A park in the shape of a quadrilateral ABCD has angle c is 90 AB 9cm BC 12m CD 5cm and AD 8cm . How much area does it occupy?
Answers
Given:-
- A park in the shape of a quadrilateral ABCD has angle c is 90 AB 9cm BC 12m CD 5cm and AD 8cm.
To find:-
- Find the area of park ....?
Solutions:-
- Let us BD is a join.
In ∆BCD,
By using Pythagoras theorem,
BD² = BC² + CD²
BD² = 12² + 5²
BD² = 144 + 25
BD² = 169
BD = √169
BD = 13m
Area of ∆BCD = 1/2 × BC × CD
= 1/2 × 12 × 5
= 6 × 5
= 30m²
∆ABD,
S = (perimeter)/2
S = (9 + 8 + 13)/2
S = 30/3
S = 15m
By heroe's formula,
Area of triangle = √s(s - a)(s - b)(s - c)
Area of ∆ABD = √15(15 - 9)(15 - 8)(15 - 13)
= √15 × 6 × 7 × 2
= √90 × 14
= √1260m²
= 6√35m²
= 6 × 5.916m²
Area of the park = Area of ∆ABD + Area of ∆BCD
= 35.496 + 30
= 65.496
= 65.5m²
Hence, the area of park is 65.5m².
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