question_answer

In an equilateral triangle the length of the altitude is 16 cm,  then find the area of triangle,

A)  \[256c{{m}^{2}}\]                             

B)  \[96\sqrt{3}c{{m}^{2}}\]

C)  \[\frac{66}{\sqrt{3}}c{{m}^{2}}\]                  

D)  \[\frac{256}{\sqrt{3}}c{{m}^{2}}\]

Correct Answer: D

Solution :

(d): \[{{a}^{2}}-\frac{{{a}^{2}}}{4}={{16}^{2}}\] \[=\frac{3{{a}^{2}}}{4}=256\] \[{{a}^{2}}=\frac{1024}{3}\] Area \[=\frac{1}{2}\times a\times 16=\frac{1}{2}\times \frac{32}{\sqrt{3}}\times 16=\frac{256}{\sqrt{3}}\]