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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?

 Apr 25, 2018
 #1
avatar+972 
+2

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:

 

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

 

The final answer is 12, 

 

I hope this helped, 

 

Gavin

 Apr 25, 2018
 #2
avatar+3987 
+3

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 \(\boxed{12}\).

smileysmiley

 Apr 25, 2018

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