Class 10 Maths Triangles questions with solutions

Q 1 :

The discus throw is an event in which an athlete aims to throw a discus as far as possible. The athlete spins counterclockwise one and a half times within a circle before releasing the discus. Upon release, the discus travels along a tangent to the circular path of the spin.

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and $∠ A B O = 30 ° .$ PQ is parallel to OA

Based on the above information answer the following questions:

(i) Find the length of AB.

$75 \sqrt{3} cm$
$75 \sqrt{2} cm$
$65 \sqrt{3} cm$
$- 75 \sqrt{3} cm$

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(1)

Since OA is radius and BA is tangent to the circle at point A

$∴ ∠ O A B = 90 ° I n Δ O A B, \tan 30 ° = \frac{O A}{A B} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{A B} \Rightarrow A B = 75 \sqrt{3} c m$

Q 2 :

The discus throw is an event in which an athlete aims to throw a discus as far as possible. The athlete spins counterclockwise one and a half times within a circle before releasing the discus. Upon release, the discus travels along a tangent to the circular path of the spin.

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and $angle A B O space equals space 30 degree. space$ PQ is parallel to OA

Based on the above information answer the following questions:

(ii) Find the length of OB.

150 cm
140
130
120

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(1)

$I n Δ O A B, w e h a v e \sin 30 ° = \frac{O A}{O B} \Rightarrow \frac{1}{2} = \frac{75}{O B} \Rightarrow O B = 150 cm$

Q 3 :

The discus throw is an event in which an athlete aims to throw a discus as far as possible. The athlete spins counterclockwise one and a half times within a circle before releasing the discus. Upon release, the discus travels along a tangent to the circular path of the spin.

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and $angle A B O space equals space 30 degree. space$ PQ is parallel to OA

Based on the above information answer the following questions:

(iii) (a) Find the length of AP

$\frac{75 \sqrt{3}}{2} cm$
$\frac{75 \sqrt{4}}{2} cm$
$- \frac{75 \sqrt{3}}{4} cm$
$\frac{76 \sqrt{3}}{2} cm$

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(1)

Since the radius of circle is perpendicular to the tangent at the point of contact $∴ ∠ O A P = 90 °$

$S i n c e P Q ∥ O A \Rightarrow ∠ O A P = ∠ Q P B = 90 ° (C o r r e s p o n d i n g a n g l e s) N o w, O B = 150 c m O Q + Q B = 150 \Rightarrow 75 + Q B = 150 \Rightarrow Q B = 150 - 75 = 75 cm I n r i g h t Δ P B Q, \cos 30 ° = \frac{P B}{Q B} \Rightarrow \frac{\sqrt{3}}{2} = \frac{P B}{75} \Rightarrow P B = \frac{75 \sqrt{3}}{2} cm A P = A B - P B = 75 \sqrt{3} - \frac{75 \sqrt{3}}{2} \Rightarrow A P = \frac{75 \sqrt{3}}{2} cm$

Q 4 :

The discus throw is an event in which an athlete aims to throw a discus as far as possible. The athlete spins counterclockwise one and a half times within a circle before releasing the discus. Upon release, the discus travels along a tangent to the circular path of the spin.

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and $angle A B O space equals space 30 degree. space$ PQ is parallel to OA

Based on the above information answer the following questions:

(iii) (b) Find the length of PQ.

35.5
37.5
36.5
33.5

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(2)

$I n r i g h t Δ Q P B, \sin \ funcapply 30 ° = \frac{P Q}{Q B} \Rightarrow \frac{1}{2} = \frac{P Q}{75} \Rightarrow P Q = \frac{75}{2} = 37.5 cm$

Q 5 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(i) If the point D is 20 m away from point A, whereas AB and AC are 80 m and 100 m respectively, find the length of AE.

20 m
15 m
25 m
10 m

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(3)

$I n Δ A B C, D E ∥ B C \frac{A D}{A B} = \frac{A E}{A C} (By BPT) \frac{20}{80} = \frac{A E}{100} \Rightarrow A E = 25 m$

Q 6 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(ii) If AD = x + 1, DB = 3x – 1, AE = x + 3, EC = 3x + 4, then find the value of x.

