1. A park, in the shape of a quadrilateral ABCD, has Z C = 90°, AB =

1.	1. A park, in the shape of a quadrilateral ABCD, has Z C = 90°, AB = 9 m, BC = 12 m,CD = 5 m and AD = 8 m. How much area does it occupy?
Answer» Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m. ER. RAVI KUMAR ROY 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= 169BD = √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

1. A park, in the shape of a quadrilateral ABCD, has Z C = 90°, AB = 9 m, BC = 12 m,CD = 5 m and AD = 8 m. How much area does it occupy?

Answer»

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

ER. RAVI KUMAR ROY

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= 169BD = √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