Question

find the area of the triangle abc. a = 106.7 m b = 71.7 m c = 76.3 m what is the area of the triangle? □ m² (round to the nearest square meter as needed.)

Explanation:

Step1: Use Heron's formula. First, find the semi - perimeter \( s \).

The formula for the semi - perimeter of a triangle with sides \( a\), \(b\), \(c\) is \( s=\frac{a + b + c}{2}\).
Given \( a = 106.7\space m\), \(b=71.7\space m\), \(c = 76.3\space m\).
\(s=\frac{106.7+71.7 + 76.3}{2}=\frac{254.7}{2}=127.35\space m\)

Step2: Apply Heron's formula for the area \( A=\sqrt{s(s - a)(s - b)(s - c)}\)

Substitute \(s = 127.35\), \(a = 106.7\), \(b = 71.7\), \(c=76.3\) into the formula:
\(A=\sqrt{127.35\times(127.35 - 106.7)\times(127.35 - 71.7)\times(127.35 - 76.3)}\)
First, calculate the values inside the square root:
\(127.35-106.7 = 20.65\)
\(127.35 - 71.7=55.65\)
\(127.35 - 76.3 = 51.05\)
Then, \(127.35\times20.65\times55.65\times51.05\)
\(127.35\times20.65=127.35\times(20 + 0.65)=127.35\times20+127.35\times0.65 = 2547+82.7775 = 2629.7775\)
\(55.65\times51.05=(55 + 0.65)\times(51+0.05)=55\times51+55\times0.05+0.65\times51 + 0.65\times0.05=2805+2.75+33.15+0.0325 = 2840.9325\)
Now, multiply the two results: \(2629.7775\times2840.9325\approx2629.7775\times2841\)
\(2629.7775\times2841=(2600 + 29.7775)\times2841=2600\times2841+29.7775\times2841\)
\(2600\times2841 = 7386600\)
\(29.7775\times2841\approx29.7775\times2800+29.7775\times41=83377+1220.8775 = 84597.8775\)
\(7386600+84597.8775 = 7471197.8775\)
Then, \(A=\sqrt{7471197.8775}\approx2733.35\approx2733\)

Answer:

\(2733\)