Question

Triangle $ABC$ is right angles at $A$. The circle with centre $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD=20$ and $DC=16$ then the length $AC$ equals ?

• A. $6\sqrt{21}$
• B. $6\sqrt{26}$
• C. $30$
• D. $32$

Answer 1

The correct option is B $6\sqrt{26}$

$\angle BAC={90}^o$

$BD=20cm$

$DC=16cm$

To find: $AC$

$AF\perp BC$

In $△ABC$ & $△AFC$

$\angle BAC=\angle AFC={90}^o\{AF\perp BC,\angle BAC={90}^o\}$

$\angle ACF=\angle ACB$ (common angle)

$\therefore △BAC\sim △DAC$ (By $AA$ similarity)

$\frac{BC}{AC}=\frac{AC}{FC}$ (By $CPCT)---(1)$

As $\perp$ draw from center of circle to a chord bisects the chord, thus

$AF\perp BC$ and $AF\perp BD$

$BF=FD$

$FD+FD=FD+BF=BD$

$2FD=20;FD=10cm---\left(2\right)$

$BC=BD+DC=20+16=36cm---\left(3\right)$

From $\left(1\right),\left(2\right),\left(3\right),$ we get

$(AC{)}^2=BC.FC$

$AC=\sqrt{36\times 26}$

$AC=6\sqrt{26}$

$\therefore$ Option $B$ is correct