: Draw DE, EF and DF.
Proof:
∠OFA = ∠OEA = 90° [Given]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 48
Now, ∠OFA + ∠OEA = 90° + 90°
∴ ∠OFA + ∠OEA = 180°
∴ □OFAE is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
∴ Points O, F, A, E are concyclic points.
∴ seg 0E subtends equal angles ∠OFE and ∠OAE on the same side of OE.
∴ ∠OFE = ∠OAE ……… (i)
∠OFB ∠ODB = 90° [Given]
Now, ∠OFB + ∠ODB = 90° + 90°
∴ ∠OFB + ∠ODB = 180°
∴ ꠸OFBD is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
∴ Points O, F, B, D are concyclic points.
∴ seg OD subtends equal angles ∠OFD and
∠OBD on the same side of OD.
∠OFD = ∠OBD ………….. (ii)
In ∆AEO and ∆BDO,
∠AEO = ∠BDO [Each angle is 90°]
∠AOE = ∠BOD [Vertically opposite angles]
∴ ∆AEO ~ ∆BDO [AA test of similarity]
∴ ∠OAE = ∠OBD …………….. (iii) [Corresponding angles of similar triangles]
∴ ∠OFE = ∠OFD [From (i), (ii) and (iii)]
∴ ray FO bisects ∠EFD.
Similarly, we can prove ray EO and ray DO bisects ∠FED and ∠FDE respectively.
∴ Point O is the intersection of angle bisectors of ∠D, ∠E and ∠F of ∆DEF.
∴ Point O is the incentre of ∆DEF.

Maharashtra Board 10th Class Maths Part 2 Practice Set 3.5 Solutions Chapter 3 Circle

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.5 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Practice Set 3.5 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Question 1.
In the adjoining figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
Solution:
i. Ray PQ is a tangent to the circle at point Q and seg PS is the secant. [Given]
∴ PR × PS = PQ2 [Tangent secant segments theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 1
∴ 8 × PS = 122
∴ 8 × PS = 144
∴ PS = \(\frac { 144 }{ 8 } \)
∴ PS = 18 units
ii. Now, PS = PR + RS [P – R – S]
∴ 18 = 8 + RS
∴ RS = 18 – 8
∴ RS = 10 units

Question 2.
In the adjoining figure, chord MN and chord RS intersect at point D.
i. If RD = 15, DS = 4, MD = 8 find DN
ii. If RS = 18, MD = 9, DN = 8 find DS
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5
Solution:
i. Chords MN and RS intersect internally at point D. [Given]
∴ MD × DN = RD × DS [Theorem of internal division of chords]
∴ 8 × DN = 15 × 4
∴ DN = \(\frac{15 \times 4}{8}\)
∴ DN = 7.5 units
ii. Let the value of RD be x.
RS = RD + DS [R – D – S]
∴ 18 = x + DS
∴ DS = 18 – x
Now, MD × DN = RD × DS [Theorem of internal division of chords]
∴ 9 × 8 = x(18 – x)
∴ 72 = 18x – x2
∴ x2 – 18x + 72 = 0
∴ x2 – 12x – 6x + 72 = 0
∴ x (x – 12) – 6 (x – 12) = 0
∴ (x – 12) (x – 6) = 0
∴ x – 12 = 0 or x – 6 = 0
∴ x = 12 or x = 6
∴ DS = 18 – 12 or DS = 18 – 6
∴ DS = 6 units or DS = 12 units

Question 3.
In the adjoining figure, O is the centre of the circle and B is a point of contact. Seg OE ⊥ seg AD, AB = 12, AC = 8, find
i. AD
ii. DC
iii. DE.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 3
Solution:
i. Line AB is the tangent at point B and seg AD is the secant. [Given]
∴ AC × AD = AB2 [Tangent secant segments theorem]
∴ 8 × AD = 122
∴ 8 × AD = 144
∴ AD = \(\frac { 144 }{ 8 } \)
∴ AD = 18 units
ii. AD = AC + DC [A – C – D]
∴ 18 = 8 + DC
∴ DC = 18 – 8
∴ DC = 10 units
iii. seg OE ⊥ seg AD [Given]
i.e. seg OE ⊥ seg CD [A – C – D]
∴ DE = \(\frac { 1 }{ 2 } \) DC [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= \(\frac { 1 }{ 2 } \) × 10
∴ DE = 5 units

Question 4.
In the adjoining figure, if PQ = 6, QR = 10, PS = 8, find TS.
Solution:
PR = PQ + QR [P-Q-R]
∴ PR = 6 + 10 = 16 units
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 4
Chords TS and RQ intersect externally at point P.
PQ × PR = PS × PT [Theorem of external division of chords]
∴ 6 × 16 = 8 × PT
∴ PT = \(\frac{6 \times 16}{8}\) = 12 units
But, PT = PS + TS [P – S – T]
∴ 12 = 8 + TS
∴ TS = 12 – 8
∴ TS = 4 units

