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°

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

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 $\Rightarrow$ RC = 11 cm - 4 cm = 7 cm

 $\Rightarrow \; CQ\; =\; 7\; cm$

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

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.

B.

10$\sqrt{2}$

$$\begin{gathered} \mathrm{Given}: \angle \mathrm{AOB} \mathrm{is} \mathrm{given} \mathrm{as} 90° \\ ∆\mathrm{AOB} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{trianglesince} \mathrm{OA}=\mathrm{OB} \\ \mathrm{Therefore} \angle \mathrm{OAB}= \angle \mathrm{OBA}= 45° \\ \mathrm{Thus} \angle \mathrm{AOP}=45° \mathrm{and} \angle \mathrm{BOP}= 45° \\ \mathrm{Hence} ∆\mathrm{AOP} \mathrm{and} ∆\mathrm{BOP} \mathrm{also} \mathrm{are} \mathrm{isosceles} \mathrm{triangles} \\ \mathrm{Thus} \mathrm{let} \mathrm{AP}=\mathrm{PB}=\mathrm{OP}=\mathrm{x} \\ \mathrm{Using} \mathrm{pythagoras} \mathrm{theorem} \\ {\mathrm{x}}^2 + {\mathrm{x}}^2 = {10}^2 \\ \mathrm{Thus} 2{\mathrm{x}}^2 = 100 \\ \mathrm{x}=5\sqrt{2} \\ \mathrm{Hence} \mathrm{length} \mathrm{of} \mathrm{chord} \mathrm{AB} = 2\mathrm{x} = 10\sqrt{2} \end{gathered}$$