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# quick...

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What is the number of square units in the area of a triangle whose sides measure 5, 5 and 6 units?

ant101  Apr 25, 2018
#1
+942
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We use Heron's formula.

$$\sqrt{s(s-a)(s-b)(s-c)}$$

Where s is half of the perimeter.

We can first solve for s, which is:

$$(5+5+6)\div2=8$$

Plugging our values into the formula, we have:

$$\sqrt{8(8-5)(8-5)(8-6)}$$

I hope this helped,

Gavin

GYanggg  Apr 25, 2018
#2
+3035
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Solution:

This triangle is isosceles, and so the altitude to the side with length 6 must hit that side at its midpoint. Thus our triangle is divided into two right triangles with hypotenuse $$5$$ and one side of length $$3$$. Thus each of these is a $$3-4-5$$  triangle, and each one has area $$\frac{3 \times 4}{2} = 6$$, for a total area of $$\boxed{12}$$.

tertre  Apr 25, 2018