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9th Standard Maths 2 Problem Set 8 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 8 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Problem Set 8 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following multiple choice questions.

i. Which of the following statements is true?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
Answer:
(A) sin θ = cos (90 – θ)

ii. Which of the following is the value of sin 90°?
(A) \( \frac { \sqrt { 3 } }{ 2 }\)
(B) 0
(C) \(\frac { 1 }{ 2 }\)
(D) 1
Answer:
(D) 1

iii. 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
2 tan 45° + cos 45° – sin
\( =2(1)+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\)
(C) 2

iv. \( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\) =?
(A) 2
(B) -1
(C) 0
(D) 1
Answer:
\( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\)

(D) 1

Question 2.
In right angled ∆TSU, TS = 5, ∠S = 90°, SU = 12, then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Solution:
i. TS = 5, SU = 12 …[Given]
In ∆TSU, ∠S = 90° … [Given]
∴ TU² = TS² + SU² …[Pythagoras theorem]
= 5² + 12² = 25 + 144 = 169
∴ TU = \(\sqrt { 169 }\) .. .[Taking square root of both sides]
= 13

Question 3.
In right angled ∆YXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:
i. XZ = 8 cm, YZ = 17 cm …[Given]
In ∆YXZ, ∠X = 90° … [Given]
∴ YZ² = XY² + XZ² .. .[Pythagoras theorem]
∴ 17² = XY² + 8²
∴ 289 = XY² + 64
∴ XY² = 289 – 64
= 225
∴ x = \(\sqrt { 225 }\) .. .[Taking square root of both sides]
= 15

Question 4.
In right angled ∆LMN, if ∠N = θ, ∠M = 90°, cos θ = \(\frac { 24 }{ 25 }\), find sin θ and tan θ. Similarly, find (sin²θ) and (cos²θ).

Solution:
i. cos θ = \(\frac { 24 }{ 25 }\)
In ∆LMN, ∠M = 90°, ∠N = θ

Let the common multiple be k.
∴ MN = 24k and LN = 25k
Now, LN²= LM² + MN² … [Pythagoras theorem]
∴ (25k)² = LM² + (24k)²
∴ 625 k² = LM² + 576k²
∴ LM² = 625k² – 576k²
∴ LM² = 49k²
∴ LM = \(\sqrt { 49{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 7k

Question 5.
Fill in the blanks.
i. sin 20° = cos
ii. tan 30° x tan  = 1
iii. cos 40° = sin
Solution:
i. sin 20° = cos (90° – 20°) …..[∵ sin θ = cos (90 – θ)]
= cos 70°

ii. tan θ x tan (90 – θ) = 1
Substituting θ = 30°,
tan 30° x tan (90 – 30)° = 1
∴ tan 30° x tan 60° = 1

iii. cos 40° = sin (90° – 40°) …[∵ COS θ = sin (90 – θ)]
= sin 50°

Maharashtra Board Class 9 Maths Solutions Chapter 8 Trigonometry Problem Set 8

Question 1.
Measuring height of a tree using trigonometric ratios. (Textbook pg. no. 101)

This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
Solution:
Rays of sunlight are parallel.
So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.
Sides of similar triangles are proportional.
∴ \(\frac { QR }{BC }\) = \(\frac { PR }{ AC }\)
∴ Height of the tree (QR) = \(\frac { BC }{ AC }\) x PR
Substituting the values of PR, BC and AC in the above equation, we can get length of QR i.e., the height of the tree.

Question 2.
It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8’O clock. Can you tell why? (Textbook pg. no. 101)
Solution:
At 8’O clock in the morning, the sunlight is not very bright. At the same time, the sun is on the horizon and the shadow would by very long. It would be extremely difficult to measure shadow in that case.
Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to measure the length of shadow.

Question 3.
Conduct the above discussed activity and find the height of a tall tree in your surrounding. If there is no tree in the premises, then find the height of a pole. (Textbook pg. no. 101)

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

10th Standard Maths 2 Practice Set 4.1 Chapter 4 Geometric Constructions Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 4 Geometric Constructions.

