How do you use Heron's formula to find the area of a triangle with sides of lengths #2 #, #2 #, and #2 #?

Question

How do you use Heron's formula to find the area of a triangle with sides of lengths #2  #, #2    #, and #2   #?

Theodore Abad · Accepted Answer

#A=sqrt3approx1.7321#

Heron's formula states that for a triangle with sides #a,b,c# and a semiperimeter #s=(a+b+c)/2#, the area of the triangle is

#A=sqrt(s(s-a)(s-b)(s-c))#

Here, we know that

#s=(2+2+2)/2=3#

which gives an area of

#A=sqrt(3(3-2)(3-2)(3-2))#

#A=sqrt3approx1.7321#

This problem could also be solved by drawing the equilateral triangle and splitting it into two right triangles with base #1# and height #sqrt3#, giving each right triangle area #sqrt3/2# and the whole triangle an area of #sqrt3#.

Violet Aldrich · Answer

undefined To use Heron's formula to find the area of a triangle with side lengths of 2, 2, and 2, we first need to calculate the semi-perimeter of the triangle, denoted by $ s $. The semi-perimeter is half the perimeter of the triangle and is calculated as:

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

where $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle.

For our triangle with sides of length 2, 2, and 2:

$$ s = \frac{{2 + 2 + 2}}{2} = \frac{6}{2} = 3 $$

Now, we can use Heron's formula to find the area $ A $ of the triangle:

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

Substitute the values of $ s $, $ a $, $ b $, and $ c $ into the formula:

$$ A = \sqrt{{3(3 - 2)(3 - 2)(3 - 2)}} $$

$$ A = \sqrt{{3 \cdot 1 \cdot 1 \cdot 1}} $$

$$ A = \sqrt{3} $$

Therefore, the area of the triangle with side lengths of 2, 2, and 2 is $ \sqrt{3} $ square units.