# Area of an isoscoles triangle

1. Sep 3, 2010

### bjnartowt

1. The problem statement, all variables and given/known data

Imagine an isoscoles triangle. The "top" vertex has an angle of 2*theta. The common side has a length of L. When you cut the isoscoles triangle in half, that common side becomes the hypotenuse of the two resulting right-triangles.

${\rm{BaseLength}} = L \cdot \sin \theta + L\sin \theta = 2L\sin \theta$

${\rm{HeightLength}} = L\cos \theta$

I also am under the impression that:

${\rm{Area = BaseLength}} \times {\rm{HeightLength}}$

...so obviously:
$A = 2{L^2}\sin \theta \cos \theta$

If this is the correct area, then I've found a typo in a book... Can you either confirm my suspicions or refute my stance?

2. Sep 3, 2010

### zhermes

Area is half base times height.

3. Sep 3, 2010

### bjnartowt

....For a right triangle, yes, area is (1/2)*base*height. And I've split my isosceles triangle into two right triangles, so the area of an isosceles triangle seems to be (1/2)*base*height + (1/2)*base*height = base*height. No?

4. Sep 3, 2010

### eumyang

For any triangle, A = (1/2)bh.

No. If you split your isosceles triangle into two right trangles, then the base of the right triangle would be half of the base of the isosceles triangle.

The base of one right triangle is
$$b = L \sin \theta$$.
So the area of one right triangle is
\begin{aligned} A &= \frac{1}{2}bh \\ &= \frac{1}{2}L^2 \sin \theta \cos \theta \end{aligned}

The base of the isosceles triangle is
$$b = 2L \sin \theta$$.
So the area of the isosceles triangle is
\begin{aligned} A &= \frac{1}{2}bh \\ &= \frac{1}{2}(2)L^2 \sin \theta \cos \theta \\ &= L^2 \sin \theta \cos \theta

The area of one right triangle is half of the area of the isosceles triangle, so it checks out.

69

5. Sep 3, 2010

### bjnartowt

Curse my tendancy to omit critical details like that! Ack.

Thanks, now I see why the author had an extra factor of 1/2 that I was confused by. ...I wonder why I love math so much if it keep blowing raspberries at me, like now?