Question

the diagram shows an isosceles triangle abc with ab = ac, bc = 24 cm and ad is perpendicular to bc. given that the area of the triangle is 108 cm², find (a) ad, (b) ac.

Explanation:

Step1: Use area formula to find AD

The area formula for a triangle is $A=\frac{1}{2}\times base\times height$. Here, the base $BC = 24$ cm and area $A = 108$ $cm^{2}$. Let $AD$ be the height. We have $A=\frac{1}{2}\times BC\times AD$. Substituting the values, we get $108=\frac{1}{2}\times24\times AD$.

Step2: Solve for AD

First, simplify the right - hand side of the equation: $\frac{1}{2}\times24\times AD = 12AD$. Then, solve the equation $108 = 12AD$ for $AD$. Divide both sides by 12: $AD=\frac{108}{12}=9$ cm.

Step3: Use Pythagorean theorem to find AC

Since $\triangle ABC$ is isosceles and $AD\perp BC$, $BD=\frac{BC}{2}=\frac{24}{2}=12$ cm. In right - triangle $ADC$, by the Pythagorean theorem $AC^{2}=AD^{2}+DC^{2}$ (and $DC = BD = 12$ cm, $AD = 9$ cm). So $AC^{2}=9^{2}+12^{2}=81 + 144=225$.

Step4: Solve for AC

Take the square root of both sides of the equation $AC^{2}=225$. Since $AC>0$, $AC = 15$ cm.

Answer:

(a) $AD = 9$ cm
(b) $AC = 15$ cm