Question 5.
In the adjoining figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE × GE = 4r2.
Given: seg EF is the diameter.
seg DF is a tangent to the circle,
radius = r
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 5
To prove: DE × GE = 4r2
Construction: Join seg GF.
Proof:
seg EF is the diameter. [Given]
∴ ∠EGF = 90° (i) [Angle inscribed in a semicircle]
seg DF is a tangent to the circle at F. [Given]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 6
∴ ∠EFD = 90° (ii) [Tangent theorem]
In ∆DFE,
∠EFD = 90 ° [From (ii)]
seg FG ⊥ side DE [From (i)]
∴ ∆EFD ~ ∆EGF [Similarity of right angled triangles]
∴ \(\frac { EF }{ GE } \) = \(\frac { DE }{ EF } \) [Corresponding sides of similar triangles]
∴ DE × GE = EF2
∴ DE × GE = (2r)2 [diameter = 2r]
∴ DE × GE = 4r2

Maharashtra Board Class 10 Maths Chapter 3 Circle Intext Questions and Activities

Question 1.
Theorem: If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc. (Textbook pg.no. 75 and 76)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 7
Given: ∠ABC is any angle, whose vertex B lies on the circle with centre M.
Line BC is tangent at B and line BA is secant intersecting the circle at point A.
Arc ADB is intercepted by ∠ABC.
To prove: ∠ABC = \(\frac { 1 }{ 2 } \) m(arc ADB)
Proof:
Case I: Centre M lies on arm BA of ∠ABC.
∠MBC = 90° [Trangnet theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 8
i.e. ∠ABC 90° (i) [A – M – B]
arc ADB is a semicircular arc.
∴ m(arc ADB) = 180° (ii) [Measure ofa semicircle is 180°]
∴ ∠ABC = \(\frac { 1 }{ 2 } \) m(arc ADB) [(From (i) and (ii)]

Case II: Centre M lies in the exterior of ∠ABC.
Draw radii MA and MB.
∴ ∠MBA = ∠MAB [Isosceles triangle theorem]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 9
Let, ∠MHA = ∠MAB =x, ∠ABC = y In ∆ABM,
∠AMB + ∠MBA + ∠MAB = 180° [Sum of the measures of all the angles of a triangle is 1800]
∴ ∠AMB + x + x = 180°
∴ ∠AMB = 180° – 2x …… (i)
Now, ∠MBC = ∠MBA + ∠ABC [Angle addition property]
∴ 90° = x + y [Tangent theorem]
∴ x = 90° – y ……(ii)
∠AMB = 180° – 2 (90° – y) [From (i) and (ii)]
∴ ∠AMB = 180° – 180° + 2y
∴ 2y = ∠AMB
∴ y = \(\frac { 1 }{ 2 } \) ∠AMB
∴ ∠ABC = \(\frac { 1 }{ 2 } \) ∠AMB
∴ ∠ABC = \(\frac { 1 }{ 2 } \) m(arc ADB) [Definition of measure of minor arc]

Case III: Centre M lies in the interior of ∠ABC.
Ray BE is the opposite ray of ray BC.
Now, ∠ABE = \(\frac { 1 }{ 2 } \) m (arc AFB) (i) [Proved in case II]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 10
∠ABC + ∠ABE = 180° [Angles in a linear pair]
∴ 180 – ∠ABC = ∠ABE
∴ 180 – ∠ABC = \(\frac { 1 }{ 2 } \) m(arc AFB) [From (i)]
= \(\frac { 1 }{ 2 } \) [360 – m (arc ADB)]
∴ 180 – ∠ABC = 180 – \(\frac { 1 }{ 2 } \) m(arc ADB)
∴ -∠ABC = – \(\frac { 1 }{ 2 } \) m(arc ADB)
∴ ∠ABC = \(\frac { 1 }{ 2 } \) m(arc ADB)

Question 2.
We have proved the above theorem by drawing seg AC and seg DB. Can the theorem be proved by drawing seg AD and seg CB, instead of seg AC and seg DB? (Textbook pg. no. 77)
Solution:
Yes, the theorem can be proved by drawing seg AD and seg CB.
Given: P is the centre of circle, chords AB and CD intersect internally at point E.
To prove: AE × EB = CE × ED
Construction: Draw seg AD and seg CB.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 11
Proof:
In ∆CEB and ∆AED,
∠CEB = ∠DEA [Vertically opposite angles]
∠CBE = ∠ADE [Angles inscribed in the same arc]
∴ ∆CEB ~ ∆AED [by AA test of similarity]
∴ \(\frac { CE }{ AE } \) = \(\frac { EB }{ ED } \) [Corresponding sides of similar triangles]
∴ AE × EB = CE × ED

