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., $CE$

Answer 1

Finding the area of $\Delta ABC$ and height from$C$to $AB$ i.e., $CE$:

Solution:

From the given figure we note that

$\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}$

$$\begin{gathered} BC=9cm \\ AD=6cm \end{gathered}$$

We know that

Area of $\Delta ABC=\frac{1}{2}\times base\times height$

$$\begin{gathered} =\frac{1}{2}\times BC\times AD \\ =\frac{1}{2}\times 9\times 6 \\ =1\times 9\times 3 \\ =27c{m}^2 \end{gathered}$$

$\therefore △\mathrm{ABC}=27{\mathrm{cm}}^2$

Also,

Area of $△\mathrm{ABC}$,

$\begin{array}{rcl}\Delta ABC& =& \frac{1}{2}\times base\times height\\ 27& =& \frac{1}{2}\times AB\times CE\\ 27& =& \frac{1}{2}\times 7.5\times CE\\ CE& =& \frac{27\times 2}{7.5}\\ CE& =& \frac{54}{7.5}\\ CE& =& 7.2cm\end{array}$

Therefore the area of triangle $△\mathrm{ABC}=27{\mathrm{cm}}^2$ and the height from$C$to $AB$ i.e., $CE$$=7.2cm$