Question

A chord of length $32\text{units}$ cuts diameter of the circle of radius $20\text{units}$ at ${90}^o$. Find the ratio in which the chord divides the diameter of circle.

• A. $4:1$
• B. $2:1$
• C. $1:2$
• D. $1:4$

Answer 1

Answer: (A), (D)

$1:4$

Let the chord be ZY.


ZY cuts the diameter XR at Q.
Given: $\angle Q={90}^o$
Hence, XR will bisect ZY at Q.
$\Rightarrow ZQ=\frac{ZY}{2}=\frac{32}{2}=16\text{units}$

Also, $△PZQ$ is a right triangle.
$\Rightarrow P{Z}^2=P{Q}^2+Z{Q}^2$

PZ is the radius of the circle.
$\Rightarrow P{Q}^2=P{Z}^2-Z{Q}^2={20}^2-{16}^2$
$\Rightarrow PQ=\sqrt{400-256}=\sqrt{144}=12\text{units}$

$XQ=XP+PQ$
XP is the radius of the circle.
$\Rightarrow XQ=20+12=32\text{units}$
And, $QR=PR-PQ=20-12=8\text{units}$

The required ratio is $XQ:QR=32:8=4:1$

When ZY is present in the upper half of the circle, the ratio is $1:4.$

$\therefore$ Both options (a) and (d) are correct.