EXERCISE 12.2
1.
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?
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Answers
Answer:
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?
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∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
Joim BD
In ΔBCD,
By applying Pythagoras theorem,
BD2 = BC2 + CD2
⇒ BD2 = 122 + 52
⇒ BD2 = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m2
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √15(15 - 13) (15 - 9) (15 - 8) m2
= √15 × 2 × 6 × 7 m2
= 6√35 m2 = 35.5 m2 (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m2 + 35.5m2 = 65.5m2
Step-by-step explanation:
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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?
∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
Joim BD
In ΔBCD,
By applying Pythagoras theorem,
BD2 = BC2 + CD2
⇒ BD2 = 122 + 52
⇒ BD2 = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m2
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √15(15 - 13) (15 - 9) (15 - 8) m2
= √15 × 2 × 6 × 7 m2
= 6√35 m2 = 35.5 m2 (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m2 + 35.5m2 = 65.5m2