Exercise 10.2 (NCERT) - Circles, Class 10
| Chapter 10 | Exercise 10.2 | Class 10 |
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Below are the solutions of every question of exercise 10.2 of chapter 10, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.
Question 1
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Given
Length of tangent, PQ = 24 cm
OQ = 25 cm
Let the radius = r = OP
Since PQ is a tangent to the circle at P
Hence ∠OPQ = 90
In right angled △OPQ
Using Pythagoras Theorem
Question 2
In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60° (B) 70° (C) 80° (D) 90°
Given
∠POQ = 110°
TP and TQ are tangents from point T to the circle.
So, ∠OPT = ∠OQT = 90° (radius ⟂ tangent)
Now, in quadrilateral POQT:
- ∠OPT = 90°
- ∠OQT = 90°
- ∠POQ = 110°
- ∠PTQ = ?
Using the angle sum property of a quadrilateral:
∠PTQ = 360° – (90° + 90° + 110°) = 70°
Answer: B
Question 3
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50° (B) 60° (C) 70° (D) 80°
Given
∠APB = 80°
PA and PB are tangents from point P to the circle.
OA = radius
Hence ∠OAP = 90°
Join OP
∠APB = 80°
Line OP bisects ∠APB
Therefore, ∠OPA = ∠OPB = 40°
In right angled triangle △OPA
sum of angles = 180°
So, ∠POA = 180° − 90° − 40° = 50°
Answer: A
Question 4
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Given
A circle with diameter AB. Tangents are drawn at points A and B.
To prove: The tangents at A and B are parallel.
Construction: Let O be the centre of the circle. Join OA and OB.
Since AB is the diameter => AOB is a straight line
Therefore, both tangents are perpendicular to the same line AOB.
Question 5
Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
Assume that ⊥ on point P, does not pass through the centre of the circle, O.
Consider it passes through point O’.
=> ∠APO’ = 90°
Also ∠BPO = 90° (∵ tangent is ⊥ to the radius through point of contact)
Also ∠APB = 180° (Straight angle)
This means
This is possible only when OP and O’P coincide.
Hence our assumption is wrong.
This implies perpendicular at the point of contact to tangent passes through the centre of the circle.
Question 6
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Given
OA = 5 cm
AP = 4 cm
To find
Radius (r) = OP
AP is tangent and OP is radius, hence
∠OPA = 90°
In right angled △OPA
Using Pythagoras Theorem
Question 7
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Let the chord AB touches the inner circle at point P. This means:
OP = 3 cm (Radius of inner circle)
OA = 5 cm (Radius of outer circle)
APB is tangent and OP is radius of inner circle, hence
∠OPA = 90°
In right angled △OPA
Using Pythagoras Theorem
Question 8
A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Let the circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
Let the lengths of tangents from each vertex to the points of contact be:
Using the above information
Using the above information
Question 9
In Fig. 10.13, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
XY and X’Y’ are two || tangents.
=> PAQ is a diameter of the circle.
∠1 = ∠2 = 90°
Because AP ⊥ OP and BQ ⊥ OQ
In quadrilateral APBQ
Now, AP and AC are two tangents from the same point, A.
This implies, OA bisects ∠PAC
Similarly, OB bisects ∠QBC
In triangle AOB
Question 10
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
To Prove: ∠1 + ∠3 = 180°
Let the circle have centre O &
A be a point outside the circle.
Draw tangents AP and AQ.
Join OP and OQ.
PA is tangent and OP is radius, hence
∠OPA = ∠4 = 90°
Similarly AQ is tangent and OQ is radius, hence
∠OQA = ∠2 = 90°
Now, OPAQ is a quadrilateral.
∠1 + ∠2 + ∠3 + ∠4 = 360°
∠1 + 90° + ∠3 + 90° = 360°
∠1 + ∠3 + 180° = 360°
∠1 + ∠3 = 180°
Hence proved.
Question 11
Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be a parallelogram circumscribing a circle.
Since tangents from a point to a circle are equal:
Using above information
This means
Now AB = CD and BC = AD because ABCD is a parallelogram. Therefore
Hence ABCD is a rhombus.
Question 12
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Given
BD = 8cm, DC = 6cm
=> BC = 14cm
Area of triangle using Heron’s Formula & inradius
Method 2
Given
BD = 8cm, DC = 6cm
=> BC = 14cm
Area of triangle ABC = Finding areas of 3 small triangles
Area of triangle ABC by Heron’s Formula
Question 13
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
To prove
Given
ABCD is a quadrilateral circumscribing a circle.
Let centre of circle is O.
Let the circles touches the quadrilateral at points P, Q, R and S as shown in the figure.
This means APB, BQC, CRD and DSA are tangents to the circle.
Join OA, OB, OC, OD, OP, OQ, OR and OS.
Use numbers (1 to 8) to name the angles as shown in the figure.
Now we know that
Also
Similarly
Hence proved.