Question

the sides of a triangle measure 18 mm, 29 mm, and 17 mm. find its area. write your answer as an integer or as a decimal rounded to the nearest tenth. \boxed{} mm²

Explanation:

Step1: Calculate the semi - perimeter (s)

The formula for the semi - perimeter of a triangle with sides \(a\), \(b\), and \(c\) is \(s=\frac{a + b + c}{2}\).
Given \(a = 18\) mm, \(b = 29\) mm, \(c = 17\) mm.
\(s=\frac{18 + 29+17}{2}=\frac{64}{2} = 32\) mm.

Step2: Apply Heron's formula to find the area (A)

Heron's formula is \(A=\sqrt{s(s - a)(s - b)(s - c)}\).
Substitute \(s = 32\), \(a = 18\), \(b = 29\), \(c = 17\) into the formula:
\(A=\sqrt{32(32 - 18)(32 - 29)(32 - 17)}\)
\(=\sqrt{32\times14\times3\times15}\)
First, calculate the product inside the square root:
\(32\times14 = 448\), \(3\times15=45\), then \(448\times45 = 20160\)
So \(A=\sqrt{20160}\approx142.0\) (rounded to the nearest tenth)

Answer:

\(142.0\)