Question
$\triangle A B C$ is right angled at $A$. Shown in figure. Find $A D$.
✓
Answer
In $\triangle A B C$, when base $=A B=12 cm$, then height $h=A C=5 cm$
$\text{Area of}~\triangle A B C =\frac{1}{2} \times A B \times A C $
$ =\frac{1}{2} \times 12 \times 5=30 cm^2$
When base $=B C=13 cm$, then height $=A D$
$\therefore \text { Area of } \triangle A B C =\frac{1}{2} \times B C \times A D $
$30 =\frac{1}{2} \times 13 \times A D $
$\Rightarrow A D=\frac{2 \times 30}{13} =4.62 cm.$
$\text{Area of}~\triangle A B C =\frac{1}{2} \times A B \times A C $
$ =\frac{1}{2} \times 12 \times 5=30 cm^2$
When base $=B C=13 cm$, then height $=A D$
$\therefore \text { Area of } \triangle A B C =\frac{1}{2} \times B C \times A D $
$30 =\frac{1}{2} \times 13 \times A D $
$\Rightarrow A D=\frac{2 \times 30}{13} =4.62 cm.$
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