Class 10 Maths Part 2 Practice Set 4.1 Chapter 1 Geometric Constructions Questions With Answers Maharashtra Board

Question 1.
∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that \(\frac { BC }{ MN } \) = \(\frac { 5 }{ 4 } \)
Solution:
Analysis:

Question 2.
∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \)
Solution:
Analysis:
As shown in the figure, Let R – P – L and R – Q – T.
∆PQR ~ ∆LTR … [Given]
∴ ∠PRQ ≅ ∠LRT … [Corresponding angles of similar triangles]

\(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) …(i)[Corresponding sides of similar triangles]
But, \(\frac { PQ }{ LT } \) = \(\frac { 3 }{ 4 } \) ….(ii) [Given]
∴ \(\frac { PQ }{ LT } \) = \(\frac { QR }{ TR } \) = \(\frac { PR }{ LR } \) = \(\frac { 3 }{ 4 } \) …[From (i) and (ii)]
∴ sides of LTR are longer than corresponding sides of ∆PQR.
If seg QR is divided into 3 equal parts, then seg TR will be 4 times each part of seg QR.
So, if we construct ∆PQR, point T will be on side RQ, at a distance equal to 4 parts from R.
Now, point L is the point of intersection of ray RP and a line through T, parallel to PQ.
∆LTR is the required triangle similar to ∆PQR.

Steps of construction:
i. Draw ∆PQR of given measure. Draw ray RS making an acute angle with side RQ.
ii. Taking convenient distance on the compass, mark 4 points R₁, R₂, R₃, and R₄, such that RR₁ = R₁R₂ = R₂R₃ = R₃R₄.
iii. Join R₃Q. Draw line parallel to R₃Q through R₄ to intersects ray RQ at T.
iv. Draw a line parallel to side PQ through T. Name the point of intersection of this line and ray RP as L.
∆LTR is the required triangle similar to ∆PQR.

Question 3.
∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm. Construct ∆RST and ∆XYZ, such that \(\frac { RS }{ XY } \) = \(\frac { 3 }{ 5 } \).
Solution:
Analysis:
∆RST ~ ∆XYZ … [Given]
∴ ∠RST ≅ ∠XYZ = 40° … [Corresponding angles of similar triangles]

Question 4.
∆AMT ~ ∆ANE. In ∆AMT, AM = 6.3 cm, ∠TAM = 500, AT = 5.6 cm. \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) Construct ∆AHE.
Solution:
Analysis:
As shown in the figure,
Let A – H – M and A – E – T.
∆AMT ~ ∆AHE … [Given]
∴ ∠TAM ≅ ∠EAH … [Corresponding angles of similar triangles]

\(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AE } \) ….. (i)[Corresponding sides of similar triangles]
But, \(\frac { AM }{ AH } \) = \(\frac { 7 }{ 5 } \) …(ii)[Given]
∴ \(\frac { AM }{ AH } \) = \(\frac { MT }{ HE } \) = \(\frac { AT }{ AH } \) = \(\frac { 7 }{ 5 } \) …[From (i) and (ii)]
∴ Sides of AAMT are longer than corresponding sides of ∆AHE.
∴ The length of side AH will be equal to 5 parts out of 7 equal parts of side AM.
So, if we construct AAMT, point H will be on side AM, at a distance equal to 5 parts from A.
Now, point E is the point of intersection of ray AT and a line through H, parallel to MT.
∆AHE is the required triangle similar to ∆AMT.

Steps of construction:
i. Draw ∆AMT of given measure. Draw ray AB making an acute angle with side AM.
ii. Taking convenient distance on the compass, mark 7 points A1, A2, A3, A4, A5, Ag and A7, such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
iii. Join A₇M. Draw line parallel to A₇M through A₅ to intersects seg AM at H.
iv. Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E.
∆AHE is the required triangle similar to ∆AMT.

