What is the area of triangle ABC if a = 19, b = 14, and c = 23?
I would use Heron's Formula
area = √(s(s-a)(s-b)(s-c)) , where s = (a+b+c)/2
check your work with this
http://www.csgnetwork.com/areacalc.html
use Heron's formula, or use law of sines/cosines to determine one of the angles. then the area is 1/2 ab*sinC
To find the area of a triangle, you can use Heron's formula. According to Heron's formula, the area (A) of a triangle with side lengths a, b, and c can be calculated using the formula:
A = √(s(s-a)(s-b)(s-c))
where s is the semi-perimeter of the triangle (half the sum of the lengths of its sides) which can be calculated as:
s = (a + b + c) / 2
Let's calculate the area of the triangle ABC step by step using the given values:
- Given:
a = 19, b = 14, c = 23
1. Calculate the semi-perimeter (s):
s = (a + b + c) / 2
= (19 + 14 + 23) / 2
= 56 / 2
= 28
2. Use Heron's formula to calculate the area (A):
A = √(s(s-a)(s-b)(s-c))
= √(28(28-19)(28-14)(28-23))
= √(28 * 9 * 14 * 5)
= √(2^2 * 3^2 * 7 * 2 * 5)
= √(2^4 * 3^2 * 5 * 7)
= √(16 * 9 * 5 * 7)
= √(16 * 45 * 7)
= √(10080)
≈ 100.4
Therefore, the area of triangle ABC is approximately 100.4 square units.