Question

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

Answer 1

You can notice that the line BD is cutting the quadrilateral in two triangles. i.e. ABD and BCD

now, for finding the area of the quadrilateral,
just find the areas of the two triangles.

area of triangle ABD = 1/2 x 9( BASE)x 8( height)

therefore, area of triangle ABD = 4x9 = 36

now, the area of BCD = 1/2 x 12x5
therefore, area of BCD = 6x5 = 30

now, as we have found the sea of the two triangles, the area of the quadrilateral = the aim of the two triangles .

therefore, area of the quadrilateral = area( ABD) + area ( BCD) = 30+36= 66m²

Hope that helps.

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