Question

Find the area of quadrilateral ABCD whose sides are 9m, 40m, 28m, and 15m. The angle between the first two sides is a right angle.

Answer 1

Step-by-step explanation:

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Answer 2

Given :-

• ABCD is a quadrilateral in which AB = 9 m , BC = 40 m , CD = 28 m , DA = 15 m

• The angle between the first two sides is a right angle.

To find :-

• The area of quadrilateral

Solution :-

By using Pythagoras theorem,

we know that,

$\rm\:AC=\sqrt{{AB}^{2}+{BC}^{2}}$

$\rm\longrightarrow\:AC=\sqrt{{(9)}^{2}+{(40)}^{2}}$

$\rm\longrightarrow\:AC=\sqrt{81+1600}$

$\rm\longrightarrow\:AC=\sqrt{1681}$

$\rm\longrightarrow\:AC=41\:cm$

Now,

Area of ∆ ABC = ½ × AB × BC

Area of ∆ ABC = ½ × 9 × 40

Area of ∆ ABC = 9 × 20

Area of ∆ ABC = 180 m²

Hence,the area of ∆ ABC will be 180 m²

Then,

In ∆ ACD

• a = 41 m
• b = 28 m
• c = 15 m

$\rm\:s=\dfrac{a+b+c}{2}$

$\rm\longrightarrow\:s=\dfrac{41+28+15}{2}$

$\rm\longrightarrow\:s=\cancel\dfrac{84}{2}$

$\rm\longrightarrow\:s=42\:m$

By using Heron's formula,

we know that,

$\rm\:Area\:of\:\triangle\:ACD=\sqrt{s(s-a)(s-b)(s-c)}$

$\rm\:Area\:of\:\triangle\:ACD=\sqrt{42(42-41)(42-28)(42-15)}$

$\rm\:Area\:of\:\triangle\:ACD=\sqrt{42\times\:1\times\:14\times\:27}$

$\rm\:Area\:of\:\triangle\:ACD=14\times\:3\times\:3$

$\rm\:Area\:of\:\triangle\:ACD=126\:{m}^{2}$

Hence,the area of ∆ ACD will be 126 m².

Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD

Area of quadrilateral ABCD = 180 + 126 m²

Area of quadrilateral ABCD = 306 m².

Hence,the area of quadrilateral ABCD will be 306 m².