Find perimeter, area and altitude of an isosceles triangle given the length of equal sides is $12\mathrm{cm}$ and the base is $18\mathrm{cm}$.

Given:$$a=12\mathrm{cm}$$$$b=18\mathrm{cm}$$

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

Solution: First lets find the perimeter of the given isosceles triangle. $$\begin{array}{rcll}\because & P& =& 2a+b\\ \therefore & P& =& 2\times 12+18\\ \therefore & P& =& 24+18\\ \therefore & P& =& 42\mathrm{cm}\end{array}$$Therefore, the perimeter of the given isosceles triangle is $42\mathrm{cm}$.

Now, lets find area of the given isosceles triangle. $$\begin{array}{lrcl}\because & A& =& \frac{b}{2}\sqrt{a^2-\frac{b^2}{4}}\\ \therefore & A& =& \frac{18}{2}\times \sqrt{{12}^2-\frac{{18}^2}{4}}\\ \therefore & A& =& 9\times \sqrt{144-\frac{324}{4}}\\ \therefore & A& =& 9\times \sqrt{144-81}\\ \therefore & A& =& 9\times \sqrt{63}\\ \therefore & A& =& 9\times \sqrt{9\times 7}\\ \therefore & A& =& 9\times \sqrt{3^2\times 7}\\ \therefore & A& =& 9\times 3\times \sqrt{7}\\ \therefore & A& =& 27\sqrt{7}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given isosceles triangle is $27\sqrt{7}{\mathrm{cm}}^2$.

Now, lets find altitude of the given isosceles triangle. $$\begin{array}{lrcl}\because & h& =& \sqrt{a^2-\frac{b^2}{4}}\\ \therefore & h& =& \sqrt{{12}^2-\frac{{18}^2}{4}}\\ \therefore & h& =& \sqrt{144-\frac{324}{4}}\\ \therefore & h& =& \sqrt{144-81}\\ \therefore & h& =& \sqrt{63}\\ \therefore & h& =& \sqrt{9\times 7}\\ \therefore & h& =& \sqrt{3^2\times 7}\\ \therefore & h& =& 3\times \sqrt{7}\\ \therefore & h& =& 3\sqrt{7}\mathrm{cm}\end{array}$$ Therefore, the altitude of the given isosceles triangle is $3\sqrt{7}\mathrm{cm}$.