Find perimeter, area and altitude of an isosceles triangle given the length of equal sides is $8\mathrm{cm}$ and the base is $12\mathrm{cm}$.

Given:$$a=8\mathrm{cm}$$$$b=12\mathrm{cm}$$

To find: Perimeter, Area and Altitude of an isosceles triangle

Solution: First lets find the perimeter of the given isosceles triangle. $$\begin{array}{rcll}\because & P& =& 2a+b\\ \therefore & P& =& 2\times 8+12\\ \therefore & P& =& 16+12\\ \therefore & P& =& 28\mathrm{cm}\end{array}$$Therefore, the perimeter of the given isosceles triangle is $28\mathrm{cm}$.

Now, lets find area of the given isosceles triangle. $$\begin{array}{lrcl}\because & A& =& \frac{b}{2}\sqrt{a^2-\frac{b^2}{4}}\\ \therefore & A& =& \frac{12}{2}\times \sqrt{8^2-\frac{{12}^2}{4}}\\ \therefore & A& =& 6\times \sqrt{64-\frac{144}{4}}\\ \therefore & A& =& 6\times \sqrt{64-36}\\ \therefore & A& =& 6\times \sqrt{28}\\ \therefore & A& =& 6\times \sqrt{4\times 7}\\ \therefore & A& =& 6\times \sqrt{2^2\times 7}\\ \therefore & A& =& 6\times 2\times \sqrt{7}\\ \therefore & A& =& 12\sqrt{7}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given isosceles triangle is $12\sqrt{7}{\mathrm{cm}}^2$.

Now, lets find altitude of the given isosceles triangle. $$\begin{array}{lrcl}\because & h& =& \sqrt{a^2-\frac{b^2}{4}}\\ \therefore & h& =& \sqrt{8^2-\frac{{12}^2}{4}}\\ \therefore & h& =& \sqrt{64-\frac{144}{4}}\\ \therefore & h& =& \sqrt{64-36}\\ \therefore & h& =& \sqrt{28}\\ \therefore & h& =& \sqrt{4\times 7}\\ \therefore & h& =& \sqrt{2^2\times 7}\\ \therefore & h& =& 2\times \sqrt{7}\\ \therefore & h& =& 2\sqrt{7}\mathrm{cm}\end{array}$$ Therefore, the altitude of the given isosceles triangle is $2\sqrt{7}\mathrm{cm}$.