What is the area of a triangle whose sides measure 13 inches, 15 inches, and 24 inches?

We can use a theorem in geometry known as "heron's formula". It states that given semi-perimeter(half of the perimeter) of triangle ABC as s, we have:

[ABC] = \(\sqrt{s(s-a)(s-b)(s-c)}\)

The semi perimeter of this triangle is:

\({13+15+24\over2} = {52\over2} = 26\)

We then have:

[ABC] = \(\sqrt{26(26-13)(26-15)(26-24)} = \sqrt{26*13*11*2} = \sqrt{26*26*11}\)

we can then take out a 26 from the square term, which gets us:

\(26\sqrt{11}\) in² as our area

Heron's formula: 

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

where \(s=\frac{a+b+c}{2}\) and \(a,b,c \) are sides of the triangle (13,15,24)

\(s=26\)

\(A=\sqrt{26(26-13)(26-15)(26-24)}\)

\(A=86.23\) inches squared.