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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, $\triangle A B C$ has sides $A B=7.5 \mathrm{~cm}, A C=6.5 \mathrm{~cm}$ and $B C=7 \mathrm{~cm}$. On base $B C$, a parallelogram, $D B C E$ of the same area as that of $\triangle A B C$ is constructed. Find the height $D F$ of the parallelogram.

  

Answers (1)

Solution:

$
\mathrm{AB}=7.5 \mathrm{~cm}, \mathrm{AC}=6.5 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}
$

Let $\mathrm{a}=7.5 \mathrm{~cm}, \mathrm{~b}=6.5 \mathrm{~cm}, \mathrm{c}=7 \mathrm{~cm}$
Now, $S=\frac{a+b+c}{2}=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5 \mathrm{~cm}$
Area of $\triangle A B C, B y$ heron's formula $=\sqrt{S(S-a)(S-b)(S-c)}$

$
\begin{aligned}
& =\sqrt{10.5(10.5-7.5)(10.5-6.5)(10.5-7)} \\
& =\sqrt{10.5 \times 3 \times 4 \times 3.5}
\end{aligned}
$

$
\begin{aligned}
& =\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 \mathrm{~cm}^2
\end{aligned}
$

Now, we find the length DF of parallelogram DBCE
Area of parallelogram $=$ base $\times$ height $=B C \times D F$
Area of parallelogram $=7 D F$
According to the question,
Area of $\triangle A B C=$ Area of parallelogram $D B C E$

$
\begin{aligned}
& 21=7 D F \\
& \frac{21}{7}=D F \\
& D F=3 \mathrm{~cm}
\end{aligned}
$

Hence the height of the parallelogram is 3 cm.

Posted by

infoexpert21

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