Category: Class 10

10th Standard Maths 2 Problem Set 5 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

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

Class 10 Maths Part 2 Problem Set 5 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks using correct alternatives.

i. Seg AB is parallel to Y-axis and co-ordinates of point A are (1, 3), then co-ordinates of point B can be _______.
(A) (3,1)
(B) (5,3)
(C) (3,0)
(D) (1,-3)
Answer: (D)
Since, seg AB || Y-axis.
∴ x co-ordinate of all points on seg AB
will be the same,
x co-ordinate of A (1, 3) = 1
x co-ordinate of B (1, – 3) = 1
∴ Option (D) is correct.

ii. Out of the following, point lies to the right of the origin on X-axis.
(A) (-2,0)
(B) (0,2)
(C) (2,3)
(D) (2,0)
Answer: (D)

iii. Distance of point (-3, 4) from the origin is _________.
(A) 7
(B) 1
(C) 5
(D) -5
Answer: (C)
Distance of (-3, 4) from origin
\(\begin{array}{l}{=\sqrt{(-3)^{2}+(4)^{2}}} \\ {=\sqrt{9+16}} \\ {=\sqrt{25}=5}\end{array}\)

iv. A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is ________.
(A) \(\frac { 1 }{ 2 } \)
(B) \(\frac{\sqrt{3}}{2}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt { 3 }\)
Answer: (C)

Question 2.
Determine whether the given points are collinear.
i. A (0, 2), B (1, -0.5), C (2, -3)
ii. P(1,2), Q(2,\(\frac { 8 }{ 5 } \)),R(3,\(\frac { 6 }{ 5 } \))
iii L (1, 2), M (5, 3), N (8, 6)
Solution:

∴ 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.

∴ 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.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.
[Note: Students can solve the above problems by using distance formula.]

Question 3.
Find the co-ordinates of the midpoint of the line segment joining P (0,6) and Q (12,20).
Solution:
P(x₁,y₁) = P (0, 6), Q(x₂, y₂) = Q (12, 20)
Here, x₁ = 0, y₁ = 6, x₂ = 12, y₂ = 20
∴ Co-ordinates of the midpoint of seg PQ

∴ The co-ordinates of the midpoint of seg PQ are (6,13).

Question 4.
Find the ratio in which the line segment joining the points A (3, 8) and B (-9, 3) is divided by the Y-axis.
Solution:
Let C be a point on Y-axis which divides seg AB in the ratio m : n.
Point C lies on the Y-axis
∴ its x co-ordinate is 0.
Let C = (0, y)
Here A (x₁,y₁) = A(3, 8)
B (x₂, y₂) = B (-9, 3)
∴ By section formula,


∴ Y-axis divides the seg AB in the ratio 1 : 3.

Question 5.
Find the point on X-axis which is equidistant from P (2, -5) and Q (-2,9).
Solution:
Let point R be on the X-axis which is equidistant from points P and Q.
Point R lies on X-axis.
∴ its y co-ordinate is 0.
Let R = (x, 0)
R is equidistant from points P and Q.
∴ PR = QR

∴ (x – 2)² + [0 – (-5)]² = [x – (- 2)]² + (0 – 9)² …[Squaring both sides]
∴ (x – 2)² + (5)² = (x + 2)² + (-9)²
∴ 4 – 4x + x² + 25 = 4 + 4x + x² + 81
∴ – 8x = 56
∴ x = -7
∴ The point on X-axis which is equidistant from points P and Q is (-7,0).

Question 6.
Find the distances between the following points.
i. A (a, 0), B (0, a)
ii. P (-6, -3), Q (-1, 9)
iii. R (-3a, a), S (a, -2a)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = a, y₁ = 0, x₂ = 0, y₂ = a
By distance formula,

∴ d(A, B) = a\(\sqrt { 2 }\) units

ii. Let P (x₁, y₁) and Q (x₂, y₂) be the given points.
∴ x₁ = -6, y₁ = -3, x₂ = -1, y₂ = 9
By distance formula,

∴ d(P, Q) = 13 units

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = -3a, y₁ = a, x₂ = a, y₂ = -2a
By distance formula,

∴ d(R, S) = 5a units

Question 7.
Find the co-ordinates of the circumcentre of a triangle whose vertices are (-3,1), (0, -2) and (1,3).
Solution:
Let A (-3, 1), B (0, -2) and C (1, 3) be the vertices of the triangle.
Suppose O (h, k) is the circumcentre of ∆ABC.

∴ (h + 3)² + (k – 1)² = h² + (k + 2)²
∴ h² + 6h + 9 + k² – 2k + 1 = h² + k² + 4k + 4
∴ 6h – 2k + 10 = 4k + 4
∴ 6h – 2k – 4k = 4 – 10
∴ 6h – 6k = – 6
∴ h – k = -1 ,..(i)[Dividing both sides by 6]
OB = OC …[Radii of the same circle]

∴ h² + (k + 2)² = (h – 1)² + (k – 3)²
∴ h² + k² + 4k + 4 = h² – 2h + 1 + k² – 6k + 9
∴ 4k + 4 = -2h + 1 – 6k + 9
∴ 2h+ 10k = 6
∴ h + 5k = 3 …(ii)
Subtracting equation (ii) from (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { -1 }{ 3 } \),\(\frac { 2 }{ 3 } \))

Question 8.
In the following examples, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle.
i. L (6, 4), M (-5, -3), N (-6, 8)
ii. P (-2, -6), Q (-4, -2), R (-5, 0)
iii. A(\(\sqrt { 2 }\),\(\sqrt { 2 }\)),B(-\(\sqrt { 2 }\),-\(\sqrt { 2 }\)),C(\(\sqrt { 6 }\),\(\sqrt { 6 }\))
Solution:
i. By distance formula,

∴ d(M, N) + d (L, N) > d (L, M)
∴ Points L, M, N are non collinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since MN ≠ LN ≠ LM
∴ ∆LMN is a scalene triangle.
∴ The segments joining the points L, M and N will form a scalene triangle.

ii. By distance formula,


∴ d(P, Q) + d(Q, R) = d (P, R) …[From (iii)]
∴ Points P, Q, R are collinear points.
We cannot construct a triangle through 3 collinear points.
∴ The segments joining the points P, Q and R will not form a triangle.

iii. By distance formula,

∴ d(A, B) + d(B, C) + d(A, C) … [From (iii)]
∴ Points A, B, C are non collinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since, AB = BC = AC
∴ ∆ABC is an equilateral triangle.
∴ The segments joining the points A, B and C will form an equilateral triangle.