Question 3.
In figure, seg PQ is a diameter of a circle with centre O. R is any point on the circle, seg RS ⊥ seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR2 = PS × SQ] (Textbook pg. no. 81)
Given: seg PQ is the diameter.
seg RS ⊥ seg PQ
To prove: SR2 = PS × SQ
Construction: Extend ray RS, let it intersect the circle at point T.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 12
Proof:
seg PQ ⊥ seg RS [Given]
∴ seg OS ⊥ chord RT [R – S – T, P – S – O]
∴ segSR = segTS (i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
Chords PQ and RT intersect internally at point S.
∴ SR × TS = PS × SQ [Theorem of internal division of chords]
∴ SR × SR = PS × SQ [From (i)]
∴ SR2 = PS × SQ

Question 4.
Theorem: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then
AE × EB = CE × ED. (Textbook pg. no. 78)
Given: Chords AB and CD of a circle intersect outside the circle in point E.
To prove: AE × EB = CE × ED
Construction: Draw seg AD and seg BC.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 13
Proof:
In ∆ADE and ∆CBE,
∠AED = ∠CEB [Common angle]
∠DAE ≅ ∠BCE [Angles inscribed in the same arc]
∴ ∆ADE ~ ∆CBE [AA testof similaritv]
∴ \(\frac { AE }{ CE } \) = \(\frac { ED }{ EB } \) [Corresponding sides of similar triangles]
∴ AE × EB = CE × ED

Question 5.
Theorem: Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA × EB = ET2.
Given: Secant through point E intersects the circle in points A and B.
Tangent drawn through point E touches the circle in point T.
To prove: EA × EB = ET2
Construction: Draw seg TA and seg TB.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 14
Proof:
In ∆EAT and ∆ETB,
∠AET ≅ ∠TEB [Common angle]
∠ETA ≅ ∠EBT [Theorem of angle between tangent and secant, E – A – B]
∴ ∆EAT ~ ∆ETB [AA test of similarity]
∴ \(\frac { EA }{ ET } \) = \(\frac { ET }{ EB } \) [Corresponding sides of similar triangles]
∴ EA × EB = ET2

Question 6.
In the figure in the above example, if seg PR and seg RQ are drawn, what is the nature of ∆PRQ. (Textbook pg. no, 81)
Answer:
seg PQ is the diameter of the circle.
∴ ∠PRQ = 90°
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 15
∴ ∆PRQ is a right angled triangle. [Angle inscribed in a semicircle]

Question 7.
Have you previously proved the property proved in the above example? (Textbook pg. no. 81)
Answer:
Yes. It is the theorem of geometric mean.
∆PSR ~ ∆RSQ [Similarity of right angled triangles]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.5 16
∴ \(\frac { PS }{ SR } \) = \(\frac { SR }{ SQ } \) [Corresponding sides of similar triangles]
∴ SR2 = PS × SQ

Maharashtra Board 10th Class Maths Part 2 Practice Set 3.4 Solutions Chapter 3 Circle

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.4 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Practice Set 3.4 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Question 1.
In the adjoining figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
i. ∠AOB
ii. ∠ACB
iii. arc AB
iv. arc ACB.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 1
Solution:
i. seg OA = seg OB = radius…… (i) [Radii of the same circle]
seg AB = radius…… (ii) [Given]
∴ seg OA = seg OB = seg AB [From (i) and (ii)]
∴ ∆OAB is an equilateral triangle.
∴ m∠AOB = 60° [Angle of an equilateral triangle]
ii. m ∠ACB = \(\frac { 1 }{ 2 } \) m ∠AOB [Measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre]
= \(\frac { 1 }{ 2 } \) × 60°
∴ m ∠ACB = 30°
iii. m(arc AB) = m ∠AOB [Definition of measure of minor arc]
∴ m(arc AB) = 60°
iv. m(arc ACB) + m(arc AB) = 360° [Measure of a circle is 360°]
∴ m(arc ACB) = 360° – m(arc AB)
= 360° – 60°
∴ m(arc ACB) = 300°

