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 ∴AreaofΔABC=√s(s−a)(s−b)(s−c) [by Heron’s formula] =√10.5(10.5−7.5)(10.5−7)(10.5−6.5) =√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.