Question 9.
Find k, if the line passing through points P (-12, -3) and Q (4, k) has slope \(\frac { 1 }{ 2 } \).
Solution:
P(x₁,y₁) = P(-12,-3),
Q(X₂,T₂) = Q(4, k)
Here, x₁ = -12, x₂ = 4, y₁ = -3, y₂ = k

But, slope of line PQ (m) is \(\frac { 1 }{ 2 } \) ….[Given]
∴ \(\frac { 1 }{ 2 } \) = \(\frac { k+3 }{ 16 } \)
∴ \(\frac { 16 }{ 2 } \) = k + 3
∴ 8 = k + 3
∴ k = 5
The value of k is 5.

Question 10.
Show that the line joining the points A (4,8) and B (5, 5) is parallel to the line joining the points C (2, 4) and D (1 ,7).
Proof:

∴ Slope of line AB = Slope of line CD
Parallel lines have equal slope.
∴ line AB || line CD

Question 11.
Show that points P (1, -2), Q (5, 2), R (3, -1), S (-1, -5) are the vertices of a parallelogram.
Proof:
By distance formula,

In ꠸PQRS,
PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
∴ ꠸ PQRS is a parallelogram.
[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 12.
Show that the ꠸PQRS formed by P (2, 1), Q (-1, 3), R (-5, -3) and S (-2, -5) is a rectangle.
Proof:
By distance formula,


In ꠸PQRS,
PQ = RS …[From (i) and (iii)]
QR = PS …[From (ii) and (iv)]
꠸PQRS is a parallelogram.
[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]

In parallelogram PQRS,
PR = QS … [From (v) and (vi)]
∴ ꠸PQRS is a rectangle.
[A parallelogram is a rectangle if its diagonals are equal]

Question 13.
Find the lengths of the medians of a triangle whose vertices are A (-1, 1), B (5, -3) and C (3,5).
Solution:

Suppose AD, BE and CF are the medians.
∴ Points D, E and F are the midpoints of sides BC, AC and AB respectively.
∴ By midpoint formula,


∴ The lengths of the medians of the triangle 5 units, 2\(\sqrt { 13 }\) units and \(\sqrt { 37 }\) units.

Question 14.
Find the co-ordinates of centroid of the triangle if points D (-7, 6), E (8, 5) and F (2, -2) are the mid points of the sides of that triangle.
Solution:

Suppose A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of the triangle.
D (-7, 6), E (8, 5) and F (2, -2) are the midpoints of sides BC, AC and AB respectively.
Let G be the centroid of ∆ABC.
D is the midpoint of seg BC.
By midpoint formula,

E is the midpoint of seg AC.
By midpoint formula,

Adding (i), (iii) and (v),
x₂ + x₃ + x₁ + x₃ + x₁ + x₂ = -14 + 16 + 4
∴ 2x₁ + 2x₂ + 2x₃ = 6
∴ x₁ + x₂ + x₃ = 3 …(vii)
Adding (ii), (iv) and (vi),
y₂ + y₃ + y₁ + y₃ + y₁ +y₂ = 12 + 10 – 4
∴ 2y₁ + 2y₂ + 2y₃ = 18
∴ y₁ + y₂ + y₃ = 9 …(viii)
G is the centroid of ∆ABC.
By centroid formula,

∴ The co-ordinates of the centroid of the triangle are (1,3).

Question 15.
Show that A (4, -1), B (6, 0), C (7, -2) and D (5, -3) are vertices of a square.
Proof:
By distance formula,


∴ □ABCD is a square.
[A rhombus is a square if its diagonals are equal]

Question 16.
Find the co-ordinates of circumcentre and radius of circumcircle of AABC if A (7, 1), B (3,5) and C (2,0) are given.
Solution:
Suppose, O (h, k) is the circumcentre of ∆ABC


∴ h² – 6h + 9 + k² – 10k + 25 = h² – 4h + 4 + k²
∴ 2h + 10k = 30
∴ h + 5k = 15 … (ii)[Dividing both sides by 2]
Multiplying equation (i) by 5, we get
25h + 5k = 115 …(iii)
Subtracting equation (ii) from (iii), we get

Substituting the value of h in equation (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { 25 }{ 6 } \),\(\frac { 13 }{ 6 } \)) and radius of circumcircle is \(\frac{13 \sqrt{2}}{6}\) units.

Question 17.
Given A (4, -3), B (8, 5). Find the co-ordinates of the point that divides segment AB in the ratio 3:1.
Solution:
Suppose point C divides seg AB in the ratio 3:1.
Here; A(x₁, y₁) = A (4, -3)
B (x₂, y₂) = B (8, 5)
By section formula,

∴ The co-ordinates of point dividing seg AB in ratio 3 : 1 are (7, 3).

Question 18.
Find the type of the quadrilateral if points A (-4, -2), B (-3, -7), C (3, -2) and D (2, 3) are joined serially.
Solution:

Slope of AB = slope of CD
∴ line AB || line CD
slope of BC = slope of AD
∴ line BC || line AD
Both the pairs of opposite sides of ∆ABCD are parallel.
∴ ꠸ ABCD is a parallelogram.
∴ The quadrilateral formed by joining the points A, B, C and D is a parallelogram.