Question 2.
In the adjoining figure, ꠸PQRS is cyclic, side PQ ≅ side RQ, ∠PSR = 110°. Find
i. measure of ∠PQR
ii. m (arc PQR)
iii. m (arc QR)
iv. measure of ∠PRQ
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 2
Solution:
i. ꠸PQRS is a cyclic quadrilateral. [Given]
∴ ∠PSR + ∠PQR = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
∴ 110° + ∠PQR = 180°
∴ ∠PQR = 180° – 110°
∴ m ∠PQR = 70°
ii. ∠PSR= \(\frac { 1 }{ 2 } \) m (arcPQR) [Inscribed angle theorem]
110°= \(\frac { 1 }{ 2 } \) m (arcPQR)
∴ m(arc PQR) = 220°
iii. In ∆PQR,
side PQ ≅ side RQ [Given]
∴ ∠PRQ = ∠QPR [Isosceles triangle theorem]
Let ∠PRQ = ∠QPR = x
Now, ∠PQR + ∠QPR + ∠PRQ = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠PQR + x + x= 180°
∴ 70° + 2x = 180°
∴ 2x = 180° – 70°
∴ 2x = 110°
∴ \(x=\frac{110^{\circ}}{2}=55^{\circ}\)
∴ ∠PRQ = ∠QPR = 55°….. (i)
But, ∠QPR = \(\frac { 1 }{ 2 } \) m(arc QR) [Inscribed angle theorem]
∴ 55° = \(\frac { 1 }{ 2 } \) m(arc QR)
∴ m(arc QR) = 110°
iv. ∠PRQ = ∠QPR =55° [From (i)]
∴ m ∠PRQ = 55°

Question 3.
□ MRPN is cyclic, ∠R = (5x -13)°, ∠N = (Ax + 4)°. Find measures of ∠R and ∠N.
Solution:
□ MRPN is a cyclic quadrilateral. [Given]
∴ ∠R + ∠N = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 3
∴ 5x – 13 + 4x + 4 = 180
∴ 9x – 9 = 180
∴ 9x = 189
∴ x = \(\frac { 189 }{ 9 } \)
∴ x = 21
∴ ∠R = 5x – 13
= 5 × 21 – 13
= 105 – 13
= 92°
∠N = 4x + 4
= 4 × 21 +4
= 84 +4
= 88°
∴ m∠R = 92° and m ∠N = 88°

Question 4.
In the adjoining figure, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠RTS is an acute angle.
Given: O is the centre of the circle, seg RS is the diameter of the circle.
To prove: ∠RTS is an acute angle.
Construction: Let seg RT intersect the circle at point P. Join PS and PT.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 4
Proof:
seg RS is the diameter. [Given]
∴ ∠RPS = 90° [Angle inscribed in a semicircle]
Now, ∠RPS is the exterior angle of ∆PTS.
∴ ∠RPS > ∠PTS [Exterior angle is greater than the remote interior angles]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 5
∴ 90° > ∠PTS
i.e. ∠PTS < 90°
i.e, ∠RTS < 90° [R – P -T]
∠RTS is an acute angle.

Question 5.
Prove that, any rectangle is a cyclic quadrilateral.
Given: ꠸ABCD is a rectangle.
To prove: ꠸ABCD is a cyclic quadrilateral.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 6
Proof:
꠸XBCD is a rectangle. [Given]
∴ ∠A = ∠B = ∠C = ∠D = 90° [Angles of a rectangle]
Now, ∠A + ∠C = 90° + 90°
∴ ∠A + ∠C = 180°
∴ ꠸ABCD is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]

Question 6.
In the adjoining figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,
i. □ WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.
Given: seg YZ ⊥ side XW
seg XT ⊥ side WY
To prove: i. □WZPT is cyclic.
ii. Points X, Z, T, Y are concyclic.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 7
Proof:
i. segYZ ⊥ side XW [Given]
∴∠PZW = 90°…… (i)
seg XT I side WY [Given]
∴ ∠PTW = 90° ……(ii)
∠PZW + ∠PTW = 90° + 90° [Adding (i) and (ii)]
∴∠PZW + ∠PTW = 180°
∴□WZPT Ls a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
ii. ∠XZY = ∠YTX = 90° [Given]
∴ Points X and Y on line XY subtend equal angles on the same side of line XY.
∴ Points X, Z, T and Y are concydic. [If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic]

Question 7.
In the adjoining figure, m (arc NS) = 125°, m(arc EF) = 37°, find the measure of ∠NMS.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 8
Chords EN and FS intersect externally at point M.
m∠NMS = \(\frac { 1 }{ 2 } \) [m (arc NS) – m(arc EF)]
= \(\frac { 1 }{ 2 } \) (125° – 37°) = \(\frac { 1 }{ 2 } \) × 88°
∴ m∠NMS = 44°

