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Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Given:
Each side of an equilateral triangle measures 12 cm.
To do:
We have to find the height of the equilateral triangle.
Solution:
In the above figure, AD is the altitude of the equilateral triangle ABC.
$AB=BC=CA=12\ cm$
In $\triangle ADB$ and $\triangle ACD$,
$\angle ADB=\angle ADC=90^o$
$AB=AC$
Therefore,
$\triangle ADB \cong\ \triangle ACD$ (By RHS congruence)
This implies,
$BD=DC=\frac{BC}{2}=\frac{12}{2}\ cm=6\ cm$ (CPCT)
In $\triangle ADB$,
$AB^2=AD^2+BD^2$ (By using Pythagoras theorem)
$(12)^2=AD^2+(6)^2$
$AD^2=144-36$
$AD^2=108$
$AD=\sqrt{108}\ cm$
$AD=\sqrt{36\times3}\ cm$
$AD=6\sqrt3\ cm$
The height of the equilateral triangle is $6\sqrt3\ cm$.
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