Question

Let triangle $ABC$ is right-angled at $A$ . The circle with centre $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD=20$ units and $DC=16$ units, then the length $AB$ is equal to

• A. $6\sqrt{21}$ units
• B. $6\sqrt{10}$ units
• C. $\sqrt{300}$ units
• D. $18$ units

Answer 1

Answer: (B)

$b^2+r^2={\left(36\right)}^2\dots \left(i\right)$
Also, $CD\cdot CB=CE\cdot CX$ ( Property of intersecting secants)
$16\cdot 36=\left(b-r\right)\left(b+r\right)$
$\therefore b^2-r^2=16\cdot 36$ $...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$2b^2=36\left(36+16\right)=36\cdot 52$
$b^2=36\cdot 26\Rightarrow b=6\sqrt{26}$ units
And $r=\sqrt{{\left(36\right)}^2-\left(36\right)\left(26\right)}=6\sqrt{10}$ units