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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 - Mathematics

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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

 

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`×๐ด๐ต×๐ต๐ถ       [โˆต๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ Δ๐‘™๐‘’=`1/2`(3๐‘ฅ+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`

๐‘๐‘œ๐‘ค ๐‘–๐‘› Δ๐ด๐ท๐ถ
Let 2s be the perimeter

∴ 2s = (AD + DC + AC)

⇒ S = `1/2`(15+14+13)=`1/2`×42=21๐‘š

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)`

∴๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘–๐‘™๐‘Ž๐‘ก๐‘’๐‘Ÿ๐‘Ž๐‘™ ๐ด๐ต๐ถ๐ท=๐‘Ž๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ (Δ๐ด๐ต๐ถ)+๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ (Δ๐ด๐ท๐ถ) = 30 + 84 = `114 m^2`

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Chapter 17: Heronโ€™s Formula - Exercise 17.2 [Page 19]

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RD Sharma Mathematics [English] Class 9
Chapter 17 Heronโ€™s Formula
Exercise 17.2 | Q 3 | Page 19

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