Question

A park in the shape of a quadrilateral ABCD ,has angle C=90 degree ,AB=9m,BC=12m,CD=5m and AD=8m. how much area does it occupy?figure ​

Answer 1

Answer:

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Answer 2

Answer:

Given,

→ABCD is a quadrilateral

→AB = 9 m , BC = 12 m , CD = 5 m , AD = 8 m and \bold{\angle C = 90}∠C=90

To Find :

Area of quadrilateral = ?

Solution :

Construction : Join BD , its help to quadrilateral divide into two triangle

⇒ In ΔBCD ,

By applying Pythagoras theorem to find Hypotenuse [ BD] :

\implies \bold{BD^{2} = BC^{2} \ + \ CD^{2}}⟹BD

2

=BC

2

+ CD

2

\implies \bold{BD^{2} = [ 12]^{2} \ + \ [5]^{2}}⟹BD

2

=[12]

2

+ [5]

2

\implies \bold{BD^{2} = 169}⟹BD

2

=169

\implies \bold{BD = 13 \ m }⟹BD=13 m

So, Now we have Hypotenuse so we find area of ΔBCD :

Area of ΔBCD = \bold{\frac{1}{2} \ x \ 12 \ x \ 5 }

2

1

x 12 x 5

Area of ΔBCD = 30 \bold{m^{2}}m

2

Now , we find the area of ΔABD ,

We have ,

a = 13 cm

b = 9 cm

c = 8 cm

So , we find firstly Semi perimeter :

→Semi perimeter of ΔABD = \bold{\frac{8 \ + \ 9 \ + \ 13 \ }{2}}

2

8 + 9 + 13

Semi perimeter of ΔABD = \bold{\frac{30}{2}} = \bold{ 15 \ m }

2

30

=15 m

Using Heron's Formula :

Area of ΔABD = \bold{\sqrt{s[s-a] \ [s-b] \ [s-c] }}

s[s−a] [s−b] [s−c]

Area of ΔABD = \bold{\sqrt{15[15 - 13] \ [15 - 13] \ [15-13] }}

15[15−13] [15−13] [15−13]

Area of ΔABD = \bold{\sqrt{15 \ x \ 2 \ x \ 6 \ x 7 }}

15 x 2 x 6 x7

Area of ΔABD = \bold{\sqrt[6]{35}}

6

35

Area of ΔABD = 35.5 \bold{m^{2}}m

2

[Approx}

Now, we find Area of Quadrilateral ABCD :

Area of ΔBCD + Area of ΔABD = Area of quadrilateral ABCD

\implies \bold{30 m^{2}} \ + \bold{ 35.5 m^{2}} = \bold{65.5 m^{2}}⟹30m

2

+35.5m

2

=65.5m

2