Question

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

Answer 1

Answer:

65.50 m²

Step-by-step explanation:

First we'll divide the quadrilateral in two triangles

ΔABD and ΔBCD

So,

Area of quadrilateral= Sum of areas of the two triangles

Area of ΔBCD, it is a right-angled triangle,

Area=½*base*height

=½*BC*CD= ½*12*5= 6*5= 30 m²

Also in this triangle, by Pythagoras Theorem,

BD²=BC²+CD²

BD=√(12)²+(5)²

=√144+25=√169=13 cm

Now, we'll find the area of ΔABD by Heron's formula,

Semi-perimeter=(9+8+13)/2=30/2=15

Area=√s(s-a)(s-b)(s-c)=√15(15-9)(15-8)(15-13)

=√15*6*7*2=√1260=35.50 m²

Therefore,

Total area of park=30+35.50=65.50 m²

Answer 2

Given :-

• ∠C = 90°
• AB = 9 m
• BC = 12 m
• CD = 5 m
• AD = 8 m

To Find :

• Area of the park .

Solution :

In Δ BCD :

Base = 5 m

Height = 12 m

Hypotenuse = ?

By Pythagorus Theorem :

$\longmapsto\tt{{(H)}^{2}={(B)}^{2}+{(P)}^{2}}$

$\longmapsto\tt{{(H)}^{2}={(5)}^{2}+{(12)}^{2}}$

$\longmapsto\tt{{(H)}^{2}=25+144}$

$\longmapsto\tt{{(H)}^{2}=169}$

$\longmapsto\tt{H=\sqrt{169}}$

$\longmapsto\tt\bf{H=13\:m}$

Now ,

a = 5 m

b = 12 m

c = 13 m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{s=\dfrac{5+12+13}{2}}$

$\longmapsto\tt{s=\cancel\dfrac{30}{2}}$

$\longmapsto\tt\bf{s=15\:m}$

$\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\longmapsto\tt{\sqrt{15(15-5)\:(15-12)\:(15-13)}}$

$\longmapsto\tt{\sqrt{15\:(10)\:(2)\:(2)}}$

$\longmapsto\tt{5\times{3}\times{2}}$

$\longmapsto\tt\bf{30\:{m}^{2}}$

Area of Δ BCD is 30 m² .

Similarly ,

In Δ ABD :

a = 9 m

b = 8 m

c = 13 m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{s=\dfrac{9+8+13}{2}}$

$\longmapsto\tt{s=\cancel\dfrac{30}{2}}$

$\longmapsto\tt\bf{s=15\:m}$

$\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\longmapsto\tt{\sqrt{15(15-8)\:(15-9)\:(15-13)}}$

$\longmapsto\tt{\sqrt{15\:(7)\:(6)\:(2)}}$

$\longmapsto\tt{3\times{2}\sqrt{5\times{7}}}$

$\longmapsto\tt\bf{6\sqrt{35}\:{m}^{2}}$

Area of Δ ABD is 30 m² .

Total Area of Park :

$\longmapsto\tt{Area\:of\:\triangle\:BCD+Area\:of\:\triangle\:ABD}$

$\longmapsto\tt{30+6\sqrt{35}}$

$\longmapsto\tt\bf{65.4\:{m}^{2}}$

So , The Area Occupied by the park is 65.4 m² .