﻿ In Fig. , PQ is tangent at point C to a circle with centre O. If

# In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA. In the given figure, In Δ ACO,
OA=OC (Radii of the same circle)
Therefore,
ΔACO is an isosceles triangle.
∠CAB = 30° (Given)
∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)
∠PCO = 90° …(radius is drawn at the point of
contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
Therefore,
∠PCA = 90° – 30° = 60°

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In Fig.2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.  In the Given figure, since tangents are drawn from an exterior point to a circle are equal in length,
AP = AS ….(1)
BP = BQ ….(2)
CR = CQ ….(3)
DR = DS ….(4)
Adding equations (1), (2), (3) and (4), we get

AP + BP + CR + DS = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + CD = BC + DA

Hence proved

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In Fig., the sides AB, BC and CA of a triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then the length of BC (in cm) is • 11

• 10

• 14

• 15

B.

10

it is know that the lengths of tangents drwan from a point outside a circle

are equal in length.

Therefore, we have;

AP = AR     ........(1)  (Tangents drawn from point A)

BP = BQ    .........(2)  (Tangents drawn from point B)

CQ = CR   ..........(3)   (Tangents drawn from point C)

Using the above equations,

AR = 4 cm          ( AP = 4 cm,  given)

BQ = 3 cm         ( BP = 3 cm,  given)

AC = 11 cm $⇒$ RC = 11 cm - 4 cm = 7 cm

$⇒ CQ = 7 cm$

Hence, BC = BQ + CQ = 3 CM + 7 CM = 10 cm.

In Fig., a circle touches the side DF of $\angle$EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of $\angle$EDF (in cm) is: • 18

• 13.5

• 12

• 9

A.

18

It is known that the tangents from an external point to the circle are equal.

$\therefore$ EK = EM,  DK = DH  and FM = FH     .....(1)

Perimeter  of $∆$EDF = ED + DF + FE

= (EK - DK) + (DH + HF) + EM - FM)

= (EK - DH) + (DH + HF) + (EM - FH)         [Using (1)]

= EK + EM

= 2EK = 2(9 CM) = 18 CM

Hence, the perimeter of $∆$EDF is 18 cm.

A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

• 5$\sqrt{2}$

• 10$\sqrt{2}$

• $\frac{5}{\sqrt{2}}$

• 10$\sqrt{3}$

B.

10$\sqrt{2}$ 