Question

In the figure, $ABCD$ is a quadrilateral in which $BC=3cm,AD=13cm,DC=12cm$ and $△ABD=△BCD={90}^o$. Calculate the length of $AB$.

• A. $4cm$
• B. $28cm$
• C. $12cm$
• D. $8cm$

Answer 1

Answer: (A)

$4cm$

In $△BDC$,

$\angle BCD={90}^{\circ }$

Thus, $B{D}^2=B{C}^2+C{D}^2$ (Pythagoras theorem)

$B{D}^2={12}^2+3^2$

$B{D}^2=144+9$

$BD=\sqrt{153}$

Now, In $△ABD$, $\angle ABD={90}^{\circ }$

$A{D}^2=B{D}^2+A{B}^2$

${13}^2=(\sqrt{153}{)}^2+A{B}^2$

$169=153+A{B}^2$

$16=A{B}^2$

$AB=4$ cm