Question 8.
In the adjoining figure, chords AC and DE intersect at B. If ∠ABE = 108°, m(arc AE) = 95°, find m (arc DC).
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4
Solution:
Chords AC and DE intersect internally at point B.
∴ ∠ABE = \(\frac { 1 }{ 2 } \) [m(arc AE) + m(arc DC)]
∴ 108° = \(\frac { 1 }{ 2 } \) [95° + m(arc DC)]
∴ 108° × 2 = 95° + m(arc DC)
∴ 95° + m(arc DC) = 216°
∴ m(arc DC) = 216° – 95°
∴ m(arc DC) = 121°

Maharashtra Board Class 10 Maths Chapter 3 Circle Intext Questions and Activities

Question 1.
Draw a sufficiently large circle of any radius as shown in the figure below. Draw a chord AB and central ∠ACB. Take any point D on the major arc and point E on the minor arc.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 10
i. Measure ∠ADB and ∠ACB and compare the measures.
ii. Measure ∠ADB and ∠AEB. Add the measures.
iii. Take points F, G, H on the arc ADB. Measure ∠AFB, ∠AGB, ∠AHB. Compare these measures with each other as well as with measure of ∠ADB.
iv. Take any point I on the arc AEB. Measure ∠AIB and compare it with ∠AEB. (Textbook pg, no. 64)
Answer:
i. ∠ACB = 2 ∠ADB.
ii. ∠ADB + ∠AEB = 180°.
iii. ∠AHB = ∠ADB = ∠AFB = ∠AGB
iv. ∠AEB = ∠AIB

Question 2.
Draw a sufficiently large circle with centre C as shown in the figure. Draw any diameter PQ. Now take points R, S, T on both the semicircles. Measure ∠PRQ, ∠PSQ, ∠PTQ. What do you observe? (Textbook pg. no.65)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 11
Answer:
∠PRQ = ∠PSQ = ∠PTQ = 90°
[Student should draw and verily the above answers.]

Question 3.
Prove that, if two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle. (Textbook pg. no. 72)
Given: Chord AB and chord CD intersect at E in the exterior of the circle.
To prove: ∠AEC = \(\frac { 1 }{ 2 } \) [m(arc AC) – m(arc BD)]
Construction: Draw seg AD.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 12
Proof:
∠ADC is the exterior angle of ∆ADE.
∴ ∠ADC = ∠DAE + ∠AED [Remote interior angle theorem]
∴ ∠ADC = ∠DAE + ∠AEC [C – D – E]
∴ ∠AEC = ∠ADC – ∠DAE ……(i)
∠ADC = \(\frac { 1 }{ 2 } \) m(arc AC) (ii) [Inscribed angle theorem]
∠DAE = \(\frac { 1 }{ 2 } \) m(arc BD) (iii) [A – B – E, Inscribed angle theorem]
∴ ∠AEC = \(\frac { 1 }{ 2 } \) m(arc AC) – \(\frac { 1 }{ 2 } \) m (arc BD) [From (i), (ii) and (iii)]
∴ ∠AEC = \(\frac { 1 }{ 2 } \) m(arc AC) – m (arc BD)

Question 4.
Angles inscribed in the same arc are congruent.
Write ‘given’ and ‘to prove’ with the help of the given figure.
Think of the answers of the following questions and write the proof.
i. Which arc is intercepted by ∠PQR ?
ii. Which arc is intercepted by ∠PSR ?
iii. What is the relation between an inscribed angle and the arc intercepted by it? (Textbook: pg. no. 68)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 13
Given: C is the centre of circle. ∠PQR and ∠PSR are inscribed in same arc PTR.
To prove: ∠PQR ≅ ∠PSR
Proof:
i. arc PTR is intercepted by ∠PQR.
ii. arc PTR is intercepted by ∠PSR.
iii. ∠PQR = \(\frac { 1 }{ 2 } \) m(arc PTR), and (i) [inscribed angle theorem]
∠PSR = \(\frac { 1 }{ 2 } \) m(arcPTR) (ii) [Inscribed angle theorem]
∴ ∠PQR ≅ ∠PSR [From (i) and (ii)]

Question 5.
Angle inscribed in a semicircle is a right angle. With the help of given figure write ‘given’, ‘to prove’ and ‘the proof. (Textbook pg. no. 68)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 14
Given: M is the centre of circle. ∠ABC is inscribed in arc ABC.
Arcs ABC and AXC are semicircles.
To prove: ∠ABC = 90°
Proof:
∠ABC = \(\frac { 1 }{ 2 } \) m(arc AXC) (i) [Inscribed angle theorem]
arc AXC is a semicircle.
∴ m(arc AXC) = 180° (ii) [Measure of semicircular arc is 1800]
∴ ∠ABC = \(\frac { 1 }{ 2 } \) × 180°
∴ ∠ABC = 90° [From (i) and (ii)]

