The base of an isosceles triangle and the altitude drawn from one of the congruent sides are equal to 18 cm and 15 cm, respectively. Find the lengths of the sides of the triangle. Please help me to solve this. Also, can you please show your solution? Thank you very much!
Let $h,x$ be the altitude drawn from the base and the length of the other side of the triangle respectively. Then considering the area of the triangle gives us $$\frac{18\times h}{2}=\frac{x\times 15}{2}\Rightarrow h=\frac{5}{6}x.$$ Also, we have $$x^2=h^2+9^2.$$
Solving these gives us $$x=\frac{54}{\sqrt{11}}.$$