Maharashtra Board Class 10 Maths Chapter 4 Geometric Constructions Intext Questions and Activities

Question 1.
If length of side AB is \(\frac { 11.6 }{ 2 } \) cm, then by dividing the line segment of length 11.6 cm in three equal parts, draw segment AB. (Textbook pg. no. 93)
Solution:

Steps of construction:
i. Draw seg AD of 11.6 cm.
ii. Draw ray AX such that ∠DAX is an acute angle.
iii. Locate points A₁, A₂ and A₃ on ray AX such that AA₁ = A₁A₂ = A₂A₃
iv. Join A₃D.
v. Through A₁, A₂ draw lines parallel to A₃D intersecting AD at B and C, wherein
AB = \(\frac { 11.6 }{ 3 } \) cm

Question 2.
Construct any ∆ABC. Construct ∆ A’BC’ such that AB : A’B = 5:3 and ∆ ABC ~ ∆ A’BC’. (Textbook pg. no. 93)
Analysis:
As shown in the figure,
Let B – A’ – A and B – C’ -C
∆ ABC – A’BC’ … [Given]
∴ ∠ABC ≅ ∠A’BC’ …[Corresponding angles of similar trianglesi


∴ Sides of ∆ABC are longer than corresponding sides of ∆A’BC’. Rough figure
∴ the length of side BC’ will be equal to 3 parts out of 5 equal parts of side BC.
So if we construct ∆ABC, point C’ will be on side BC, at a distance equal to 3 parts from B.
Now A’ is the point of intersection of AB and a line through C’, parallel to CA.
Solution:
Let ∆ABC be any triangle constructed such that AB = 7 cm, BC = 7 cm and AC = 6 cm.

Question 3.
Construct any ∆ABC. Construct ∆A’BC’ such that AB: A’B = 5:3 and ∆ABC ~ ∆A’BC’.
∆A’BC’ can also be constructed as shown in the adjoining figure. What changes do we have to make in steps of construction in that case? (Textbook pg. no. 94)

Solution:
Let ∆ABC be any triangle constructed such that AB = 5cm,
BC = 5.5 cm and AC = 6 cm.

i. Steps of construction:
Construct ∆ABC, extend rays AB and CB.
Draw line BM making an acute angle with side AB.
Mark 5 points B₁, B₂, B₃, B₄, B₅ starting from B at equal distance.
Join B₃C” (ie 3rd part)
Draw a line parallel to AB₅ through B₃ to intersect line AB at C”
Draw a line parallel to AC through C” to intersect line BC at A”
ii. Extra construction:
With radius BC” cut an arc on extended ray CB at C’ [C’ – B – C]
With radius BA” cut an arc on extended ray AB at A’ [A’ – B – A]
∆A’BC’ is the required triangle.

Maharashtra State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

9th Standard Maths 2 Practice Set 8.2 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.2 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = \(\sqrt { 2ghK }\)² … [Taking square root of both sides]
= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]
In right angled ∆ABC, ∠C = θ.


Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.



Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.
Solution:
sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.
Solution:
cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ


Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

10th Standard Maths 2 Problem Set 3 Chapter 3 Circle Textbook Answers Maharashtra Board

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

Class 10 Maths Part 2 Problem Set 3 Chapter 3 Circle Questions With Answers Maharashtra Board

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

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°.

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)

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.

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]

∴ 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]

∴ 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°

∴ 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]

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?

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.

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.

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.

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]

∴ 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.

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]

∴ 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.

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.

∴ ∠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.

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.

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]

∴ ∠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.

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 ]

∴ 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.

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]

∴ 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]

∴ 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.

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]

∴ 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.

∴ 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.

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).

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.

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.

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.

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.

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]

∴ ∠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.

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]

∴ 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.

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.

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]

∴ 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°.

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°]

∴ ∠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.

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)]

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.

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,

∠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.

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]

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 State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 3.5 Chapter 3 Circle Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 3.5 Chapter 3 Circle Questions With Answers Maharashtra Board

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]

∴ 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

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.

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

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

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]

∴ ∠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)

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]

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]

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]

∠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.

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.

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.

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.