7
5
4
3

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(1)

Given, AD = x + 1, DB = 3x – 1, AE = x + 3, EC = 3x + 4.

$S i n c e, i n Δ A B C, D E ∥ B C \frac{A D}{D B} = \frac{A E}{E C} (Using BPT) \frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4} C r o s s m u l t i p l y i n g : (x + 1) (3 x + 4) = (3 x - 1) (x + 3) 3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3 7 x + 4 = 8 x - 3 \Rightarrow x = 7$

Q 7 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(iii) (a) State whether $\frac{B D}{A D} = \frac{A E}{E C}$ or not. Give reason

$\frac{B D}{A D} \neq \frac{A E}{E C}$
$\frac{A D}{D B} \neq \frac{A E}{E C}$
$\frac{B D}{A E} \neq \frac{D E}{E C}$
$\frac{B D}{A D} = \frac{A E}{E C}$

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(1)

$N o, b e c a u s e w h e n i n Δ A B C, D E ∥ B C \frac{A D}{D B} = \frac{A E}{E C} (Using BPT) \Rightarrow \frac{1}{A D / D B} = \frac{1}{A E / E C} \Rightarrow \frac{B D}{A D} = \frac{E C}{A E} H e n c e, \frac{B D}{A D} \neq \frac{A E}{E C}$

Q 8 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(iii) (b) If P and Q are the midpoints of sides YZ and XZ respectively in $Δ Z Y X$ , then state whether $P Q ∥ Y X$ or not. Give reason.

$P Q ∥ Y X$
$P X ∥ Y Q$
$Q X ∥ Y P$
$- P X ∥ Y Q$

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(1)

$I n Δ Z Y X,$ P and Q are midpoints of side YZ and XZ respectively.

$\frac{Z P}{P Y} = 1 and \frac{Z Q}{Q X} = 1 \Rightarrow \frac{Z P}{P Y} = \frac{Z Q}{Q X} \Rightarrow P Q ∥ Y X (By converse of BPT) Y e s, P Q ∥ Y X .$

Q 9 :

Vijay is trying to estimate the height of a tower near his house using the properties of similar triangles. His house, which is 20 m tall, casts a 10 m long shadow on the ground.

At the same time, the tower casts a 50 m long shadow on the ground, and Ajay’s house casts a 20 m long shadow on the ground.

Based on the above information, answer the following questions:

(i) What is the height of the tower?

100 m
150 m
120 m
140 m

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(1)

The figure can be drawn as Fig. 6.35.

Let CD = h m be the height of the tower, BE = 20 m be the height of Vijay’s house, and GF be the height of Ajay’s house.

Since, $∆ A C D ~ ∆ A B E \frac{A C}{A B} = \frac{C D}{E B} \frac{50}{10} = \frac{h}{20} h = 100 m$

Q 10 :

Vijay is trying to estimate the height of a tower near his house using the properties of similar triangles. His house, which is 20 m tall, casts a 10 m long shadow on the ground.

At the same time, the tower casts a 50 m long shadow on the ground, and Ajay’s house casts a 20 m long shadow on the ground.

Based on the above information, answer the following questions:

(ii) What will be the length of the shadow of the tower when Vijay’s house casts a 12 m long shadow?

50 m
60 m
40 m
30 m

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(2)

Given AB = 12 m, let AC = x.

$I n s i m i l a r Δ A B E a n d Δ A C D : \frac{A B}{A C} = \frac{B E}{C D} \frac{12}{x} = \frac{20}{100} x = \frac{12 \times 100}{20} = 12 \times 5 = 60 m$

Topic Question Set

Q 6 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(ii) If AD = x + 1, DB = 3x – 1, AE = x + 3, EC = 3x + 4, then find the value of x.

Q 7 :

A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.

Based on the above information answer the following questions:

(iii) (a) State whether $\frac{B D}{A D} = \frac{A E}{E C}$ or not. Give reason

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