Question 6.
Theorem: Opposite angles of a cyclic quadrilateral are supplementry.
Fill in the blanks and complete the following proof. (Textbook pg. no. 68)
Given: □ ABCD is cyclic.
To prove: ∠B + ∠D = 180°
∠A + ∠C = 180°
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 15
Proof:
arc ABC is intercepted by the inscribed angle ∠ADC.
∴ ∠ADC m(arcABC) (i) [Inscribed angle theorem]
Similarly, ∠ABC is an inscribed angle. It intercepts arc ADC.
∴ ABC = \(\frac { 1 }{ 2 } \) m(arc ADC) (ii) [Inscribed angle theorem]
∴ ∠ADC + ∠ABC
= \(\frac { 1 }{ 2 } \) m(arcABC) + \(\frac { 1 }{ 2 } \) m(arc ADC) [Adding (i) and (ii)]
∴ ∠D + ∠B = \(\frac { 1 }{ 2 } \) m(areABC) + m(arc ADC)]
∴ ∠B + ∠D = \(\frac { 1 }{ 2 } \) × 360° [arc ABC and arc ADC constitute a complete circle]
= 180°
∴ ∠B + ∠D = 180°
Similarly we can prove,
∠A + ∠C = 180°

Question 7.
In the above theorem, after proving ∠B + ∠D = 180°, can you use another way to prove ∠A + ∠C = 180°? (Textbook pg. no. 69)
Proof:
Yes, we can prove ∠A + ∠C = 180° by another way.
∠B + ∠D = 180°
In ꠸ABCD,
∠A + ∠B + ∠C + ∠D = 360° [Sum of the measures of all angles of a quadrilateral is 360°.]
∴ ∠A + ∠C + 180° = 360°
∴ ∠A + ∠C = 360° – 180°
∴ ∠A + ∠C = 180°

Question 8.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. (Textbook pg. no. 69)
Given: ꠸ABCD is a cyclic quadrilateral.
∠BCE is the exterior angle of ꠸ABCD.
To prove: ∠BCE ≅ ∠BAD
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 16
Proof:
∠BCE + ∠BCD = 180°…… (i) [Angles in a linear pair]
꠸ABCD is a cyclic quadrilateral. [Given]
∠BAD + ∠BCD = 180°………. (ii) [Opposite angles of a cyclic quadrilateral are supplementary]
∴ ∠BCE + ∠BCD = ∠BAD + ∠BCD [From (i) and (ii)]
∴ ∠BCE = ∠BAD

Question 9.
Theorem : If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic. (Textbook pg. no. 69)
Given: In ꠸ABCD, ∠A + ∠C = 180°
To prove: ꠸ABCD is a cyclic quadrilateral.
Proof:
(Indirect method)
Suppose ꠸ABCD is not a cyclic quadrilateral.
We can still draw a circle passing through three non collinear points A, B, D.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 17
Case I: Point C lies outside the circle.
Then, take point E on the circle
such that D – E – C.
∴ ꠸ABED is a cyclic quadrilateral.
∠DAB + ∠DEB = 180° (i) [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB + ∠DCB = 180° (ii) [Given]
∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (i) and (ii)]
∴ ∠DEB = ∠DCB
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 18
But, ∠DEB ≠ ∠DCB as ∠DEB is the exterior angle of ∆BEC.
∴ Our supposition is wrong.
∴ ꠸ABCD is a cyclic quadrilateral.
Case II: Point C lies inside the circle.
Then, take point E on the circle such that
D – C – E
∴ □ABED is a cyclic quadrilateral.
∠DAB + ∠DEB = 180° (iii) [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB + ∠DCB = 180° (iv) [Given]
∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (iii) and (iv)]
∴ ∠DEB = ∠DCB
But ∠DEB ≠ ∠DCB as ∠DCB is the exterior angle of ∆BCE.
∴ Our supposition is wrong.
∴ □ABCD is a cyclic quadrilateral.

Question 10.
Theorem: If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic. (Textbook pg. no. 70)
Given: Points B and C lie on the same side of the line AD.
∠ABD = ∠ACD
To prove: Points A, B, C, D are concyclic. i.e., □ABCD is a cyclic quadrilateral.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 19
Proof:
Suppose points A, B, C, D are not concyclic points.
We can still draw a circle passing through three non collinear points A, B, D.
Case I: Point C lies outside the circle.
Then, take point E on the circle such that
A – E – C.
∠ABD ≅ ∠AED (i) [Angles inscribed in the same arc]
∠ABD ≅ ∠ACD (ii) [Given]
∴ ∠AED ≅ ∠ACD [From (i) and (ii)]
∴ ∠AED ≅ ∠ECD [A – E – C]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 20
But, ∠AED ≅ ∠ECD as ∠AED is the exterior angle of ∆ECD.
∴ Our supposition is wrong.
∴ Points A, B, C, D are concyclic points.
Case II: Point C lies inside the circle. Then, take point E on the circle such that A – C – E.
∠ABD ≅ ∠AED (iii) [Angles inscribed in the same arc]
∠ABD ≅ ∠ACD (iv) [Given]
∴ ∠AED ≅ ∠ACD [From (iii) and (iv)]
∴ ∠CED ≅ ∠ACD [A – C – E]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.4 21
But, ∠CED ≅ ∠ACD as ∠ACD is the exterior angle of ∆ECD.
∴ Our supposition is wrong.
∴ Points A, B, C, D are concyclic points.

