Question

In the given figure, AB = 15 cm, BC = 20 cm and CD = 7 cm. If $\angle ABC$ = $\angle ADC={90}^{\circ }$, then AD =____________cm

Answer 1

In $△$ ABC, Using Pythagoras theorem we have,

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

$A{C}^2={15}^2+{20}^2$

$A{C}^2=225+400$

$A{C}^2=625$

$AC=25\text{cm}$

$\text{In}△ADC,\text{Using Pythagoras theorem we have,}$

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

$A{D}^2=A{C}^2-C{D}^2$

$A{D}^2={25}^2-7^2$

$A{D}^2=625-49$

$A{D}^2=576$

$AD=24cm....(7-24-25\text{Pythagoras Triplet})$