A park in the shape of a quadrilateral ABCD has ∠C = 90^o. AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
How much area does it occupy?

Given: A park in the shape of a quadrilateral ABCD has $\angle C\ =\ 90^{o} ,\ AB\ =\ 9\ m,\ BC\ =\ 12\ m,\ CD\ =\ 5\ m\ and\ AD\ =\ 8\ m$.

To find: Here we have to find the area occupied by the quadrilateral ABCD.

Solution:

Above figure shows the quadrilateral park ABCD. 

Area of ABCD  $=$  Area of $\vartriangle$BCD  $+$  Area of $\vartriangle$ABD

$\vartriangle$BCD is a right angle triangle so we can calculate its area easily, but to calculate the area of $\vartriangle$ABD we need side BD whose length can also be calculated using $\vartriangle$BCD.

Calculating length of side BD:

Using Pythagoras theorem in right $\vartriangle$BCD;

$BD^2\ =\ BC^2\ +\ DC^2$  

$BD^2\ =\ 12^2\ +\ 5^2$  

$BD^2\ =\ 144\ +\ 25$  

$BD^2\ =\ 169$  

$BD\ =\ \sqrt{169}$  

$\mathbf{BD\ =\ 13\ m}$ 

Now,

$Area\ of\ triangle\ BCD\ =\ \frac{1}{2} \ \times \ base\ \times \ height$

$Area\ of\ triangle\ BCD\ =\ \frac{1}{2} \ \times \ 12\ \times \ 5$

$Area\ of\ triangle\ BCD\ =\ 6\ \times \ 5\ =\ 30\ m^{2}$

Using the heron's formula to calculate the area of triangle ABD:  

Considering side of $\vartriangle$ABD as, $a\ =\ 9\ m,\ b\ =\ 13\ m\ and\ c\ =\ 8\ m$. So,

$Semi\ perimeter\ =\ \frac{a\ +\ b\ +\ c}{2}$

$Semi\ perimeter\ =\ \frac{9\ +\ 13\ +\ 8}{2} \ =\ 15\ m$

Therefore, semi perimeter $=\ 15\ m$.

Applying the heron's formula:

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

$Area\ of\ triangle\ ABD\ =\ \sqrt{15( 15-9)( 15-13)( 15-8)}$

$Area\ of\ triangle\ ABD\ =\ \sqrt{15(6)(2)( 7)}$

$Area\ of\ triangle\ ABD\ =\ \sqrt{1260} \ =\ 35.5\ m^{2}$

Therefore,

Area of ABCD $=$ Area of $\vartriangle$BCD $+$ Area of $\vartriangle$ABD

Area of ABCD $=\ 30\ +\ 35.50\ m^2$   

Area of ABCD $=\ \mathbf{65.50\  m^2}$   

So, the area occupied by the quadrilateral ABCD is 65.50 m$^2$.

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Updated on: 10-Oct-2022

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