Geometry & Mensuration
Aptitude

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Q.

The perimeter of an equilateral $\triangle$ is \(72\sqrt{3}\) meters.

Find its height.

 A.

63 metres

 B.

55 metres

 C.

40 metres

 D.

36 metres

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Solution:
Option(D) is correct

Answer image for Mensuration, Aptitude:2169-1

Let one side of the \(\triangle\) be \(=a\)

Perimeter of equilateral \(\triangle=3a\)

$\Rightarrow 3a=72\sqrt{3}$

$\Rightarrow a=24\sqrt{3}$

Height \(=AC\) by pythagoras theorem

\(AC^2=a^2-\left(\dfrac{a}{2}\right)^2\)

$AC^2=a^2\times \left[1-\left(\dfrac{1}{2}\right)^2\right]$

$AC^2=a^2\times \left[1- \dfrac{1}{4}\right]$

$AC^2=a^2\times \dfrac{3}{4}$

Now, putting, $a=24\sqrt{3}$

$\Rightarrow AC^2 = 24^2 \times 3 \times \dfrac{3}{4} $

$\Rightarrow AC^2 = \dfrac{24^2\times 3^2}{2^2} $

$\Rightarrow AC = \dfrac{24\times 3}{2} $

\(AC=36\text{ cm}\)

Edit: For an alternative solution with less calculation, check comment from Thulasi.


(3) Comment(s)


Thulasi
 ()

$3a=72\sqrt{3}$

$\Rightarrow a=24\sqrt{3}$

Now, area of an equilateral triangle, $\dfrac{3}{4} \times (\text{side})^2$ -------- (1)

Also, we know that area of any triangle is, $\dfrac{1}{2} \times \text{base} \times \text{height}$ -------- (2)

From (1) and (2),

$\Rightarrow \dfrac{3}{4}\times a^2 = \dfrac{1}{2} \times a \times h$

$\dfrac{3}{4} \times 24 \sqrt{3} = \dfrac{1}{2} \times h$

$h = \textbf{36 cms}$



KARTIK
 ()

please make necessary corrections.


Deepak
 ()

Typos have been corrected. Thank you letting me know.