Q7 Q30.) Find the area of the triangle specified below.

Use Heron's formula to find the area of the triangle since you're given all the side lengths.

Heron's formula: [tex] \sqrt{s(s-a)(s-b)(s-c)} [/tex], where s = [tex] \frac{a + b + c}{2} [/tex].

Find s.
[tex] \frac{19 + 21 + 15}{2} [/tex] = 27.5

Now plug in s and a, b, and c into the formula.

[tex] \sqrt{27.5(27.5-19)(27.5 - 21)(27.5 - 15)} [/tex] = 137.8

The area of the triangle is 138 square meters.

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THEBRAINLY2 THEBRAINLY2 · Answer 2 · 2018-04-26T06:14:19+02:00

Let the sides of the triangle be a = 19 m, b = 21 m, c = 15 m

[tex]p = \frac{a + b + c}{2} [/tex]
[tex]p = \frac{19 + 21 + 15}{2} [/tex]
[tex]p = \frac{40 + 15}{2} [/tex]
[tex]p = \frac{55}{2} [/tex]
[tex]p = 27.50[/tex]

By Heron's Formula, area of triangle
[tex] s = \sqrt{p \times (p - a) \times (p - b)\times (p - c)}[/tex]
[tex]s = \sqrt{27.50 \times (27.50 - 19) \times (27.50 - 21) \times (27.50 - 15)} [/tex]
[tex]s = \sqrt{27.50 \times 8.50 \times 6.50 \times 12.50} [/tex]
[tex]s = \sqrt{233.75 \times 6.50 \times 12.50} [/tex]
[tex]s = \sqrt{1519.38 \times 12.50} [/tex]
[tex]s = \sqrt{18992.25} [/tex]
[tex]s = 137.81[/tex]
[tex]A = 138 \: (rounded \: to \: the \: nearest \: integer)[/tex]