Question

find the area of following figure:- quadrilateral AB=9m,BC=40m,CD=15m,DA=28m ABC=90° hint:-first find AC by using Pythagoras theorem and then find area of ∆ADC ​

Answer 1

Given:

• AB = 9m
• BC = 40m
• CD = 15m
• AD = 28m
• ∠ABC = 90°

To Find:

• Find the area of the Quadrilateral ABCD?

Solution:

$\\ {\sf{ Area \: of \: right \: \Delta ABC = \sf\purple \left( \dfrac{1}{2} \times base \times height \right) }} \\ \\ \implies{\sf{ \left( \dfrac{1}{2} \times 40 \times 9 \right) m^2 }} \\ \implies\sf{ 180 \: m^2}$

Also, From ΔABC;

We get

AC² = AB² + BC²

= [ (9)² + (40)² ] m²

= ( 81 + 1600 ) m²

= 1681 m²

AC = $\sf\green{\sqrt{1681} m = 41 \: m}$

In ΔACD, we have;

AC = 41 m

CD = 15 m

AD = 28 m

Let a = 41m , b = 15 m , and c = 28 m

Then,

We can use Heron's Formula to Find area of ACD;

$\\ \bullet{\boxed{\sf \pink{ \sqrt{s(s-a)(s-b)(s-c) } }}}$

Where,

S = ½ (a + b + c)

= ½ ( 84)

= 42 m

∴ $\sf\red{Area \ of \ \Delta ACD = \sqrt{ s(s-a)(s-b)(s-c)}}$

$\implies \sf{ = \sqrt{ 42(42-41)(42-15)(42-28)}} \\ \\ \implies\sf{ \sqrt{ 42 \times 1 \times 27 \times 14 } m^2 } \\ \\ \implies\sf{ ( 14 \times 9) m^2 } \\ \implies\sf{ 126 \: m^2}$

Area of the Quadrilateral ABCD = Area (ΔABC) + area (ΔACD)

= (180 + 126) m²

= 306 m²

Therefore:

The area of the Quadrilateral is 306 m².