Question

$\Delta ABC$ is isosceles with $AB=AC=7.5cm$ and $BC=9cm$ (Figure). The height$AD$ from $A$ to$BC$, is $6cm$. Find the area of $\Delta ABC$. What will be the height from$C$to $AB$ i.e., $CE$

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Solution

Finding the area of $\Delta ABC$ and height from$C$to $AB$ i.e., $CE$:Solution:From the given figure we note that$\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}$$BC=9cm\phantom{\rule{0ex}{0ex}}AD=6cm$We know thatArea of $\Delta ABC=\frac{1}{2}×base×height$ $=\frac{1}{2}×BC×AD\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×9×6\phantom{\rule{0ex}{0ex}}=1×9×3\phantom{\rule{0ex}{0ex}}=27c{m}^{2}$$\therefore △\mathrm{ABC}=27{\mathrm{cm}}^{2}$Also,Area of $△\mathrm{ABC}$, $\begin{array}{rcl}\Delta ABC& =& \frac{1}{2}×base×height\\ 27& =& \frac{1}{2}×AB×CE\\ 27& =& \frac{1}{2}×7.5×CE\\ CE& =& \frac{27×2}{7.5}\\ CE& =& \frac{54}{7.5}\\ CE& =& 7.2cm\end{array}$Therefore the area of triangle $△\mathrm{ABC}=27{\mathrm{cm}}^{2}$ and the height from$C$to $AB$ i.e., $CE$$=7.2cm$

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