Question

Calculate the area of quadrilateral ABCD in which AB=32 cm,AD=24cm, angle A=90° and BC=CD=52cm

Answer 1

Hope the answer is correct

Answer 2

Area of Quadrilateral ABCD = $\bold {1344\ cm^2}$

Step-by-step explanation:

In a quadrilateral ABC ,∠A = 90°,

AB=32 cm,AD=24 cm,BC=CD = 52 cm

Construction:

Join BD.

In ΔABD,∠A= 90°

As we know Pythagorean triplets (24,32,40)

So,BD = 40 cm

Also,$Area\ of\ \Delta ABD = \frac{1}{2}\times Base\times Height$

                               =$\frac{1}{2}\times 32\times 24$

                               = $384\ cm^2$

Now,For ΔBCD,As it is an isoceles triangle

Length of perpendicular drawn from point C on BD of ΔBCD

$Perpendicular = \sqrt{(52)^2-\frac{(40)^2}{4}$

                        =$\sqrt{2704-400}$

                       = $\sqrt{2304}$

                       = 48 cm

Now$,\bold {Area\ of\ \Delta BCD = \frac{1}{2}\times Base\times Perpendicular}$

                                 =$\frac{1}{2}\times 40\times 48$

                                 = $960\ cm^2$

Area of quadrilateral ABCD = $384\ cm^2+960\ cm^2$

                                              =$1344\ cm^2$

Hence,Area of Quadrilateral ABCD = $\bold {1344\ cm^2}$