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  • In Figure, DeltaABC 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 DeltaABC is constructed. Find the height DF of the parallelogram.

In Figure, \DeltaABC 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 \DeltaABC is constructed. Find the height DF of the parallelogram.

                                                                

Answers (1)

[3 cm]

AB = 7.5 cm, AC = 6.5 cm, BC = 7 cm

Let a = 7.5 cm, b = 6.5 cm, c = 7 cm

Now, S=\frac{a+b+c}{2}=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5\; cm

Area\; of\; \Delta ABC,\; By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

=\sqrt{10.5\left ( 10.5-7.5 \right )\left ( 10.5-6.5 \right )\left ( 10.5-7 \right )}

=\sqrt{10.5\times 3\times 4\times 3.5}

=\sqrt{\frac{105}{10}\times 3\times 4\times \frac{35}{10}}

=\sqrt{21\times 3\times 7}=\sqrt{3\times 7\times 3\times 7}=3\times 7=21\; cm^{2}

Now,\; we\; find\; the \; length\; DF \; of\; parallelogram\: DBCE

Area\; of\; parallelogram = base \times height = BC \times DF

Area\; of\; parallelogram = 7DF

According to question,

Area\; of\; \Delta ABC = Area \; of\; parallelogram DBCE

21 = 7DF

\frac{21}{7}= DF

DF = 3\; cm

Hence\; height\; of\; parallelogram\; is\; 3\; cm.

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