Question 19.
The line segment AB is divided into five congruent parts at P, Q, R and S such that A-P-Q-R-S-B. If point Q (12, 14) and S (4, 18) are given, find the co-ordinates of A, P, R, B.
Solution:

Points P, Q, R and S divide seg AB in five congruent parts.
Let A (x₁, y₁), B (x₂, y₂), P (x₃, y₃) and
R (x₄, y₄) be the given points.
Point R is the midpoint of seg QS.
By midpoint formula,
x co-ordinate of R = \(\frac { 12+4 }{ 2 } \) = \(\frac { 16 }{ 2 } \) = 8
y co-ordinate of R = \(\frac { 14+18 }{ 2 } \) = \(\frac { 32 }{ 2 } \) = 16
∴ co-ordinates of R are (8, 16).
Point Q is the midpoint of seg PR.
By midpoint formula,

∴ 28 = y₃ + 16
∴ y₃ = 12
∴ P(x₃,y₃) = (16, 12)
∴ co-ordinates of P are (16, 12).
Point P is the midpoint of seg AQ.
By midpoint formula,

∴ co-ordinates of A are (20, 10).
Point S is the midpoint of seg RB.
By midpoint formula,

∴ 36 = y₂ + 16
∴ y₂ = 20
∴ B(x₂, y₂) = (0, 20)
∴ co-ordinates of B are (0, 20).
∴ The co-ordinates of points A, P, R and B are (20, 10), (16, 12), (8, 16) and (0, 20) respectively.

Question 20.
Find the co-ordinates of the centre of the circle passing through the points P (6, -6), Q (3, -7) and R (3,3).
Solution:
Suppose O (h, k) is the centre of the circle passing through the points P, Q and R.

∴ (h – 6)² + (k + 6)² = (h – 3)² + (k + 7)²
∴ h² – 12h + 36 + k² + 12k + 36
= h² – 6h + 9 + k² + 14k + 49
∴ 6h + 2k = 14
∴ 3h + k = 7 …(i)[Dividing both sides by 2]
OP = OR …[Radii of the same circle]

∴ (h – 6)² + (k + 6)² = (h – 3)² + (k – 3)²
∴ h² – 12h + 36 + k² + 12k + 36
= h² – 6h + 9 + k² – 6k + 9
∴ 6h – 18k = 54
∴ 3h – 9k = 27 …(ii)[Dividing both sides by 2]
Subtracting equation (ii) from (i), we get

Substituting the value of k in equation (i), we get
3h – 2 = 7
∴ 3h = 9
∴ h = \(\frac { 9 }{ 3 } \) = 3
∴ The co-ordinates of the centre of the circle are (3, -2).

Question 21.
Find the possible pairs of co-ordinates of the fourth vertex D of the parallelogram, if three of its vertices are A (5, 6), B (1, -2) and C (3, -2).
Solution:

Let the points A (5, 6), B (1, -2) and C (3, -2) be the three vertices of a parallelogram.
The fourth vertex can be point D or point Di or point D₂ as shown in the figure.
Let D(x₁,y₁), D, (x₂, y₂) and D₂ (x₃,y₃).
Consider the parallelogram ACBD.
The diagonals of a parallelogram bisect each other.
∴ midpoint of DC = midpoint of AB


Co-ordinates of point D(x₁, y₁) are (3, 6).
Consider the parallelogram ABD₁C.
The diagonals of a parallelogram bisect each other.
∴ midpoint of AD₁ = midpoint of BC

∴ Co-ordinates of D₁(x₂,y₂) are (-1,-10).
Consider the parallelogram ABCD₂.
The diagonals of a parallelogram bisect each other.
∴ midpoint of BD₂ = midpoint of AC

∴ co-ordinates of point D₂ (x₃, y₃) are (7, 6).
∴ The possible pairs of co-ordinates of the fourth vertex D of the parallelogram are (3, 6), (-1,-10) and (7,6).

Question 22.
Find the slope of the diagonals of a quadrilateral with vertices A (1, 7), B (6,3), C (0, -3) and D (-3,3).
Solution:
Suppose ABCD is the given quadrilateral.

∴ The slopes of the diagonals of the quadrilateral are 10 and 0.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 1.3 Chapter 1 Similarity Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

Class 10 Maths Part 2 Practice Set 1.3 Chapter 1 Similarity Questions With Answers Maharashtra Board

Practice Set 1.3 Question 1.
In the adjoining figure, ∠ABC = 75°, ∠EDC = 75°. State which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Solution:
In ∆ABC and ∆EDC,
∠ABC ≅ ∠EDC [Each angle is of measure 75°]
∠ACB ≅ ∠ECD [Common angle]
∴ ∆ABC ~ ∆EDC [AA test of similarity]
One to one correspondence is
ABC ↔ EDC

Similarity Class 10 Practice Set 1.3 Question 2.
Are the triangles in the adjoining figure similar? If yes, by which test?

Solution:
In ∆PQR and ∆LMN,
\(\frac { PQ }{ LM } \) = \(\frac { 6 }{ 3 } \) = \(\frac { 2 }{ 1 } \) (i)
\(\frac { QR }{ MN } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 2 }{ 1 } \) (ii)
\(\frac { PR }{ LN } \) = \(\frac { 10 }{ 5 } \) = \(\frac { 2 }{ 1 } \) (iii)
∴ \(\frac { PQ }{ LM } \) = \(\frac { QR }{ MN } \) = \(\frac { PR }{ LN } \) [From (i), (ii) and (iii)]
∴ ∆PQR – ∆LMN [SSS test of similarity]

Similarity Practice Set 1.3 Question 3.
As shown in the adjoining figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m, then how long will be the shadow of the bigger pole at the same time?

Solution:
Here, AC and PR represents the bigger and smaller poles, and BC and QR represents their shadows respectively.
Now, ∆ACB – ∆PRQ [ ∵ Vertical poles and their shadows form similar figures]
∴ \(\frac { CB }{ RQ } \) = \(\frac { AC }{ PR } \) [Corresponding sides of similar triangles]
∴ \(\frac { x }{ 6 } \) = \(\frac { 8 }{ 4 } \)
∴ \(x=\frac{8 \times 6}{4}\)
∴ x = 12 m
∴ The shadow of the bigger pole will be 12 metres long at that time.

Practice Set 1.3 Geometry 10th Maharashtra Board Question 4.
In ∆ABC, AP ⊥ BC, BQ ⊥ AC, B – P – C, A – Q – C, then prove that ∆CPA – ∆CQB. If AP = 7, BQ = 8, BC = 12, then find AC.

