Question

innopark of a quadrilateral shape ABCD has angle c equal to 90 degree AB equal to 9 M BC equal to 12 m and CD is equal to 5 M and AD
equal to 8 M how much does area it occupy

Answer 1

Given, ∠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

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