A park, in the shape of a quadrilateral ABCD has ∠C=90⁰, AB = 4.8 m,
BC = 8√2 m, CD = 5 m and AD = 2√2 m. how much area does it occupy ?
Answers
Answer:
To find the area of a quadrilateral we divide the quadrilateral into 2 triangular parts and use Heron’s formula or any other suitable formula to calculate the area of the triangular parts.
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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
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Hope this will help you...
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