A park, in the shape of a quadrilateral ABCD,has angle C=90', AB=9m,BC=12m,CD=5m and AD=8m how much area deos it - Maths - Heron\s Formula

Question

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

Shridhar Kaushik · Accepted Answer

In right angled triangle BCD,

$(BD)^{2} = {(5 m)}^{2} + {(12 m)}^{2} = 169 m^{2} \Rightarrow BD = 13 m$

Area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD ...... (1)

Now, Area of ΔBCD = $\frac{1}{2}$ × Base × Height

$= \frac{1}{2} \times 5 \times 12 m^{2} = 30 m^{2} ........ (2)$

In ΔABD, AB = 9m

BD = 13m

and AD = 8m

$Now, s = \frac{9 + 13 + 8}{2} m = 15 m$

By Heron's Formula,

$Area of Δ ABD = \sqrt{15 (15 - 9) (15 - 13) (15 - 8)} m^{2} = \sqrt{15 \times 6 \times 2 \times 7} m^{2} = \sqrt{1260} m^{2} = 35.497 m^{2}$

Hence from (1),

Area of Quadrilateral ABCD = (30 + 35.497)m²

= 65.497 m²

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