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A park,in the shape of a quadrilateral ABCD ,has angle c =90 ,AB=9m,BC=12m,CD=5m,AD=8m How much area does it occupy

Posted by Charanjit Singh 4 years, 10 months ago

CBSE > Class 09 > Mathematics

2 answers

Smita Malvankar 4 years, 10 months ago

0Thank You

Yogita Ingle 4 years, 10 months ago

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

0Thank You

ANSWER

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CBSE > Class 09 > Mathematics