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°

∴ ∆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]

∴ \(\frac { PS }{ SR } \) = \(\frac { SR }{ SQ } \) [Corresponding sides of similar triangles]
∴ SR² = PS × SQ

Maharashtra State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

9th Standard Maths 2 Practice Set 8.1 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.1 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.

i. sin P
ii. cos Q
iii. tan P
iv. tan Q
Solution:

Question 2.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.

i. sin x
ii. tan z
iii. cos x
iv. tan x.
Solution:

Question 3.
In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.

i. sin 50°
ii. cos 50°
iii. tan 40°
iv. cos 40°
Solution:

Question 4.
In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
i. sin α, cos α , tan α
ii. sin θ, cos θ, tan θ

Solution:
i. In ∆PQR,

ii. In ∆PQS,

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1.
In the figure gIven below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)

Solution:
In right angled ∆PQR,
i. side opposite to ∠P = QR
ii. side opposite to ∠R = PQ
iii. side adjacent to ∠P = PQ
iv. side adjacent to ∠R = QR

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

10th Standard Maths 2 Practice Set 3.4 Chapter 3 Circle Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 3.4 Chapter 3 Circle Questions With Answers Maharashtra Board

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.

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

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]

∴ 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.

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]

∴ 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.

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.

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:

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).

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.

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)

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.

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)

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)

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°

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

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.

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

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.

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]

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]

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 State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

10th Standard Maths 2 Practice Set 3.3 Chapter 3 Circle Textbook Answers Maharashtra Board

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.

Class 10 Maths Part 2 Practice Set 3.3 Chapter 3 Circle Questions With Answers Maharashtra Board

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).

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°.

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.

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

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

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

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

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.

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

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

Maharashtra State Board Class 10 Maths Solutions Part 2

• Circle Practice Set 3.1 Class 10 Maths Solutions
• Circle Practice Set 3.2 Class 10 Maths Solutions
• Circle Practice Set 3.3 Class 10 Maths Solutions
• Circle Practice Set 3.4 Class 10 Maths Solutions
• Circle Practice Set 3.5 Class 10 Maths Solutions
• Circle Problem Set 3 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.1 Class 10 Maths Solutions
• Geometric Constructions Practice Set 4.2 Class 10 Maths Solutions
• Geometric Constructions Problem Set 4 Class 10 Maths Solutions

9th Standard Maths 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Class 9 Maths Part 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following questions.

i. What is the form of co-ordinates of a point on the X-axis?
(A) (b,b)
(B) (0, b)
(C) (a, 0)
(D) (a, a)
Answer:
(C) (a, 0)

ii. Any point on the line y = x is of the form _____.
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, -a)
Answer:
(A) (a, a)

iii. What is the equation of the X-axis ?
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y
Answer:
(B) y = 0

iv. In which quadrant does the point (-4, -3) lie ?
(A) First
(B) Second
(C) Third
(D) Fourth
Answer:
(C) Third

v. What is the nature of the line which includes the points (-5, 5), (6, 5), (-3, 5), (0, 5)?
(A) Passes through the origin
(B) Parallel to Y-axis
(C) Parallel to X-axis
(D) None of these
Answer:
The y co-ordinate of all the points is the same.
∴ The line which passes through the given points is parallel to X-axis.
(C) Parallel to X-axis

vi. Which of the points P(-1, 1), Q(3, -4), R( -1, -1), S(-2, -3), T (-4, 4) lie in the fourth quadrant?
(A) P and T
(B) Q and R
(C) only S
(D) P and R
Answer:
(B) Q and R

Question 2.
Some points are shown in the adjoining figure. With the help of it answer the following questions :

i. Write the co-ordinates of the points Q and R.
ii. Write the co-ordinates of the points T and M.
iii. Which point lies in the third quadrant ?
iv. Which are the points whose x and y co-ordinates are equal ?
Solution:
i. Q(-2, 2) and R(4, -1)
ii. T(0, -1) and M(3, 0)
iii. Point S lies in the third quadrant.
iv. The x and y co-ordinates of point O are equal.

