Question

One side of an equilateral triangle measures $8$ cm. Find its area using Heron's formula. What is its altitude

• A. $12\sqrt{3}{\text{cm}}^2,3\sqrt{3}$ cm
• B. $16\sqrt{3}{\text{cm}}^2,4\sqrt{3}$ cm
• C. $14\sqrt{3}{\text{cm}}^2,\frac{7}{2}\sqrt{3}$ cm
• D. $20\sqrt{3}{\text{cm}}^2,5\sqrt{3}$ cm

Answer 1

Answer: (B)

$16\sqrt{3}{\text{cm}}^2,4\sqrt{3}$ cm

According to the Heron's formula, Area (A) of the triangle having sides $a,b,c$ units is

$A=\sqrt{s(s-a)(s-b)(s-c)}$

where

$s=\frac{a+b+c}{2}$

For the given triangle,

$a=b=c=8cm$

$s=\frac{8+8+8}{2}=12$

$A=\sqrt{12(12-8)(12-8)(12-8)}$

$A=\sqrt{3\times 4\times 4^3}$

$=4\times 4\sqrt{3}$

$=16\sqrt{3}$

Area of the triangle =$\frac{1}{2}\times base\times altitude=16\sqrt{3}$

$\frac{1}{2}\times 8\times$ altitude$=16\sqrt{3}$

Altitude $=\frac{16\sqrt{3}}{4}=4\sqrt{3}$