Question

In the above figure, CM is a tangent at point C. ABCD is a square. B is the centre of the circle. BM = 10 cm, CM = 8 cm. What is the value of AD?

• A. 6 cm
• B. 4 cm
• C. 9 cm
• D. 8 cm

Answer 1

The correct option is A

6 cm

Given, CM is a tangent at C and B is the centre of the circle.

$\therefore \angle BCM={90}^{\circ }$ [Tangent at any point of a circle and the radius through this point are perpendicular to each other]

Consider $\Delta BCM$. It is a right angled $\Delta$ with right angle at C.

$B{C}^2+C{M}^2=B{M}^2$ (Pythagoras theorem)
$B{C}^2+8^2={10}^{2}B{C}^2+64=100B{C}^2=36$
BC = 6

As ABCD is a square, BC = AD [All sides of a square are equal]

BC = AD = 6 cm