Question 3.
Without plotting the points on a graph, state in which quadrant or on which axis do the following points lie.
i. (5, -3)
ii. (-7, -12)
iii. (-23, 4)
iv. (-9, 5)
v. (0, -3)
vi. (-6, 0)
Solution:

Question 4.
Plot the following points on one and the same co-ordinate system.
A(1, 3), B(-3, -1), C(1, -4), D(-2, 3), E(0, -8), F(1, 0)
Solution:

Question 5.
In the graph alongside, line LM is parallel to the Y-axis.

i. What is the distance of line LM from the Y-axis?
ii. Write the co-ordinates of the points P, Q and R.
iii. What is the difference between the x co-ordinates of the points L and M?
Solution:
i. Distance of line LM from the Y-axis is 3 units.
ii. P(3, 2), Q (3, -1), R(3, 0)
iii. x co-ordinate of point L = 3
x co-ordinate of point M = 3
∴ Difference between the x co-ordinates of the points L and M = 3 – 3
= 0

Question 6.
How many lines are there which are parallel to X-axis and having a distance 5 units?
Solution:
The equation of a line parallel to the X-axis is y = b.
There are 2 lines which are parallel to X-axis and at a distance of 5 units.
Their equations are y = 5 and y = -5.

Question 7.
If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a?
Solution:
Equation of Y-axis is x = 0.
Since, ‘a’ is a real number, there are two possibilities.
Case I: a > 0
Case II: a < 0 ∴ Distance between the Y-axis and the line x = a = a-0 = a Since, |a| = a, a > 0
= – a, a < 0
∴ Distance between the Y-axis and the line x = a is |a|.

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Problem Set 7 Intext Questions and Activities

Question 1.
As shown in the adjoining figure, ask girls to sit in lines so as to form the X-axis and Y-axis.
i. Ask some boys to sit at the positions marked by the coloured dots in the four quadrants.
i. Now, call the students turn by turn using the initial letter of each student’s name. As his or her initial is called, the student stands and gives his or her own co-ordinates. For example Rajendra (2, 2) and Kirti (-1, 0)
iii. Even as they have fun during this field activity, the students will leam how to state the position of a point in a plane. (Textbook pg. no. 92)

Maharashtra Board Class 9 Maths Solutions

• Co-ordinate Geometry Practice Set 7.1 Class 9 Maths Solutions
• Co-ordinate Geometry Practice Set 7.2 Class 9 Maths Solutions
• Co-ordinate Geometry Problem Set 7 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

Class 9 Maths Part 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
On a graph paper plot the points A(3, 0), B(3, 3), C(0, 3). Join A, B and B, C. What is the figure formed?
Soiution:

d(O, A) = 3 cm, d(A, B) = 3 cm, d(B, C) = 3 cm, d(O, C) = 3 cm and each angle of □ OABC is 90°
∴ □ OABC is a square.

Question 2.
Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Since, the line is at a distance of 7 units to the left of Y-axis,
∴ a = -7
∴ x = -1 is the equation of the required line.

Question 3.
Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.
Solution:
The equation of a line parallel to the X-axis is y = b.
Since, the line is at a distance of 5 units below the X-axis.
∴ b = -5
∴ y = -5 is the equation of the required line.

Question 4.
The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Here, a = -3
∴ x = -3 is the equation of the required line.

Question 5.
Y-axis and line x = – 4 are parallel lines. What is the distance between them?
Solution:
Equation of Y-axis is x = 0.
Equation of the line parallel to the Y-axis is x = – 4. … [Given]
∴ Distance between the Y-axis and the line x = – 4 is 0 – (- 4) … [0 > -4]
= 0 + 4 = 4 units
∴ The distance between the Y-axis and the line x = – 4 is 4 units.
[Note: The question is modified as X-axis cannot be parallel to the line x = – 4.]

Question 6.
Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis? [1 Mark each]
i. x = 3
ii. y – 2 = 0
iii. x + 6 = 0
iv. y = -5
Solution:
i. The equation of a line parallel to the Y-axis is x = a.
∴ The line x = 3 is parallel to the Y-axis.

ii. y – 2 = 0
∴ y = 2
The equation of a line parallel to the X-axis is y = b.
∴ The line y – 2 = 0 is parallel to the X-axis.

iii. x + 6 = 0
∴ x = -6
The equation of a line parallel to the Y-axis is x = a.
∴ The line x + 6 = 0 is parallel to the Y-axis.

iv. The equation of a line parallel to the X-axis is y = b.
∴ The line y = – 5 is parallel to the X-axis.

