Question

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

Answer 1

GIVEN:

• Sides of quadrilateral ABCD are 9m, 40m, 28m and 15m
• The angle between first two sides is a right angle.

TO FIND : Area of quadrilateral.

$\bigstar$We will join A and C in order to divide the quadrilateral into two triangles.

• The two triangles are ∆ABC and ∆ADC.

Now, $\boxed{\sf\:Area\:of\:quadrilateral= Area\:of \triangle ABC + Area\:of\triangle ADC}$

In ∆ABC, AB=9m and BC = 40m

$\sf\: \angle ABC = 90°$

$\longrightarrow \sf\: By \:pythagoras\:theorem$

$\longrightarrow \sf\: AC =\sqrt{{40}^{2}+{9}^{2}}\\ \\ \longrightarrow \sf\: AC=\sqrt{1600+81}\\ \\ \longrightarrow \sf\: AC=\sqrt{1681}\\ \\ \implies \sf\: AC= 41m$

Area of ∆ABC = $\boxed{\sf\: \frac{1}{2}\times Base \times Height}$

$\longrightarrow \sf\: \frac{1}{2} \times AB \times BC\\ \\ \longrightarrow \sf\: \frac{1}{2} \times 9 \times 40\\ \\ \implies \sf\: 180 {m}^{2}$

And by using heron's formula,

$\boxed{\sf\:Area\:of \triangle = \sqrt{s(s-a)(s-b)(s-c)}{m}^{2}} \\ \\ \longrightarrow \sf\: \sqrt{42(42-15)(42-28)(42-41)}{m}^{2}\\ \\ \longrightarrow \sf\: \sqrt{42 \times 27 \times 14 \times 1}{m}^{2}\\ \\ \implies \sf\: 126 {m}^{2}$

$\therefore$ Area of quadrilateral ABCD,

$\longrightarrow \sf\: 180{m}^{2} + 126{m}^{2}\\ \\ \implies \sf\:306 {m}^{2}$

Area = $\sf\:306{m}^{2}$