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?
Answers
Answer:
30 + 6√35 cm²
Step-by-step explanation:
BC = 12 cm
CD = 5 cm
∠C = 90°
=> BD² = BC² + CD² = 12² + 5² = 144 + 25 = 169
=> BD = 13 cm
Area of Δ BCD = (1/2) * BC * CD = (1/2) * 12 * 5 = 30 cm²
Area of Δ ABD using hero formula
AB = 9 cm , BD = 13 cm , AD = 8 cm
s = (9 + 13 + 8)/2 = 15
Area of Δ ABD = √15(15-9)(15-13)(15-8) = √15 * 6 * 2 * 7
=> Area of Δ ABD = 6 √35 cm²
Area of Park quadrilateral ABCD = Area of Δ BCD + Area of Δ ABD
= 30 + 6√35 cm²
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