Question 7.
On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If these points are collinear, then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.
Solution:

From the graph, the line drawn intersects the X-axis at D(5, 0) and the Y-axis at C(0, 5).

Question 8.
Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.
x + 4 = 0,
y – 1 = 0,
2x + 3 = 0,
3y – 15 = 0
Solution:
i. x + 4 = 0
∴ x = – 4

ii. y – 1 = 0
∴ y = 1

iii. 2x + 3 = 0
∴2x = -3
∴ x = \(\frac { -3 }{ 2 }\)
∴ x = -1.5

iv. 3y- 15 = 0
3y = 15
y = \(\frac { 15 }{ 3 }\)
∴ y = 5

The co-ordinates of the point of intersection of x + 4 = 0 and y – 1 = 0 are A(-4, 1).
The co-ordinates of the point of intersection ofy – 1 = 0 and 2x + 3 = 0 are B(-1.5, 1).
The co-ordinates of the point of intersection of 3y – 15 = 0 and 2x + 3 = 0 are C(-1.5, 5).
The co-ordinates of the point of intersection of x + 4 = 0 and 3y – 15 = 0 are D(-4, 5).

Question 9.
Draw the graphs of the equations given below.
i. x + y = 2
ii. 3x – y = 0
iii. 2x + y = 1
Solution:
i. x + y = 2
∴ y = 2 – x
When x = 0,
y = 2 – x
= 2 – 0
= 2
When x = 1,
y = 2 – x
= 2 – 1
= 1
When x = 2,
y = 2 – x
= 0

ii. 3x – y = 0
∴ y = 3x
When x = 0,
y = 3x
= 3(0)
= 0

When x = 1,
y = 3x
= 3(1)
= 3

When x = -1,
y = 3x
= 3(-1)
= -3

iii. 2x + y = 1
∴ y = 1 – 2x
When x = 0,
y = 1 – 2x
= 1 – 2(0)
= 1 – o
When x = 1,
y = 1 – 2x
= 1- 2(1)
= 1 – 2
= -1
When x = -1,
y = 1 – 2x
= 1 – 2(-1)
= 1 + 2
= 3

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.2 Intext Questions and Activities

Question 1.
i. Can we draw a line parallel to the X-axis at a distance of 6 unIts from It and below the X-axis?
ii. Will all of the points (-3,-6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) be on that line?
iii. What would be the equation of this line?(Textbook pg. no. 94)
Solution:

i. Yes.
This line will pass through the point (0,-6).

ii. Yes.
Here, y co-ordinate of the points (-3, -6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) is the same, which is -6.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 6 units below the X-axis.
∴ b = -6
∴ Equation of the line is y = -6.

Question 2.
i. Can we draw a line parallel to the Y – axis at a distance of 2 units from ¡t and to its right?
ii. Will all of the points (2, 10), (2, 8), (2, -) be on that line?
iii. What would be the equation of this line? (Textbook pg. no. 95)
Solution:

i. Yes.
(2, 10)
This line will pass through the point (2, 0).
(2,8)
ii. Yes.
Here, x co-ordinate of the points (2, 10), (2, 8), (2,-\(\frac { 1 }{ 2 }\) ) is the same, which is 2.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 2 units to the right of Y-axis.
a = 2
∴ Equation of the line is x = 2.

Question 3.
On a graph paper, plot the points (0, 1), (1, 3), (2, 5). Are they collinear? If so, draw the line that passes through them.
i. Through which quadrants does this line pass ?
ii. Write the co-ordinates of the point at which it intersects the Y-axis.
iii. Show any point in the third quadrant which lies on this line. Write the co-ordinates of the point. (Textbook pg. no. 96)
Solution:
i. The line passes through the quadrants I, II and III.
ii. The line intersects the Y-axis at (0, 1).
iii. (-1,-1)

Maharashtra Board Class 9 Maths Solutions

• Co-ordinate Geometry Practice Set 7.1 Class 9 Maths Solutions
• Co-ordinate Geometry Practice Set 7.2 Class 9 Maths Solutions
• Co-ordinate Geometry Problem Set 7 Class 9 Maths Solutions