Solution:
In ∆CPA and ∆CQB,
∠CPA ≅ ∠CQB [Each angle is of measure 90°]
∠ACP ≅ ∠BCQ [Common angle]
∴ ∆CPA ~ ∆CQB [AA test of similarity]
∴\(\frac { AC }{ BC } \) = \(\frac { AP }{ BQ } \) [Corresponding sides of similar triangles]
∴ \(\frac { AC }{ 12 } \) = \(\frac { 7 }{ 8 } \)
∴ AC = \(x=\frac{12 \times 7}{8}\)
∴ AC = 10.5 Units

10th Geometry Practice Set 1.3 Question 5.
Given: In trapezium PQRS, side PQ || side SR, AR = 5 AP, AS = 5 AQ, then prove that SR = 5 PQ.

Solution:
side PQ || side SR [Given]
and seg SQ is their transversal.
∴ ∠QSR = ∠SQP [Altemate angles]
∴ ∠ASR = ∠AQP (i) [Q – A – S]
In ∆ASR and ∆AQP,
∠ASR = ∠AQP [From (i)]
∠SAR ≅ ∠QAP [Vertically opposite angles]
∆ASR ~ ∆AQP [AA test of similarity]
∴ \(\frac { AS }{ AQ } \) = \(\frac { SR }{ PQ } \) (ii) [Corresponding sides of similar triangles]
But, AS = 5 AQ [Given]
∴ \(\frac { AS }{ AQ } \) = \(\frac { 5 }{ 1 } \) (iii)
∴ \(\frac { SR }{ PQ } \) = \(\frac { 5 }{ 1 } \) [From (ii) and (iii)]
∴ SR = 5 PQ

Practice Set 1.3 Geometry 10th Question 6.
Id trapezium ABCD (adjoining figure), side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15, then find OD.

Solution:
side AB || side DC [Given]
and seg BD is their transversal.
∴ ∠DBA ≅ ∠BDC [Alternate angles]
∴ ∠OBA ≅ ∠ODC (i) [D – O – B]
In ∆OBA and ∆ODC
∠OBA ≅ ∠ODC [From (i)]
∠BOA ≅ ∠DOC [Vertically opposite angles]
∴ ∆OBA ~ ∆ODC [AA test of similarity]
∴ \(\frac { OB }{ OD } \) = \(\frac { AB }{ DC } \) [Corresponding sides of similar triangles]
∴ \(\frac { 15 }{ OD } \) = \(\frac { 20 }{ 6 } \)
∴ OD = \(x=\frac{15 \times 6}{20}\)
∴ OD = 4.5 units

Class 10 Geometry Practice Set 1.3 Question 7.
꠸ ABCD is a parallelogram. Point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.

Solution:
Proof:
꠸ ABCD is a parallelogram. [Given]
∴ side AB || side CD [Opposite sides of a parallelogram]
∴ side AT || side CD [A – B – T]
and seg DT is their transversal.
∴ ∠ATD ≅ ∠CDT [Alternate angles]
∴ ∠BTE ≅ ∠CDE (i) [A – B – T, T – E – D]
In ∆BTE and ∆CDE,
∠BTE ≅ ∠CDE [From (i)]
∠BET ≅ ∠CED [Vertically opposite angles]
∴ ∆BTE ~ ∆CDE. [AA test of similarity]
∴ \(\frac { TE }{ DE } \) = \(\frac { BE }{ CE } \) [Corresponding sides of similar triangles]
∴ DE × BE = CE × TE

Geometry Practice Set 1.3 Question 8.
In the adjoining figure, seg AC and seg BD intersect each other in point P and \(\frac { AP }{ CP } \) = \(\frac { BP }{ DP } \) Prove that, ∆ABP ~ ∆CDP

Solution:
Proof:
In ∆ABP and ∆CDP,
\(\frac { AP }{ CP } \) = \(\frac { BP }{ DP } \) [Given]
∠APB ≅ ∠CPD [Vertically opposite angles]
∴ ∆ABP ~ ∆CDP [SAS test of similarity]

Math 2 Practice Set 1.3 Question 9.
In the adjoining figure, in ∆ABC, point D is on side BC such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD,

Solution:
Proof:
In ∆BAC and ∆ADC,
∠BAC ≅ ∠ADC [Given]
∠BCA ≅ ∠ACD [Common angle]
∴ ∆BAC ~ ∆ADC [AA test of similarity]
∴ \(\frac { CA }{ CD } \) = \(\frac { CB }{ CA } \) [Corresponding sides of similar triangles]
∴ CA × CA = CB × CD
∴ CA² = CB × CD

Question 1.
In the adjoining figure, BP ⊥ AC, CQ ⊥ AB, A – P – C, A – Q – B, then prove that ∆APB and ∆AQC are similar. (Textbook pg. no. 20)

Solution:

2. SAS test for similarity of triangles:
For a given correspondence, if two pairs of corresponding sides are in the same proportion and the angle between them is congruent, then the two triangles are similar.

In the given figure, if \(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \), and ∠B ≅∠Q, then ∆ABC ~ ∆PQR

3. SSS test for similarity of triangles:
For a given correspondence, if three sides of one triangle are in proportion with the corresponding three sides of the another triangle, then the two triangles are similar.

In the given figure, if \(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \) = \(\frac { AC }{ PR } \), then ∆ABC ~ ∆PQR

Properties of similar triangles:

1. Reflexivity: ∆ABC ~ ∆ABC
2. Symmetry : If ∆ABC ~ ∆DEF, then ∆DEF ~ ∆ABC.
3. Transitivity: If ∆ABC ~ ∆DEF and ∆DEF ~ ∆GHI, then ∆ABC ~ ∆GHI.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Similarity Practice Set 1.1 Class 10 Maths Solutions
• Similarity Practice Set 1.2 Class 10 Maths Solutions
•  Similarity Practice Set 1.3 Class 10 Maths Solutions
• Similarity Practice Set 1.4 Class 10 Maths Solutions
• Similarity Problem Set 1 Class 10 Maths Solutions
• Pythagoras Theorem Practice Set 2.1 Class 10 Maths Solutions
• Pythagoras Theorem Practice Set 2.2 Class 10 Maths Solutions
• Pythagoras Theorem Problem Set 2 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 7.4 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.4 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.4 Chapter 7 Mensuration Questions With Answers Maharashtra Board

Practice Set 7.4 Geometry Class 10 Question 1. In the adjoining figure, A is the centre of the circle. ∠ABC = 45° and AC = 7\(\sqrt { 2 }\) cm. Find the area of segment BXC, (π = 3.14)

Solution:
In ∆ABC,
AC = AB … [Radii of same circle]
∴ ∠ABC = ∠ACB …[Isosceles triangle theorem]
∴ ∠ABC = ∠ACB = 45°
In ∆ABC,
∠ABC + ∠ACB + ∠BAC = 180° … [Sum of the measures of angles of a triangle is 180° ]
∴ 45° + 45° + ∠BAC = 180°
∴ 90° + ∠BAC = 180°
∴ ∠BAC = 90°
Let ∠BAC = θ = 90°

∴ The area of segment BXC is 27.93 cm².