Question 11.
The above theorem is converse of a certain theorem. State it. (Textbook pg. no. 70)
Answer:
If four points are concyclic, then the line joining any two points subtend equal angles at the other two points which are on the same side of that line.

Maharashtra Board 10th Class Maths Part 2 Practice Set 3.3 Solutions Chapter 3 Circle

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Practice Set 3.3 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Question 1.
In the adjoining figure, points G, D, E, F are concyclic points of a circle with centre C.
∠ECF = 70°, m(arc DGF) = 200°. Find m(arc DE) and m(arc DEF).
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 1
Solution:
m(arc EF) = m∠ECF [Definition of measure of minor arc]
∴ m(arc EF) = 70°
i. m(arc DE) + m(arc DGF)
+ m(arc EF) = 360° [Measure of a circle is 360°]
∴ m(arc DE) = 360° – m(arc DGF) – m(arc EF)
= 360° – 200° – 70°
∴ m(arc DE) = 90°
ii. m(arc DEF) = m(arc DE) + m(arc EF) [Arc addition property]
= 90° + 70°
∴ m(arc DEF) = 160°

Question 2.
In the adjoining figure, AQRS is an equilateral triangle. Prove that,
i. arc RS ≅ arc QS ≅ arc QR
ii. m(arc QRS) = 240°.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 2
Solution:
Proof:
i. ∆QRS is an equilateral triangle, [Given]
∴ seg RS ≅ seg QS ≅ seg QR [Sides of an equilateral triangle]
∴ arc RS ≅ arc QS ≅ arc QR [Corresponding arcs of congruents chords of a circle are congruent]
ii. Let m(arc RS) = m(arc QS)= m(arc QR) = x
m(arc RS) + m(arc QS) + m(arc QR) = 360° [Measure of a circle is 360°, arc addition property]
∴ x + x + x = 360°
∴ 3x = 360°
∴ x = \(\frac{360^{\circ}}{3}=120^{\circ}\)
∴ m(arc RS) = m(arc QS) = m(arc QR) = 120° (i)
Now, m(arc QRS) = m(arc QR) + m(arc RS) [Arc addition property]
= 120° + 120° [From (i)]
∴ m(arc QRS) = 240°

Question 3.
In the adjoining figure, chord AB ≅ chord CD. Prove that, arc AC = arc BD.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 3
Solution:
Proof:
chord AB ≅ chord CD [Given]
∴ arc AB ≅ arc CD [Corresponding arcs of congruents chords of a circle are congruent]
∴ m(arc AB) = m(arc CD)
∴ m(arc AC) + m(arc BC) = m(arc BC) + m(arc BD) [Arc addition property]
∴ m(arc AC) = m(arc BD)
∴ arc AC ≅ arc BD

Maharashtra Board Class 10 Maths Chapter 3 Circle Intext Questions and Activities

Question 1.
Theorem : The chords corresponding to congruent arcs of a circle (or congruent circles) are congruent. (Textbook pg. no. 61)
Given: B is the centre of circle.
arc APC ≅ arc DQE
To prove: chord AC ≅ chord DE
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 5
Proof:
[m(arc APC) = ∠ABC (i) [Definition of measure of
m(arc DQE) = ∠DBE] (ii) minor arc]
arc APC ≅ arc ∠DQE (iii) [Given]
∴ ∠ABC ≅ ∠DBE [From (i), (ii) and (iii)]
In ∆ABC and ∆DBE,
side AB ≅ side DB [Radil of the same circle]
side [CB] side [EB] [Radii of the same circle]
∠ABC ≅∠DBE [From (iii), Measures of congruent arcs]
∴ ∆ABC ≅ ∆DBE [SAS test of congruency]
∴ chord AC ≅ chord DE [c.s.c.t]

