Question

BD - 160 m, we have
DE 160 m2 80 m
DE + CE-DC, which gives
CE - VDC - DE
160 m
And
D
CE - 100 - 80m=60 m
Fig. 12.14
1
Therefore, area of A BCD --x160 x 60 m² = 4800 m²
2
EXERCISE 12.2
1. A park, in the shape of a quadrilateral ABCD, has 2 C = 90°, AB = 9 m, BC =
CD-5 m and AD = 8 m. How much area does it occupy?
2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD =
DA=5 cm and AC=5 cm.​

Answer 1

Answer:

BD - 160 m, we have

DE 160 m 2 80 m

DE + CE-DC, which gives

CE - VDC - DE

160 m

And

D

CE - 100 - 80m=60 m

Fig. 12.14

1

Therefore, area of A BCD --x160 x 60 m² = 4800 m²

2

EXERCISE 12.2

1. A park, in the shape of a quadrilateral ABCD, has 2 C = 90°, AB = 9 m, BC =

CD-5 m and AD = 8 m. How much area does it occupy?

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD =

DA=5 cm and AC=5 cm.

Answer 2

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