Question

13. In the given circle with centre O, AD = DB=
16 cm and DE = 8 cm. Find the radius of the
circle.​

Answer 1

In any circle if two chords are intersected by an arbitrary point and is divided into two parts each by that point, then the products of lengths of the two parts of each chord will be the same.

Thus, from the fig., if C is the other end of the chord that passes through O from E, we have,

$\longrightarrow\sf{AD\times DB=DE\times CD\quad\quad\dots(1)}$

Here,

• $\sf{AD=DB=16\ cm}$

• $\sf{DE=8\ cm}$

Then (1) becomes,

$\longrightarrow\sf16\times16=8\times CD}$

$\longrightarrow\sf{CD=\dfrac{16\times16}{8}}$

$\longrightarrow\sf{CD=32\ cm}$

Let $\sf{r}$ be the radius of the circle. In the circle CE is one of its diameter. Then we have,

$\longrightarrow\sf{r=\dfrac{CE}{2}}$

$\longrightarrow\sf{r=\dfrac{CD+DE}{2}}$

$\longrightarrow\sf{r=\dfrac{32+8}{2}}$

$\longrightarrow\underline{\underline{\sf{r=20\ cm}}}$

Hence 20 cm is the answer.