10th Class Geometry Practice Set 7.4 Question 2. In the adjoining figure, O is the centre of the circle.
m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region.
(π = 3.14, \(\sqrt { 3 }\) = 1.73)

Given: m(arc PQR) = 60°, radius (r) = OP = 10 cm
To find: Area of shaded region.
Solution:
∠POR = m (arc PQR) …[Measure of central angle]
∴ ∠POR = θ = 60°


∴ The area of the shaded region is 9.08 cm².

7.4 Class 10 Question 3. In the adjoining figure, if A is the centre of the circle, ∠PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)

Given: Central angle (θ) = ∠PAR = 30°,
radius (r) = AP = 7.5
To find: Area of segment PQR.
Solution:
Let ∠PAR = θ = 30°

∴ The area of segment PQR is 0.65625 sq. units.

Chapter 7 Maths Class 10 Question 4. In the adjoining figure, if O is the centre of the circle, PQ is a chord, ∠POQ = 90°, area of shaded region is 114 cm2, find the radius of the circle, (π = 3.14)

Given: Central angle (θ) = ∠POQ= 90°,
A (segment PRQ) = 114 cm²
To find: Radius (r).
Solution:

…[Taking square root of both sides]
∴ r = 20 cm
∴ The radius of the circle is 20 cm.

Mensuration Questions for Class 10 Question 5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, \(\sqrt { 3 }\) = 1.73)
Given: Radius (r) =15 cm, central angle (θ) = 60°
To find: Areas of major and minor segments.
Solution:
Let chord PQ subtend ∠POQ = 60° at centre.
∴ θ = 60°

= 225 [0.0908]
= 20.43 cm²
∴ area of minor segment = 20.43 cm²
Area of circle = πr²
= 3.14 × 15 × 15
= 3.14 × 225
= 706.5 cm²
Area of major segment
= Area of circle – area of minor segment
= 706.5 – 20.43
= 686.07 cm²
Area of major segment 686.07 cm²
∴ The area of minor segment Is 20.43 cm² and the area of major segment is 686.07 cm².

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 7.3 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.3 Chapter 7 Mensuration Questions With Answers Maharashtra Board

Practice Set 7.3 Geometry Class 10 Question 1.
Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)
Given : Radius (r) = 10 cm,
Measure of the arc (θ) = 54°
To find : Area of the sector.
Solution:
Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)
= \(\frac { 54 }{ 360 } \) × 3.14 × (10)²
= \(\frac { 3 }{ 20 } \) × 3.14 × 100
= 3 × 3.14 × 5
= 15 × 3.14
= 47.1 cm²
∴ The area of the sector is 47.1 cm².

Mensuration Practice Set 7.3 Question 2.
Measure of an arc of a circle is 80° and its radius is 18 cm. Find the length of the arc. (π = 3.14)
Given: Radius (r) = 18 cm,
Measure of the arc (θ) = 80°
To find: Length of the arc.
Solution:
Length of arc = \(\frac{\theta}{360} \times 2 \pi r\) × 2πr
= \(\frac { 8 }{ 360 } \) × 2 × 3.14 × 18
= \(\frac { 2 }{ 9 } \) × 2 × 3.14 × 18
= 2 × 2 × 3.14 × 2 = 25.12 cm
∴ The length of the arc is 25.12 cm.

Practice Set 7.3 Geometry Question 3.
Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Solution:
Given: Radius (r) = 3.5 cm,
length of arc (l) = 2.2 cm
To find: Area of the sector.
Solution:
Area of sector = \(\frac{l \times \mathrm{r}}{2}\)
= \(\frac{2.2 \times 3.5}{2}\)
= 1.1 × 3.5 = 3.85 cm²
∴ The area of the sector is 3.85 cm².

Question 4.
Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector, (π = 3.14)
Given: Radius (r) = 10 cm,
area of minor sector =100 cm²
To find: Area of maj or sector.
Solution:
Area of circle = πr²
= 3.14 × (10)²
= 3.14 × 100 = 314 cm²
Now, area of major sector
= area of circle – area of minor sector
= 314 – 100
= 214 cm²
∴ The area of the corresponding major sector is 214 cm².

Question 5.
Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector.
Given: Radius (r) =15 cm,
area of sector = 30 cm²
To find: Length of the arc (l).
Solution:

∴ The length of the arc is 4 cm.

Practice Set 7.3 Question 6.
In the adjoining figure, radius of the circle is 7 cm and m (arc MBN) = 60°, find
i. Area of the circle.
ii. A(O-MBN).
iii. A(O-MCN).

Given: radius (r) = 7 cm,
m(arc MBN) = θ = 60°
Solution:
i. Area of circle = πr²
= \(\frac { 22 }{ 7 } \) × (7)²
= 22 × 7
= 154 cm²
∴ The area of the circle is 154 cm²

ii. Central angle (θ) = ∠MON = 60°
Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)
∴ A(O – MBN) = \(\frac { 60 }{ 360 } \) × \(\frac { 22 }{ 7 } \) × (7)²
\(\frac { 1 }{ 6 } \) × 22 × 7
= 25.67 cm²
= 25.7 cm²
∴ A(O-MBN) = 25.7 cm²

iii. Area of major sector = area of circle – area of minor sector
∴ A(O-MCN) = Area of circle – A(O-MBN)
= 154 – 25.7
∴ A(O-MCN) = 128.3 cm²

Question 7.
In the adjoining figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).
Given: Radius (r) = 3.4 cm,
perimeter of sector 12.8 cm
To find: A(P-ABC)

