Find perimeter, area and altitude of an isosceles triangle given the length of equal sides is $10\mathrm{cm}$ and the base is $12\mathrm{cm}$.

Given:$$a=10\mathrm{cm}$$$$b=12\mathrm{cm}$$

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

Solution: First lets find the perimeter of the given isosceles triangle. $$\begin{array}{rcll}\because & P& =& 2a+b\\ \therefore & P& =& 2\times 10+12\\ \therefore & P& =& 20+12\\ \therefore & P& =& 32\mathrm{cm}\end{array}$$Therefore, the perimeter of the given isosceles triangle is $32\mathrm{cm}$.

Now, lets find area of the given isosceles triangle. $$\begin{array}{lrcl}\because & A& =& \frac{b}{2}\sqrt{a^2-\frac{b^2}{4}}\\ \therefore & A& =& \frac{12}{2}\times \sqrt{{10}^2-\frac{{12}^2}{4}}\\ \therefore & A& =& 6\times \sqrt{100-\frac{144}{4}}\\ \therefore & A& =& 6\times \sqrt{100-36}\\ \therefore & A& =& 6\times \sqrt{64}\\ \therefore & A& =& 6\times \sqrt{8^2}\\ \therefore & A& =& 6\times 8\\ \therefore & A& =& 48{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given isosceles triangle is $48{\mathrm{cm}}^2$.

Now, lets find altitude of the given isosceles triangle. $$\begin{array}{lrcl}\because & h& =& \sqrt{a^2-\frac{b^2}{4}}\\ \therefore & h& =& \sqrt{{10}^2-\frac{{12}^2}{4}}\\ \therefore & h& =& \sqrt{100-\frac{144}{4}}\\ \therefore & h& =& \sqrt{100-36}\\ \therefore & h& =& \sqrt{64}\\ \therefore & h& =& \sqrt{64\times 1}\\ \therefore & h& =& \sqrt{8^2\times 1}\\ \therefore & h& =& 8\times \sqrt{1}\\ \therefore & h& =& 8\mathrm{cm}\end{array}$$ Therefore, the altitude of the given isosceles triangle is $8\mathrm{cm}$.