Question 2.
Theorem: Corresponding arcs of congruent chords of a circle (or congruent circles) are congruent (Textbook pg. no. 61)
Given: O is the centre of circle, chord PQ = chord RS
To prove: arc PMQ = arc RNS
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 4
Proof:
In ∆POQ and ∆ROS,
[side PO ≅ side RO
side OQ ≅ side OS] [Radii of the same circle]
chord PQ ≅ chord RS [Given]
∴ ∆POQ ≅ ∆ROS [SSS test of congruency]
∴ ∠POQ ≅ ∠ROS (i) [c.a.c.t.]
m(arc PMQ) = ∠POQ (ii)
m(arc RNS) = ∠ROS (iii) [Definition of measure of minor arc]
∴ arc PMQ ≅ arc RNS [From (i), (ii) and (iii)]

Question 3.
Prove the two theorems on textbook pg.no.61 for congruent circles. (Textbook pg. no. 62)
Theorem : The chords corresponding to congruent arcs of congruent circles are congruent
Given: In congruent circles with centres B and R,
arc APC ≅ arc DQE
To prove: chord AC ≅ chord DE
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 6
Proof:
[m(arc APC) = ∠ABC (i)
m(arc DQE) = ∠DRE] (ii) [Definition of measure of minor arc]
arc APC ≅ arc DQE (iii) [Given]
∴ ∠ABC = ∠DRE (iv) [From (i), (ii) and (iii)]
In ∆ABC and ∆DRE,
[side AB ≅ side DR [Radii of congruent circles]
side CB ≅ side ER] [From (iv)]
∠ABC ≅ ∠DRE
∴ ∆ABC ≅ ∆DRE [SAS test of congruency]

Question 4.
While proving the first theorem of the two, we assume that the minor arc APC and minor arc DQE are congruent. Can you prove the same theorem by assuming that corresponding major arcs congruent? (Textbook pg. no. 62)
Statement:
The chords corresponding to congruent major arcs of a circle are congruent.
Given: B is the centre of circle.
arc AXC ≅ arc DXE
To prove: chord AC ≅ chord DE
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 7
Proof:
m(major arc) = 360° – m(minor arc)
∴ m(arc AXC) = 360° – m(arc APC) (i)
m(arc DXE) = 360° – m(arc DQE) (ii)
m(arc AXC) = m(arc DXE) (iii) [Given]
∴ 360° – m(arc APC) = 360°- m(arc DQE) [From (i), (ii) and (iii)]
∴ m(arc APC) = m(arc DQE) (iv)
∴ m(arc APC) = ∠ABC (v) [Definition of measure of minor arc]
m(arc DQE) = ∠DBE (vi)
∴ ∠ABC = ∠DBE (vii) [From (iv), (v) and (vi)]
In ∆ABC and ∆DBE,
[side AB ≅ side DB
Side CB ≅ side EB] [Radii of the same circle]
∠ABC ≅ ∠DBE [From (vii)]
∴ ∆ABC ≅ ∆DBE [SAS test of congruency]
∴ chord AC ≅ chord DE [c.s.c.t.]

Question 5.
i. In the second theorem, are the major arcs corresponding to congruent chords congruent?
ii. Is the theorem true, when the chord PQ and chord RS are diameters of the circle? (Textbook pg. no. 62)
Solution:
i. Yes, the major arcs corresponding to congruent chords are congruent.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 8
Proof:
In ∆POQ and ∆ROS,
seg OP ≅ seg OR [Radii of the same circle]
seg OQ ≅ seg OS [Radii of the same circle]
seg PQ ≅ seg RS [Given]
∴ ∆POQ ≅ ∆ROS [SSS test of congruence]
∴ ∠POQ ≅ ∠SOR (i) [c.a.c.t]
[ m(arc PMQ) = ∠POQ (ii)
m(arc RNS) = ∠SOR ] (iii) [Definition of measure of minor arc]
∴ m(arc PMQ) = m(arc RNS)
m(minor arc) = 360° – m(major arc) (iv) [From (i), (ii) and (iii)]
m(arc PMQ) = 360° – m(arc PXQ) (v)
and m(arc RNS) = 360° – m(arc RXS) (vi)
∴ 360°- m(arc PXQ) = 360°- m(arc RXS) [From (iv), (v) and (vi)]
∴ m(arc PXQ) = m(arc RXS)

ii. Yes, the major arcs corresponding to congruent chords (diameters) are congruent.
Given: O is the centre of circle.
seg PQ and seg RS are the diameters.
To prove: arc PYQ ≅ arc RYS
Maharashtra Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.3 9
Proof:
seg PQ and seg RS are the diameters of the same circle. [Given]
∴ arc PYQ and arc RYS are semicircular arcs.
∴ m(arc PYQ) = m(arc RYS) = 180° [Measure of a semicircular arc is 180°]
∴ arc PYQ ≅ arc RYS