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Areas of Triangles and Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas of Triangles and Quadrilaterals4 Marks
Question
In the given figure, a $\triangle\text{ABC}$ has been given in which $AB = 7.5\ cm, AC = 6.5\ cm$ and $BC = 7\ cm$. On base $BC$, a parallelogram $DBCE$ of the same area as that of $\triangle\text{ABC}$ is constructed. Find the height $DL$ of the parallelogram.
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Answer


In $\triangle\text{ABC},$ $AB = 7.5\ cm, BC = 7\ cm$ and $AC = 6.5\ cm$
Let $a = 7.5\ cm, b = 7\ cm$ and $c = 6.5\ cm$
Semi-perimeter, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{7.5+7+6.5}{2}=\frac{21}{2}=10.5\text{cm}$
$\therefore$ Area of $\triangle\text{ABC}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$ $=\sqrt{10.5\times3\times3.5\times4}$
$=\sqrt{441}$ $=21\text{cm}^2$
Now, Area of parallelogram $\text{DBCE}=\text{Area}\ \text{of}\ \triangle\text{ABC}$
$\Rightarrow\text{BC}\times\text{DL}=21$
$\Rightarrow7\times\text{DL}=21$
$\Rightarrow\text{DL}=3\text{cm}$

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