Solution:
Perimeter of sector
= Iength of arc ABC + AP + CP

∴ 12.8 = l + 3.4 + 3.4
∴ 12.8 = l + 6.8
∴ l = 12.8 – 6.8 = 6cm

∴ A(P-ABC) = 10.2 cm²

7.3 Class 10 Question 8.
In the adjoining figure, O is the centre of the sector. ∠ROQ = ∠MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and (π = \(\frac { 22 }{ 7 } \))

Given: ∠ROQ = ∠MON = 60°,
radius (r) = OR = 7 cm, radius (R) = OM = 21 cm
To find: Lengths of arc RXQ and arc MYN.
Solution:
i. Length of arc RXQ = \(\frac{\theta}{360} \times 2 \pi r\)
= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 7
= \(\frac { 1 }{ 6 } \) × 2 × 22
= 7.33 cm
ii. Length of arc MYN = \(\frac{\theta}{360} \times 2 \pi R\)
= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 21
= \(\frac { 1 }{ 6 } \) × 2 × 22 × 3
= 22 cm
∴ The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.

Question 9.
In the adjoining figure, if A(P-ABC) = 154 cm², radius of the circle is 14 cm, find
i. ∠APC,
ii. l(arc ABC).

Given: A(P – ABC) = 154 cm²,
radius (r) = 14 cm
Solution:
i. Let ∠APC = θ

Std 10 Geometry Mensuration Question 10.
Radius of a sector of a circle is 7 cm. If measure of arc of the sector is
i. 30°
ii. 210°
iii. three right angles, find the area of the sector in each case.
Given: Radius (r) = 7 cm
To find: Area of the sector.
Solution:
i. Measure of the arc (θ) = 30°

∴ Area of the sector is 12.83 cm².
ii. Measure of the arc (θ) = 210°

∴ Area of the sector is 89.83 cm².
iii. Measure of the arc (θ) = 3 right angle
= 3 × 90° = 270°

∴ Area of the sector is 115.50 cm².

Mensuration Practice Question 11.
The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle.
Given: Area of minor sector = 3.85 cm²,
central angle (θ) = 36°
To find: Radius of the circle (r).
Solution:
Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)

∴ The radius of the circle ¡s 3.5 cm.

10th Geometry Practice Set 7.3 Question 12.
In the given figure, ꠸PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Given: In rectangle PQRS,
PQ = 14 cm, QR = 21 cm
To find: Areas of the parts x, y and z.
Solution:

∠Q = ∠R = θ = 90° …[Angles of a rectangle]

For the sector (Q – PA),
PQ = QA …[Radii of the same circle]
∴ QA = 14 cm
Now, QR = QA + AR … [Q – A – R]
∴ 21 = 14 + AR
∴ AR = 7 cm

Area of rectangle = length × breadth
area of ꠸PQRS = PQ × QR
= 14 × 21
= 294 cm²
Area of part z = area of ꠸PQRS
– area of part x – area of part y
= 294 – 154 – 38.5
= 101.5 cm²
∴ The area of part x is 154 cm², the area of part y is 38,5 cm² and the area of part z is 101.5 cm².

Question 13.
∆ALMN is an equilat triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find,

i. A (∆ LMN).
ii. Area of any one of the sectors.
iii. Total area of all the three sectors.
iv. Area of the shaded region. (\(\sqrt { 3 }\) = 1.732 )
Given: In equilateral triangle LMN, LM =14 cm,
radius of sectors (r) = 7 cm
Solution:
i. ∆LMN is an equilateral triangle.

ii. Central angle (θ) = 60° …[Angle of an equilateral triangle]

∴ Area of one sector = 25.67 cm²
iii. Total area of all three sectors
= 3 × Area of one sector
= 3 × 25.67
= 77.01 cm²
∴ Total area of all three sectors = 77.01 cm²
iv. Area of shaded region
= A(∆LMN) – total area of all three sectors
= 84.87 – 77.01
= 7.86 cm²
∴ Area of shaded region = 7.86 cm

Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities

Mensuration Practice Set 7.3 Question 1.
Complete the following table with the help of given figure. (Textbook pg. no. 149)


Solution:

Question 2.
Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. (Textbook pg. no. 150)

Solution:

Thus, if measure of an arc of a circle is θ, then
Area of sector (A) = \(\frac{\theta}{360}\) × Area of circle
∴ Area of sector (A) = \(\frac{\theta}{360}\) × πr²

Mensuration In Maths Question 3.
In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. (Textbook pg. no. 151)

Solution:

Thus, if the measure of an arc of a circle is 0, then
Length of arc (l) = \(\frac{\theta}{360}\) × circumference of circle
∴ Length of arc (l) = \(\frac{\theta}{360}\) × 2πr

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 7.2 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.2 Chapter 7 Mensuration Questions With Answers Maharashtra Board

Question 1.
The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm³)
Given: Radii (r₁) = 14 cm, and (r₂) = 7 cm,
height (h) = 30 cm
To find: Amount of water the bucket can hold.
Solution:
Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r₁² + r₂² + r₁ × r₂)

∴ The bucket can hold 10.78 litres of water.

Question 2.
The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
i. curved surface area,
ii. total surface area,
iii. volume, (π = 3.14)
Given: Radii (r₁) = 14 cm, and (r₂) = 6 cm,
height (h) = 6 cm
Solution:

i. Curved surface area of frustum
= πl (r₁ + r₂)
= 3.14 × 10(14 + 6)
= 3.14 × 10 × 20 = 628 cm²
∴ The curved surface area of the frustum is 628 cm².

ii. Total surface area of frustum
= πl (r₁+ r₂) + πr₁² + πr₂²
= 628 + 3.14 × (14)² + 3.14 × (6)²
= 628 + 3.14 × 196 + 3.14 × 36
= 628 + 3.14(196 + 36)
= 628 + 3.14 × 232
= 628 + 728.48
= 1356.48 cm²
∴ The total surface area of the frustum is 1356.48 cm².

iii. Volume of frustum
= \(\frac { 1 }{ 3 } \) πth(r₁² +r₂² + r₁ × r₂)
= \(\frac { 1 }{ 3 } \) × 3.14 × 6(14² + 6² + 14 × 6)
= 3.14 × 2(196 + 36 + 84)
= 3.14 × 2 × 316
= 1984.48 cm³
∴ The volume of the frustum is 1984.48 cm³.

