Question

The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and $\angle ABD=\angle BCD={90}^{\circ }$. Then the length of AB is ____.

Answer 1

Given, AD = 13 cm, DC = 12 cm, BC = 3 cm and $\angle ABD=\angle BCD={90}^{\circ }$.

In the right-angled triangle $\Delta BCD$,

$B{D}^2=B{C}^2+C{D}^2$ [By Pythagoras theorem]

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

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

$\Rightarrow B{D}^2=153cm$

Now, in the right-angled triangle $\Delta ABD$

$A{D}^2=A{B}^2+B{D}^2$ [By Pythagoras theorem]

${13}^2=A{B}^2+153$ [Since $B{D}^2=153$]

$\Rightarrow A{B}^2=169-153$

$\Rightarrow A{B}^2=16$

$\Rightarrow AB=\sqrt{16}=4cm$