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

$\huge\underline \mathfrak \color {Aqua}{❥ \: Answer : }$

306 m ²

$\huge\underline\mathfrak \color {deeppink}{ ❥Explanation : }$

☆To find :-

The area of the park in the shape of a quadrilateral ABCD.

☆Given:-

The quadrilateral park has angle B=90° .

Length of sides : AB=9m,BC=40m,CD=15 m & DA=28 m

☆Construction: -

Let ABCD be the given quadrilateral.

Join AC in quadrilateral ABCD

☆Formula used :-

$\Rightarrow \color{lime}Hypotenuse ²= base ² + height ²$

$\Rightarrow\color{yellow}Area \: of \triangle \: = \frac{1}{2} \times base \times height$

$\small \Rightarrow \color{pink}Area \: of Scalen \:\triangle = \sqrt{s(s - a)(s - b)(s - c)}$

☆Solution:

Now we will calculate the length of AC by Pythagoras Formula .

$\sf \therefore {AC} = \sqrt{ {40}^{2} + {9}^{2} } \\ \: \: \qquad\quad \Rightarrow \sqrt{1600 + 81} \\ \quad \Rightarrow \sqrt{1681} \\ \: \: \color{red} \Rightarrow41m \: \:$

The area of quadrilateral ABCD is equal to sum of area of Triangle ABC and Triangle ACD.

$\small\sf \: Now \: the \: area \: of \triangle \: ABC \: = \frac{1}{2} \times 40 \times 9 \\ \qquad\qquad\qquad\qquad\quad \Rightarrow20 \times 9 \\ \qquad\qquad\qquad\qquad\quad \: \: \Rightarrow \color{green}180 {m}^{2} \:$

Now,we will calculate the area of triangle ACD by Heron's Formula .

$\sf \therefore \small\: semi \: perimeter \: (s) = \frac{1}{2} \times (28 + 15 + 41) \\ \qquad \Rightarrow \frac{84}{2} \\ \qquad \quad \: \color{violet} \Rightarrow42m \: \: \:$

$\sf \therefore \small Area \: of \triangle \: ACD = \sqrt{42(42 - 41)(42 - 28)(42 - 15)} \\ \qquad \: \Rightarrow \sqrt{42 \times 1 \times 14 \times 27} \\ \qquad \quad\: \: \Rightarrow \sqrt{3 \times 14 \times 14 \times 3 \times 9} \\ \qquad \Rightarrow 3 \times 14 \times 3 \qquad \quad \: \: \\ \Rightarrow \color{deeppink} 126m^{2} \quad \quad \: \: \: \: \:$

$\sf \therefore \small Area \: of \: quadrilateral \: ABCD \: = (126 + 180) \\\qquad \qquad\qquad \qquad \qquad \Rightarrow\bold \color{yellow}306 {m}^{2}$

☆Hence :-

The area of Quadrilateral shape park is

306m²

___________________________

☆Extra information:

• Perimeter of rectangle =2(l+b)

• Area of rectangle =lb

• Perimeter of square = 4a

• Area of square = a²

• Area of trapezium =1/2(sum of || sides)(height)

• Area of parallelogram =Base ×Height

• Circumference of Circle= 2pie r

where pie =22/7 and r= Radius

€€_______________________€€

hope it helps..... :)