The sides of a quadrilateral, taken in order are 5, 12, 14 and 15 meters respectively, and the angle contained by the first two sides is a right angle. Find its are

#### Solution

Given that sides of quadrilateral are AB = 5 m, BC = 12 m, CD = 14 m and DA = 15 m

AB = 5m, BC = 12m, CD = 14 m and DA = 15 m

Join AC

Area of ΔABC = `1/2`×π΄π΅×π΅πΆ [β΅π΄πππ ππ Δππ=`1/2`(3π₯+1)]

= `1/2×5×12`

= 30 cm2

In ΔABC By applying Pythagoras theorem.

`AC^2=AB^2+BC^2`

`⇒AC=sqrt(5^2+12^2)`

`⇒sqrt(25+144)`

`⇒sqrt169=13m`

πππ€ ππ Δπ΄π·πΆ

Let 2s be the perimeter

∴ 2s = (AD + DC + AC)

⇒ S = `1/2`(15+14+13)=`1/2`×42=21π

By using Heron’s formula

∴ Area of ΔADC = `sqrt(S(S-AD)(S-DC)(S-AC))`

`=sqrt(21(21-15)(21-14)(21-13))`

`sqrt(21xx6xx7xx8)`

∴π΄πππ ππ ππ’πππππππ‘ππππ π΄π΅πΆπ·=ππππ ππ (Δπ΄π΅πΆ)+π΄πππ ππ (Δπ΄π·πΆ) = 30 + 84 = `114 m^2`