Question

A park, in the shape of a quadrilateral ABCD has angle B=900 , AB=9m, BC=40m, CD=15m, DA=28m. How much area does it occupy?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given a quadrilateral ABCD such that

• ∠ B = 90°.

• AB = 9 m

• BC = 40 m

• CD = 15 m

• DA = 28 m.

Construction :- Join diagonal AC.

Now,

In right triangle ABC,

AB = 9 m

BC = 40 m

Using Pythagoras Theorem, we have

$\begin{gathered}\star\;{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}*$

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

$\rm :\longmapsto\: {AC}^{2} = {9}^{2} + {40}^{2}$

$\rm :\longmapsto\: {AC}^{2} = 81 + 1600$

$\rm :\longmapsto\: {AC}^{2} = 1681$

$\bf\implies \:AC = 41 \: m$

Now,

Area of triangle ABC is

$\rm :\longmapsto\:Area_{ \triangle \: ABC} \: = \dfrac{1}{2} \times AB \times BC$

$\rm \: = \: \:\dfrac{1}{2} \times 9 \times 40$

$\rm \: = \: \:9 \times 20$

$\rm \: = \: \:180 \: {m}^{2}$

$\purple{\boxed{ \quad \bf{ \: Area_{ \triangle \: ABC} = 180 \: {m}^{2} \quad}}}$

Now,

In triangle ACD

We have

AC = 41 m

CD = 15 m

AD = 28 m

Let assume that,

a = AC = 41 m

b = CD = 15 m

c = AD = 28 m

So, we know that

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle=a+b+c}}$

and

$\underline{\boxed{\sf Semi \ perimeter, \: s \: =\dfrac{Perimeter}{2} = \dfrac{a + b + c}{2} }}$

Thus,

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

$\rm :\longmapsto\:s = \dfrac{84}{2}$

$\bf\implies \:s = 42$

We know that,

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

So, on substituting the values,

s = 42 m

a = 41 m

b = 15 m

c = 28 m

Hence,

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

$\rm \: = \: \: \sqrt{42 \times 1 \times 27 \times 14}$

$\rm \: = \: \: \sqrt{14 \times 3 \times 3 \times 3 \times 3 \times 14}$

$\rm \: = \: \:14 \times 3 \times 3$

$\rm \: = \: \:126 \: {m}^{2}$

$\purple{\boxed{ \quad \bf{ \: Area_{ \triangle \: ACD} = 126 \: {m}^{2} \quad}}}$

Therefore,

$\rm :\longmapsto\:Area_{ABCD} = Area_{ \triangle \: ABC} + Area_{ \triangle \: ACD}$

$\rm \: = \: \:180 + 126$

$\rm \: = \: \:306 \: {m}^{2}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \purple{ \underbrace{\boxed{ \quad \bf{ \: Area_{ABCD} = 306 \: {m}^{2} \quad}}}}$