CAT 2018

Slot 1

QA

Question & Solution

Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively. 

AM = 3 cm and CD = 2 cm. Let 'OM' be x cm. 

In right angle triangle AMO,

$AO^2 = AM^2 + OM^2$ 

$\Rightarrow$ $AO^2 = 3^2 + x^2$  ... (1)

In right angle triangle CNO,

$CO^2 = CN^2 + ON^2$ 

$\Rightarrow$ $CO^2 = 2^2 + (OM+MN)^2$ 

$\Rightarrow$ $CO^2 = 2^2 + (x+1)^2$  ... (2)

We know that both AO and CO are the radius of the circle. Hence $AO = CO$

Therefore, we can equate equation (1) and (2)

$3^2+x^2$=$2^2+(x+1)^2$

$\Rightarrow$ x = 2 cm

Therefore, the radius of the circle 

$AO = \sqrt{AM^2 + OM^2}$ 

$\Rightarrow$ $AO=\sqrt{3^2+2^2}=\sqrt{13}$. Hence, option A is the correct answer.