Question

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

[3 Marks]

Answer 1

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

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

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

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

$A{C}^2=625$

$AC=25\text{cm}$

[1.5 Marks]

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

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

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

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

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

$A{D}^2=576$

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

[1.5 Marks]