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Question 6
In the figure, ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7cm. On base BC a parallelogram DBCE of same area as that of ΔABC is constructed. Find the height DF of the parallelogram.

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Solution

The sides of a triangle are
AB = a = 7.5cm, BC = b = 7cm, and CA = c = 6.5cm
Now, semi-perimeter of ΔABC ,
s=a+b+c2=7.5+7+6.52=212=10.5cm
Area of ΔABC=s(sa)(sb)(sc) [by Heron’s formula]
=10.5(10.57.5)(10.57)(10.56.5)
=10.5×3×3.5×4=441=21cm2 (i)
Now, area of parallelogram BCED = Base × height
=BC×DF=7×DF . . . . . .(ii)

According to the question,
Area of ΔABC = Area of parallelogram BCED
21=7×DF [from Eqs.(i) and (ii)]
DF=217=3cm
Hence, the height of parallelogram is 3cm.

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