Question

In the adjoining figure, AE = DC = 13 cm BE = 5 cm, $\angle ABC={90}^{\circ }$ & AD = EC = x cm. Calculate the length of AB & the value of x

Answer 1

In $ABE=A{B}^2+B{E}^2=A{E}^2$ (image)

$A{B}^2+25=169$

$A{B}^2=169-25=144$

$AB=12$

In $DBC,D{C}^2=D{B}^2+B{C}^2$

$\Rightarrow 169={\left(AB-x\right)}^2+{\left(5+x\right)}^2$

$169=A{B}^2+x^2-2xAB\times x+25+x^2+10X$

$=196=144+x^2-24x+25+x^2+10x$

$\Rightarrow 2x^2-14x=0$

$\Rightarrow x\left(x-7\right)=0\Rightarrow x=7$

$\left[x=7\right]$