In $ \triangle \mathrm{ABC}, \mathrm{AD} $ is a median. If $ \mathrm{AB}=18 \mathrm{~cm} $. $ \mathrm{AC}=14 \mathrm{~cm} $ and $ \mathrm{AD}=14 \mathrm{~cm} $, find the perimeter of $ \triangle \mathrm{ABC} $.
Given:
In \( \triangle \mathrm{ABC}, \mathrm{AD} \) is a median.
\( \mathrm{AB}=18 \mathrm{~cm} \), \( \mathrm{AC}=14 \mathrm{~cm} \) and \( \mathrm{AD}=14 \mathrm{~cm} \)
To do:
We have to find the perimeter of \( \triangle \mathrm{ABC} \).
Solution:
We know that,
The length of the median to side BC $= \frac{1}{2}\sqrt{2AB^{2}+2AC^{2}-BC^{2}}$
Therefore,
$AD=\frac{1}{2}\sqrt{2AB^{2}+2AC^{2}-BC^{2}}$
$2(14)=\sqrt{2(18)^2+2(14)^2-BC^2}$
$(28)^2=2(324)+2(196)-BC^2$
$BC^2=648+392-784$
$BC^2=1040-784$
$BC^2=256$
$\Rightarrow BC=\sqrt{256}=16\ cm$
The perimeter of $\triangle \mathrm{ABC}=AB+BC+CA$
$=18+16+14$
$=48\ cm$
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