+885
What is the number of square units in the area of a triangle whose sides measure 5, 5 and 6 units?
+983 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
+4622 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 \(\boxed{12}\).
