Question

A Park is in the shape of a quadrilateral ABCD, where the ∠= 90 and length of sides are: AB = 11 m, BC = 3m, CD= 4 m and AD= 8 m . Then find the area of the park

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

Area of a triangle with sides a unit, b unit, c unit is

$= \sqrt{s(s - a)(s - b)(s - c)} \: \: \: sq \: unit$

Where

$\displaystyle \: s = \frac{a + b + c}{2}$

2.

Area of an right angle triangle

$= \displaystyle \: \frac{1}{2} \times base \times height$

EVALUATION

First check the attachment for figure

Now the quadrilateral ABCD is divided into two triangles by the line BD

SO Area of ABCD = Area of BCD + Area of ABD

AREA OF BCD

Here BCD is a right angled triangle

BC = 3 m , CD = 4 m

So Area of the triangle BCD

$\displaystyle \: = \frac{1}{2} \times 3 \times 4 = 6 \: \: {metre}^{2}$

AREA OF ABD

Here BC = 3 m , CD = 4 m

So

$BD = \sqrt{ {3}^{2} + {4}^{2} } = \sqrt{25} = 5$

ABD is a triangle with sides

AB = 11 m , AD = 8 m , BD = 5 m

So

$\displaystyle \:s = \frac{11 + 8 + 5}{2} = \frac{24}{2} = 12$

So area of the triangle ABD

$= \displaystyle \: \sqrt{12 \times (12 - 11) \times (12 - 8) \times (12 - 4)}$

$= \sqrt{12 \times 4\times 8} \: \: \: {metre}^{2}$

$= 8 \sqrt{6} \: \: \: {metre}^{2}$

RESULT

So the required area of the park is

$= 6 + 8 \sqrt{6} \: \: \: {metre}^{2}$