Question

In the given figure, Find the value of AB ? [2 Marks]

Answer 1

$\text{Using Pythagoras' theorem in right-angled triangle ABD,}A{B}^2=A{D}^2+B{D}^2.---\left(i\right)\left[0.5Mark\right]\text{Using it in}△ADC,\text{we get}A{C}^2=A{D}^2+D{C}^2.\Rightarrow A{D}^2=A{C}^2-D{C}^2=4^2-3^2=16-9=7\left[0.5Mark\right]$
Substituting this value of $A{D}^2$ in (i), we get
$A{B}^2=7+5^2=7+25=32.$
$\Rightarrow AB=\sqrt{32}=4\sqrt{2}$ units ~~~[1 Mark]