Question

In figure given below, ABCD is a quadrilateral in which AB = AD $\angle A={90}^o=\angle C$ BC = 8 cm and CD =6 cm Find AB and calculate the area of $\Delta ABD$

Answer 1

Given $AB=AD,A=90°=C,BC=8$ cm and $CD=6$ cm
$BCD$ is a right triangle.
$B{D}^2=B{C}^2+D{C}^2$[Pythagoras theorem]
$B{D}^2=8^2+6^2$$B{D}^2=64+36$
$B{D}^2=100$
Taking square root on both sides,
$BD=10cm$
$ABD$ is a right triangle.
$B{D}^2=A{B}^2+A{D}^2$[Pythagoras theorem]
${10}^2=2A{B}^2[\because AB=AD]$
$100=2A{B}^2$
$A{B}^2=100/2$
$A{B}^2=50$
Taking square root on both sides,
$AB=\surd 50$
$AB=\surd (2\times 25)$
$AB=5\surd 2cm$
Hence the length of $AB$ is $5\surd 2cm.$