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Question
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.
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Solution
The sides of a triangle DCE are
DC = 15 cm, CE = 13 cm, ED = 14 cm
Let h be the height of parallelogram ABCD
Given,
Perimeter of ΔDCE
2s = DC + CE + ED
`⇒S=1/2(15+13+4)`
`⇒s=1/2(42)`
`⇒s=21cm`
๐ด๐๐๐ ๐๐ Δ๐ท๐ถ๐ธ = `sqrt(s(s-dc)(s-ce)(s-ed))`
`=sqrt(21(21-15)(21-13)(21-14))`
`=sqrt(21xx7xx8xx6)`
`=sqrt(84xx84)`
`84 cm^2`
Given that
Area of Δ๐๐ ๐ท๐ถ๐ธ= ๐๐๐๐ ๐๐ ๐ด๐ต๐ถ๐ท
= Area of parallelogram ABCD = =`84cm^2`
⇒ 24×โ=84 [∴ Area of parallelogram = base × height]
⇒ h = 6 cm
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