Question 3.
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity. (π = \(\frac { 22 }{ 7 } \))
Solution:
Circumference1 = 27πr₁ = 132 cm

Curved surface area of frustum = π (r₁ + r₂) l
= π (21 + 14) × 25
=π × 35 × 35
= \(\frac { 22 }{ 7 } \) × 35 × 25
= 2750 cm²

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 7.1 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.1  Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Class 10 Maths Part 2 Practice Set 7.1 Chapter 7  Mensuration Questions With Answers Maharashtra Board

Practice Set 7.1 Geometry 10th Question 1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Given: For the cone,
radius (r) = 1.5 cm,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:
Volume of cone = \(\frac { 1 }{ 3 } \) πr²h

∴ The volume of the cone is 11.79 cm³.

Mensuration Practice Set 7.1 Question 2. Find the volume of a sphere of diameter 6 cm. [π = 3.14]
Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3 cm
Volume of sphere = \(\frac { 4 }{ 3 } \) πr²
= \(\frac { 4 }{ 3 } \) × 3.14 × (3)³
= 4 × 3.14 × 3 × 3
= 113.04 cm³
∴ The volume of the sphere is 113.04 cm³.

Practice Set 7.1 Geometry Class 10 Question 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. [π = 3.14]
Given: For the cylinder,
radius (r) = 5 cm,
height (h) = 40 cm
To find: Total surface area of the cylinder.
Solution:
Total surface area of cylinder = 2πr (r + h)
= 2 × 3.14 × 5 (5 + 40)
= 2 × 3.14 × 5 × 45
= 1413 cm²
The total surface area of the cylinder is 1413 cm².

Practice Set 7.1 Geometry Question 4. Find the surface area of a sphere of radius 7 cm.
Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:
Surface area of sphere = Aπr²
= 4 × \(\frac { 22 }{ 7 } \) × (7)²
= 88 × 7
= 616 cm²
∴ The surface area of the sphere is 616 cm².

Practice Set 7.1 Question 5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Given: For the cuboid,
length (l) = 44 cm, breadth (b) = 21 cm,
height (h) = 12 cm
For the cone, height (H) = 24 cm
To find: Radius of base of the cone (r).
Solution:
Volume of cuboid = l × b × h
= 44 × 21 × 12 cm³
Volume of cone = \(\frac { 1 }{ 3 } \) πr²H
= \(\frac { 1 }{ 3 } \) × \(\frac { 22 }{ 7 } \) × r² × 24 cm³
Since the cuboid is melted to form a cone,
∴ volume of cuboid = volume of cone

∴ r² = 21 × 21
∴ r = 21 cm …[Taking square root of both sides]
∴ The radius of the base of the cone is 21 cm.

10th Class Geometry Practice Set 7.1 Question 6. Observe the measures of pots in the given figures. How many jugs of water can the cylindrical pot hold?

Given: For the conical water jug,
radius (r) = 3.5 cm, height (h) = 10 cm
For the cylindrical water pot,
radius (R) = 7 cm, height (H) = 10 cm
To find: Number of jugs of water the cylindrical pot can hold.
Solution:
Volume of conical jug = \(\frac { 1 }{ 3 } \) πr²h

∴ The cylindrical pot can hold 12 jugs of water.

Mensuration Class 10 Practice Set 7.1 Question 7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm². The cone is placed up on the cylinder. Volume of the solid figure so formed is 500 cm³. Find the total height of the figure

Given: For the cylindrical part,
height (h) = 3 cm,
area of the base (πr²)= 100 cm²
Volume of the entire figure = 500 cm³
To find: Total height of the figure.
Solution:
A cylinder and a cone have equal bases.
∴ they have equal radii.
radius of cylinder = radius of cone = r
Area of base = 100 cm²
∴ πr² =100 …(i)
Let the height of the conical part be H.
Volume of the entire figure
= Volume of the entire + Volume of cone

∴ Height of conical part (H) =6 cm
Total height of the figure = h + H
= 3 + 6
= 9 cm
∴ The total height of the figure is 9 cm.

10th Geometry Practice Set 7.1 Question 8. In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Given: For the conical Part,
height (h) = 4 cm, radius (r) = 3 cm
For the cylindrical part,
height (H) = 40 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Total area of the toy.
Solution:
Slant height of cone (l) = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)
= \(\sqrt{\mathrm{4}^{2}+\mathrm{3}^{2}}\)
= \(\sqrt { 16+9 }\)
= \(\sqrt { 25 }\) = 5 cm
Curved surface area of cone = πrl
= π × 3 × 5
= 15π cm²
Curved surface area of cylinder = 2πrH
= 2 × π × 3 × 40
= 240π cm²
Curved surface area of hemisphere = 2πr²
= 2 × π × 3²
= 18π cm²
Total area of the toy
= Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
= 15π + 240π + 18π
= 2737 π cm²
∴ The total area of the toy is 273π cm².

7.1.8 Practice Question 9. In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Given: For the cylindrical tablets,
radius (r) = 7 mm,
thickness = height(h) = 5 mm
For the cylindrical wrapper,
diameter (D) = 14 mm, height (H) = 10 cm
To find: Number of tablets that can be wrapped.
Solution:
Radius of wrapper (R) = \(\frac { Diameter }{ 2 } \)
= \(\frac { 14 }{ 2 } \) = 7 mm
Height of wrapper (H) = 10 cm
= 10 × 10 mm
= 100 mm
Volume of a cylindrical wrapper = πR²H
= π(7)² × 100
= 4900π mm³
Volume of a cylindrical tablet = πr²h
= π(7)² × 5
= 245 π mm³
No. of tablets that can be wrapped

∴ 20 tables can be wrapped in the wrapper

Class 10 Maths 7.1 Question 10. The given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.
(π = 3.14)
Given: For the conical part,
height (h) = 4 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Volume and surface area of the toy.
Solution:

Now, volume of the toy
= Volume of cone + volume of hemisphere
= 12π + 18π
= 30π
= 30 × 3.14
= 94.20 cm³
Also, surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
= 15π + 18π
= 33π
= 33 × 3.14
= 103.62 cm²
∴ The volume and surface area of the toy are 94.20 cm³ and 103.62 cm² respectively.

