Given that
[tex]\begin{gathered} AB=AC=16\operatorname{cm} \\ BC=20\operatorname{cm} \\ BD=\frac{BC}{2}=\frac{20\operatorname{cm}}{2}=10\operatorname{cm} \\ CD=\frac{BC}{2}=\frac{20\operatorname{cm}}{2}=10\operatorname{cm} \\ AD=h \end{gathered}[/tex]
To calculate the height of the triangle, we will use the Pythagoras theorem
With the Pythagorean theorem, we will have
[tex]\begin{gathered} \text{hypotenus}^2=\text{opposite}^2+\text{adjacent}^2 \\ \text{where,} \\ \text{HYPOTENUS}=AC=16\operatorname{cm} \\ \text{opposite}=DC=10\operatorname{cm} \\ \text{adjacent}=AD=h \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} \text{hypotenus}^2=\text{opposite}^2+\text{adjacent}^2 \\ 16^2=10^2+h^2 \\ 256=100+h^2 \\ \text{substract 100 from both sides} \\ 256-100=100-100+h^2 \\ 156=h^2 \\ \text{square root both sides} \\ \sqrt[]{h^2}=\sqrt[]{156} \\ h=\sqrt[]{4}\times\sqrt[]{39} \\ h=2\sqrt[]{39} \\ or\text{ } \\ h=12.39\operatorname{cm}\approx to\text{ 1 d.p=} \\ h=12.5\operatorname{cm} \end{gathered}[/tex]
Therefore,
The height of the triangle is = 12.5 cm
