Math, asked by mdfaisal4395gmailcom, 5 hours ago

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?​

Answers

Answered by mathdude500
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}}}}

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