Question 11.
Find the surface area and the volume of a beach ball shown in the figure.

Given: For the spherical ball,
diameter (d) = 42 cm
To find: Surface area and volume of the beach ball.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 42 }{ 2 } \) = 21 cm
Surface area of sphere= 4πr²
= 4 × 3.14 × (21)²
= 4 × 3.14 × 21 × 21
= 5538.96 cm²
Volume of sphere = \(\frac { 4 }{ 3 } \) πr³
= \(\frac { 4 }{ 3 } \) × 3.14 × (21)³
= 4 × 3.14 × 7 × 21 × 21
= 38772.72 cm³
∴ The surface area and the volume of the beach ball are 5538.96 cm² and 38772.72 cm³ respectively.

Question 12.
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

Given: For the metal sphere,
diameter (d) = 2 cm
For the cylindrical glass, diameter (D) =14 cm,
height of water in the glass (H) = 30 cm
To find: Volume of water in the glass.
Solution:
Let the radii of the sphere and glass be r and R respectively.

Volume of water with sphere in it = πR²H
= π × (7)² × 30
= 1470π cm³
Volume of water in the glass
= Volume of water with sphere in it – Volume of sphere

∴ The volume of the water in the glass is 1468.67 π cm³ (i.e. 4615.80 cm³).

Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities

Question 1.
The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm³) (Textbook pg. no.141)

Given: For the cuboidal can,
length (l) = 20 cm,
breadth (b) = 20 cm,
height (h) = 30 cm
To find: Oil that can be contained in the can.
Solution:
Volume of cuboid = l × b × h
= 20 × 20 × 30
= 12000 cm³
= \(\frac { 12000 }{ 1000 } \) litres
= 12 litres
∴ The oil can will contain 12 litres of oil.

Question 2.
The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap? (Textbook pg. no. 141)

Given: For the conical cap,
radius (r) = 10 cm,
slant height (l) = 21 cm
To find: Cloth required to make the cap.
Solution:
Cloth required to make the cap
= Curved surface area of the conical cap
= πrl = \(\frac { 22 }{ 7 } \) × 10 × 21
=22 × 10 × 3
= 660 cm²
∴ 660 cm² of cloth will be required to make the cap.

Question 3.
As shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is ‘r’,

i. what is the ratio of the radii of the sphere and the cylinder ?
ii. what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
iii. what is the ratio of the volumes of the cylinder and the sphere? (Textbook pg. no. 141)
Solution:
Radius of base of cylinder = radius of sphere
∴ Radius of sphere = r
Also, height of cylinder = diameter of sphere
∴ h = d
∴ h = 2r …(i)

∴ radius of sphere : radius of cylinder = 1 : 1.

∴ curved surface area of cylinder : surface area of sphere = 1:1.

∴ volume of cylinder : volume of sphere = 3 : 2.

Question 4.
Finding volume of a sphere using cylindrical beaker and water. (Textbook, pg. no. 142)

i. Take a ball and a beaker of the same radius.
ii. Cut a strip of paper of length equal to the diameter of the beaker.
iii. Draw two lines on the strip dividing it into three equal parts.
iv. Stick this strip on the beaker straight up from the bottom.
v. Fill the water in the beaker upto the first mark of the strip from bottom.
vi. Push the ball in the beaker so that it touches the bottom.
Observe how much water level rises.
You will notice that the water level has risen exactly upto the total height of the strip. Try to obtain the formula for volume of sphere using the volume of the cylindrical beaker.
Solution:
Suppose volume of beaker upto height 2r is V.
V = πr² h
∴ V = πr²(2r) …[∵ h = 2r]
∴ V = 2πr³
But, V = volume of the ball + volume of water in the beaker
∴ 2πr³ = Volume of the ball + \(\frac { 1 }{ 3 } \) × 2πr³
∴ Volume of the ball = 2πr³ – \(\frac { 2 }{ 3 } \) πr³
= \(\frac{6 \pi r^{3}-2 \pi r^{3}}{3}\)
∴ Volume of the ball = \(\frac { 4 }{ 3 } \) πr³
∴ Volume of a sphere = \(\frac { 4 }{ 3 } \) πr³

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

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

∴ 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

∴ 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

∴ 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

∴ 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

∴ 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

∴ 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:


∴ 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.

∴ 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.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ 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.

∴ 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.

∴ 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:

∴ 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:


∴ 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

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

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:

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 State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

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,

∴ 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,

∴ 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,

∴ 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,

∴ 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,

∴ 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

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ 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,

∴ 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,

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,

∴ 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,

∴ 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,

∴ 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,

∴ 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,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 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

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

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.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ 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

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ 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,

∴ 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.

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 State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 6.2 Chapter 6 Trigonometry Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 6.2 Chapter 6 Trigonometry Questions With Answers Maharashtra Board

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.

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

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]

∴ 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.

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.

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,

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.

∠ACB = 60°
AB = 60 m
In right angled ∆ABC,
sin 60° = \(\frac { AB }{ AC } \) … [By definition]

∴ AC = 40 \(\sqrt { 3 }\) = 40 × 1.73 = 69.20 m
∴ The length of the string is 69.20 m.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

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,

∴ 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,

∴ 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,

∴ 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,

∴ 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,

∴ 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.

∴ 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,


∴ 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,

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,

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,

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

∴ (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

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

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


∴ 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,

∴ 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

∴ 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)

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 State Board Class 10 Maths Solutions Part 2

• Co-ordinate Geometry Practice Set 5.1 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.2 Class 10 Maths Solutions
• Co-ordinate Geometry Practice Set 5.3 Class 10 Maths Solutions
• Co-ordinate Geometry Problem Set 5 Class 10 Maths Solutions
• Trigonometry Practice Set 6.1 Class 10 Maths Solutions
• Trigonometry Practice Set 6.2 Class 10 Maths Solutions
• Trigonometry Problem Set 6 Class 10 Maths Solutions
• Mensuration Practice Set 7.1 Class 10 Maths Solutions
• Mensuration Practice Set 7.2 Class 10 Maths Solutions
• Mensuration Practice Set 7.3 Class 10 Maths Solutions
• Mensuration Practice Set 7.4 Class 10 Maths Solutions
• Mensuration Problem Set 7 Class 10 Maths Solutions