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

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

Enter your answer, in simplified radical form, in the box.

_____ inches squared

Apr 17, 2020

### 3+0 Answers

#1
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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

Apr 17, 2020
#3
+1991
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i remember this now! it makes sense thank you for your explaination . it really helped!

jjennylove  Apr 17, 2020
#2
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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.

Apr 17, 2020