A park, in the shape of a quadrilateral ABCD, has angle C = 90° l, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
[NCERT MATHS CLASS 9TH CH 12 - HERON's FORMULA EX 12.2 Q2]
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Answers
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}}⟼(H)
2
=(B)
2
+(P)
2
\longmapsto\tt{{(H)}^{2}={(5)}^{2}+{(12)}^{2}}⟼(H)
2
=(5)
2
+(12)
2
\longmapsto\tt{{(H)}^{2}=25+144}⟼(H)
2
=25+144
\longmapsto\tt{{(H)}^{2}=169}⟼(H)
2
=169
\longmapsto\tt{H=\sqrt{169}}⟼H=
169
\longmapsto\tt\bf{H=13\:m}⟼H=13m
Now ,
a = 5 m
b = 12 m
c = 13 m
\longmapsto\tt{s=\dfrac{a+b+c}{2}}⟼s=
2
a+b+c
\longmapsto\tt{s=\dfrac{5+12+13}{2}}⟼s=
2
5+12+13
\longmapsto\tt{s=\cancel\dfrac{30}{2}}⟼s=
2
30
\longmapsto\tt\bf{s=15\:m}⟼s=15m
\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}⟼Area=
s(s−a)(s−b)(s−c)
\longmapsto\tt{\sqrt{15(15-5)\:(15-12)\:(15-13)}}⟼
15(15−5)(15−12)(15−13)
\longmapsto\tt{\sqrt{15\:(10)\:(2)\:(2)}}⟼
15(10)(2)(2)
\longmapsto\tt{5\times{3}\times{2}}⟼5×3×2
\longmapsto\tt\bf{30\:{m}^{2}}⟼30m
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}}⟼s=
2
a+b+c
\longmapsto\tt{s=\dfrac{9+8+13}{2}}⟼s=
2
9+8+13
\longmapsto\tt{s=\cancel\dfrac{30}{2}}⟼s=
2
30
\longmapsto\tt\bf{s=15\:m}⟼s=15m
\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}⟼Area=
s(s−a)(s−b)(s−c)
\longmapsto\tt{\sqrt{15(15-8)\:(15-9)\:(15-13)}}⟼
15(15−8)(15−9)(15−13)
\longmapsto\tt{\sqrt{15\:(7)\:(6)\:(2)}}⟼
15(7)(6)(2)
\longmapsto\tt{3\times{2}\sqrt{5\times{7}}}⟼3×2
5×7
\longmapsto\tt\bf{6\sqrt{35}\:{m}^{2}}⟼6
35
m
2
Area of Δ ABD is 30 m² .
Total Area of Park :
\longmapsto\tt{Area\:of\:\triangle\:BCD+Area\:of\:\triangle\:ABD}⟼Areaof△BCD+Areaof△ABD
\longmapsto\tt{30+6\sqrt{35}}⟼30+6
35
\longmapsto\tt\bf{65.4\:{m}^{2}}⟼65.4m
2
So , The Area Occupied by the park is 65.4 m² ..
