Question

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 side of the triangle.

Answer 1

Consider Isosceles triangle $ABC:AB=BC.$ Let $AD$ be the altitude drawn from one of the congruent sides.

Given $AB=18\ cm, AD=15\ cm.$

Let $BD=x\ cm, CD=y\ cm.$ Then $BC=AB=(x+y)\ cm.$

Consider right triangle $CAD.$ The Pythagorean Theorem

$$18^2=15^2+y^2$$

Consider right triangle $ABD.$ The Pythagorean Theorem

$$(x+y)^2=15^2+x^2$$

$$\begin{matrix} y^2=99 \\ x^2+2xy+y^2=225+x^2 \end{matrix}$$

$$y=3\sqrt{11}$$ $$2x(3\sqrt{11})=126$$

$$x=\dfrac{21}{11}\sqrt{11}$$

$$y=3\sqrt{11}$$

$$AB=BC=\dfrac{21}{11}\sqrt{11}+3\sqrt{11}=\dfrac{54}{11}\sqrt{11}$$

$\dfrac{54}{11}\sqrt{11}\ cm, \dfrac{54}{11}\sqrt{11}\ cm, 18\ cm.$