How do you use Heron's formula to determine the area of a triangle with sides of that are 14, 16, and 17 units in length?

Question

Grace Ashcraft · Accepted Answer

#=104.324# square units

The area of a triangle, according to Heron, is: #A = sqrt(s(s-a)(s-b)(s-c)#, where #s# is the semi-pertimeter. #=>s=(a+b+c)/2# Here, #a=14#, #b=16# and #c=17#. First find #s#: #s=(a+b+c)/2# #=(14+16+17)/2=47/2# Let's compute the area now: #A = sqrt(s(s-a)(s-b)(s-c)# #= sqrt(47/2(47/2-14)(47/2-16)(47/2-17)# #= sqrt(47/2((47-28)/2)((47-32)/2)((47-34)/2)# #= sqrt(47/2(19/2)(15/2)(13/2)# #=sqrt((47xx19xx15xx13)/16)# #=1/4sqrt(47xx19xx15xx13)# #=1/4sqrt(174135)# #=1/4xx417.294# #=104.324#

Adam Aguirre · Answer

undefined To use Heron's formula, first calculate the semi-perimeter of the triangle by adding the lengths of all three sides and dividing by 2. In this case, the semi-perimeter is (14 + 16 + 17) / 2 = 23. Then, apply Heron's formula, which states that the area $ A $ of a triangle with side lengths $ a $, $ b $, and $ c $ is given by:

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

where $ s $ is the semi-perimeter, and $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle.

So, plugging in the values:

$$ A = \sqrt{23(23-14)(23-16)(23-17)} $$
$$ A = \sqrt{23(9)(7)(6)} $$
$$ A = \sqrt{23 \times 9 \times 7 \times 6} $$
$$ A = \sqrt{2646} $$
$$ A ≈ 51.51 \, \text{units}^2 $$

Therefore, the area of the triangle is approximately $ 51.51 \, \text{units}^2 $.