A park in the shape of quadrilateral ABCD has ∠C=90°,AB=9m BC=12m CD=5m and AD=8m. How much area does it occupy ?
Answers
ANSWER
Join BD in ΔBCD, BC and DC are given.
So, we can calculate BD by applying Pythagoras theorem
⇒BD=
BC
2
+CD
2
=
12
2
+5
2
=
144+25
=13 m=BD
⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD
⇒Area of ΔBCD
=
2
1
×b×h=
2
1
×12×5
=30 m
2
⇒Area of ΔABD
=
s(s−a)(s−b)(s−c)
(Heron's formula)
⇒2S=9+8+13, S=
2
30
⇒S=15 m
⇒Area of ΔABD
=
15(15−9)(15−8)(15−13)
=
15×6×7×2
=
630×2
=6
1260
=35.49m
2
⇒Area of Park = Quad ABCD
=30+35.49
=65.49 m
2
≈65.5 m
2
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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