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Question
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
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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`×๐ด๐ต×๐ต๐ถ [โต๐ด๐๐๐ ๐๐ Δ๐๐=`1/2`(3๐ฅ+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`
๐๐๐ค ๐๐ Δ๐ด๐ท๐ถ
Let 2s be the perimeter
∴ 2s = (AD + DC + AC)
⇒ S = `1/2`(15+14+13)=`1/2`×42=21๐
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)`
∴๐ด๐๐๐ ๐๐ ๐๐ข๐๐๐๐๐๐๐ก๐๐๐๐ ๐ด๐ต๐ถ๐ท=๐๐๐๐ ๐๐ (Δ๐ด๐ต๐ถ)+๐ด๐๐๐ ๐๐ (Δ๐ด๐ท๐ถ) = 30 + 84 = `114 m^2`
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