Question

In
quadrilateral ABCD ,B =D=90.AB=7cm.Ac =25 cm,CD=20cm.find the area of ABCD. Answer=234.​

Answer 1

In quadrilateral ABCD the area of the whole figure is equal to 234

Answer 2

Step-by-step explanation:

In △ABC,

AC^2=AB^2+BC^2AC

2

=AB

2

+BC

2

25^2=7^2+BC^225

2

=7

2

+BC

2

625-49=BC^2625−49=BC

2

$\implies\;BC^2=576⟹BC </p><p>2</p><p> =576$

$\implies\;BC=\sqrt{576}⟹BC= </p><p>576$

$\implies\;BC=24\;\text{cm}⟹BC=24cm$

$\textbf{In \triangle ACD,}In △ACD,$

AC^2=AD^2+CD^2AC

2

=AD

2

+CD

2

25^2=AD^2+20^225

2

=AD

2

+20

2

625-400=AD^2625−400=AD

2

$\implies\;AD^2=225⟹AD </p><p>2</p><p> =225$

$\implies\;AD=\sqrt{225}⟹AD= </p><p>225$

$\implies\;AD=15\;\text{cm}$⟹AD=15cm

$\textbf{Area of \triangle ABC}$Area of △ABC

$=\frac{1}{2}{\times}AB{\times}$BC=

2

1

×AB×BC

$=\frac{1}{2}{\times}7{\times}$24=

2

1

×7×24

$=7{\times}12=7×12$

$=84\;cm^2=84cm$

2

$\textbf{Area of \triangle ACD}$Area of △ACD

$=\frac{1}{2}{\times}AD{\times}$CD=

2

1

×AD×CD

$=\frac{1}{2}{\times}15{\times}$20=

2

1

×15×20

$=15{\times}10=15×10$

=150\;cm^2=150cm

2

$\therefore\textbf{Area of quadrilateral ABCD}∴Area of quadrilateral ABCD$

$=\text{Area of triangle ABC}+\text{Area of ACD}$=Area of triangle ABC+Area of ACD

=84+150=84+150

=234\;cm^2=234cm

2