Math Is Fun Forum

parseconds

Member

Registered: 2008-05-13

Posts: 3

Isosceles Triangle

One angle of an isosceles triangle is 150 degrees.  If the area of the triangle is 9 square feet, what is the perimeter of the triangle?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Isosceles Triangle

The 150 degree angle must be the one that there aren't two of, because otherwise the angles' sum would exceed 180. It is therefore next to the two matching sides.

The area of a triangle can be found by taking the product of two sides and the sine of the angle between them, then halving that.

Calling one of the isoceles triangle's matching sides a, we have a²/2 x sin150 = 9
∴ a = √(18/sin150) = 6.

Now we can use the sine rule to find the length of the remaining unknown side:

b/sin 150 = 6/sin 15
b = 6sin 150/sin 15 ≈ 11.6.

Therefore, the triangle's perimeter is ~6 + 6 + 11.6 = 23.6 feet.

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Why did the vector cross the road?
It wanted to be normal.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Isosceles Triangle

It is possible to solve it exactly too.
as mathsy said, the area is a²sin150/2=9 -> a=√(18/sin150)=6
the ramaining side is 2*sin75*a (drawing the height gives two right triangles), so the perimeter is p=2*sin75*a +a +a = 2a(sin75+1)
but

so

Last edited by Kurre (2008-05-14 08:42:29)