Find perimeter, area and altitude of an isosceles triangle given the length of equal sides is $9\mathrm{cm}$ and the base is $12\mathrm{cm}$.

Given:$$a=9\mathrm{cm}$$$$b=12\mathrm{cm}$$

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

Solution: First lets find the perimeter of the given isosceles triangle. $$\begin{array}{rcll}\because & P& =& 2a+b\\ \therefore & P& =& 2\times 9+12\\ \therefore & P& =& 18+12\\ \therefore & P& =& 30\mathrm{cm}\end{array}$$Therefore, the perimeter of the given isosceles triangle is $30\mathrm{cm}$.

Now, lets find area of the given isosceles triangle. $$\begin{array}{lrcl}\because & A& =& \frac{b}{2}\sqrt{a^2-\frac{b^2}{4}}\\ \therefore & A& =& \frac{12}{2}\times \sqrt{9^2-\frac{{12}^2}{4}}\\ \therefore & A& =& 6\times \sqrt{81-\frac{144}{4}}\\ \therefore & A& =& 6\times \sqrt{81-36}\\ \therefore & A& =& 6\times \sqrt{45}\\ \therefore & A& =& 6\times \sqrt{9\times 5}\\ \therefore & A& =& 6\times \sqrt{3^2\times 5}\\ \therefore & A& =& 6\times 3\times \sqrt{5}\\ \therefore & A& =& 18\sqrt{5}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given isosceles triangle is $18\sqrt{5}{\mathrm{cm}}^2$.

Now, lets find altitude of the given isosceles triangle. $$\begin{array}{lrcl}\because & h& =& \sqrt{a^2-\frac{b^2}{4}}\\ \therefore & h& =& \sqrt{9^2-\frac{{12}^2}{4}}\\ \therefore & h& =& \sqrt{81-\frac{144}{4}}\\ \therefore & h& =& \sqrt{81-36}\\ \therefore & h& =& \sqrt{45}\\ \therefore & h& =& \sqrt{9\times 5}\\ \therefore & h& =& \sqrt{3^2\times 5}\\ \therefore & h& =& 3\times \sqrt{5}\\ \therefore & h& =& 3\sqrt{5}\mathrm{cm}\end{array}$$ Therefore, the altitude of the given isosceles triangle is $3\sqrt{5}\mathrm{cm}$.