a park in the shape of a quadrilateral ABCD has C =90 ab =9m bc=12m cd = 5m and ad = 8m .how much area does it occupy
Answers
Solution:
Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.
Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.
In ΔBCD,
By applying Pythagoras Theorem
BD²=BC² +CD²
BD²= 12²+ 5²= 144+25
BD²= 169
BD = √169= 13m
∆BCD is a right angled triangle.
Area of ΔBCD = 1/2 ×base× height
=1/2× 5 × 12= 30 m²
For ∆ABD,
Let a= 9m, b= 8m, c=13m
Now,
Semi perimeter of ΔABD,(s) = (a+b+c) /2
s=(8 + 9 + 13)/2 m
= 30/2 m = 15 m
s = 15m
Using heron’s formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √15(15 – 9) (15 – 9) (15 – 13)
= √15 × 6 × 7× 2
=√5×3×3×2×7×2
=3×2√35
= 6√35= 6× 5.92
[ √6= 5.92..]
= 35.52m² (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD
= 30+ 35.5= 65.5 m²
Hence, area of the park is 65.5m²
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