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प्रश्न
In the following figure, ∆ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ABC is constructed. Find the height DF of the parallelogram.
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उत्तर
Now, first determine the area of ∆ABC.
The sides of a triangle are AB = a = 7.5 cm, BC = b = 7 cm and CA = c = 6.5 cm
Now, semi-perimeter of a triangle,
`s = (a + b + c)/2`
= `(7.5 + 7 + 6.5)/2`
= `21/2`
= 10.5 cm
∴ Area of ΔABC = `sqrt(s(s - a)(s - b)(s - c))` ...[By Heron’s formula]
= `sqrt(10.5(10.5 - 7.5)(10.5 - 7)(10.5 - 6.5))`
= `sqrt(10.5 xx 3 xx 3.5 xx 4)`
= `sqrt(441)`
= 21 cm2 ...(i)
Now, area of parallelogram BCED = Base × Height
= BC × DF
= 7 × DF ...(ii)
According to the question,
Area of ∆ABC = Area of parallelogram BCED
⇒ 21 = 7 × DF ...[From equations (i) and (ii)]
⇒ DF = `21/4` = 3 cm
Hence, the height of parallelogram is 3 cm.
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