Question
In $\triangle\text{ABC}$ 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 $\triangle\text{ABC}$ is constructed. Find the height $DF$ of the parallelogram.
✓
Answer
Now, first determine the area of $\triangle\text{ABC}$
The sides of a traingle are $\text{AB} = \text{a} = 7.5\text{cm,} \text{ BC} = \text{b} = 7\text{cm} \text{ and}\text{ CA} = 6.5\text{cm}$
Now, semi-perimeter of a traingle, $\text{S} = \frac{\text{a+b+c}}{2} = \frac{\text{7.5+7+6.5}}{2}=\frac{21}{2}=10.5\text{cm}$
$\therefore\ \text{Area of }\triangle\text{ABC}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula] $=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$
$=\sqrt{10.5\times3\times3.5\times4}$
$=\sqrt{441}=21\text{cm}^2$
Now, area of parallelogram $BCED$ = Base \times Height $=\text{BC}\times\text{DF}=7\times\text{DF}$
According to the question, Area of $\triangle\text{ABC}$ = Area of parallelogram $BCED$
$\Rightarrow\ 21=7\times\text{DF}$ [from eqs. $(i)$ and $(ii)]$
$\Rightarrow\text{DF}=\frac{21}{4}=3\text{cm}$
Hence, the height of parallelogrom is $3\ cm.$
The sides of a traingle are $\text{AB} = \text{a} = 7.5\text{cm,} \text{ BC} = \text{b} = 7\text{cm} \text{ and}\text{ CA} = 6.5\text{cm}$
Now, semi-perimeter of a traingle, $\text{S} = \frac{\text{a+b+c}}{2} = \frac{\text{7.5+7+6.5}}{2}=\frac{21}{2}=10.5\text{cm}$
$\therefore\ \text{Area of }\triangle\text{ABC}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula] $=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$
$=\sqrt{10.5\times3\times3.5\times4}$
$=\sqrt{441}=21\text{cm}^2$
Now, area of parallelogram $BCED$ = Base \times Height $=\text{BC}\times\text{DF}=7\times\text{DF}$
According to the question, Area of $\triangle\text{ABC}$ = Area of parallelogram $BCED$
$\Rightarrow\ 21=7\times\text{DF}$ [from eqs. $(i)$ and $(ii)]$
$\Rightarrow\text{DF}=\frac{21}{4}=3\text{cm}$
Hence, the height of parallelogrom is $3\ cm.$
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