Find perimeter, area and altitude of an isosceles triangle given the length of equal sides is $12\mathrm{cm}$ and the base is $16\mathrm{cm}$.

Given:$$a=12\mathrm{cm}$$$$b=16\mathrm{cm}$$

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

Solution: First lets find the perimeter of the given isosceles triangle. $$\begin{array}{rcll}\because & P& =& 2a+b\\ \therefore & P& =& 2\times 12+16\\ \therefore & P& =& 24+16\\ \therefore & P& =& 40\mathrm{cm}\end{array}$$Therefore, the perimeter of the given isosceles triangle is $40\mathrm{cm}$.

Now, lets find area of the given isosceles triangle. $$\begin{array}{lrcl}\because & A& =& \frac{b}{2}\sqrt{a^2-\frac{b^2}{4}}\\ \therefore & A& =& \frac{16}{2}\times \sqrt{{12}^2-\frac{{16}^2}{4}}\\ \therefore & A& =& 8\times \sqrt{144-\frac{256}{4}}\\ \therefore & A& =& 8\times \sqrt{144-64}\\ \therefore & A& =& 8\times \sqrt{80}\\ \therefore & A& =& 8\times \sqrt{16\times 5}\\ \therefore & A& =& 8\times \sqrt{4^2\times 5}\\ \therefore & A& =& 8\times 4\times \sqrt{5}\\ \therefore & A& =& 32\sqrt{5}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given isosceles triangle is $32\sqrt{5}{\mathrm{cm}}^2$.

Now, lets find altitude of the given isosceles triangle. $$\begin{array}{lrcl}\because & h& =& \sqrt{a^2-\frac{b^2}{4}}\\ \therefore & h& =& \sqrt{{12}^2-\frac{{16}^2}{4}}\\ \therefore & h& =& \sqrt{144-\frac{256}{4}}\\ \therefore & h& =& \sqrt{144-64}\\ \therefore & h& =& \sqrt{80}\\ \therefore & h& =& \sqrt{16\times 5}\\ \therefore & h& =& \sqrt{4^2\times 5}\\ \therefore & h& =& 4\times \sqrt{5}\\ \therefore & h& =& 4\sqrt{5}\mathrm{cm}\end{array}$$ Therefore, the altitude of the given isosceles triangle is $4\sqrt{5}\mathrm{cm}$.