Thread: Area of an Isoscles Triangle

1. Area of an Isoscles Triangle

Reeeeally need help with this:
Thanks!!

2. Re: Area of an Isoscles Triangle

$h = s \cos\left(\dfrac{\theta}{2}\right)$

$\dfrac{b}{2} = s \sin\left(\dfrac{\theta}{2}\right)$

$A = \dfrac{bh}{2} = s \sin\left(\dfrac{\theta}{2}\right) \cdot s \cos\left(\dfrac{\theta}{2}\right)$

$A = s^2 \sin\left(\dfrac{\theta}{2}\right)\cos\left(\dfrac {\theta}{2}\right)$

$A = s^2 \cdot \dfrac{1}{2} \cdot 2 \sin\left(\dfrac{\theta}{2}\right)\cos\left(\dfrac {\theta}{2}\right)$

recall the double angle identity for sine ...

$A = \dfrac{1}{2}s^2 \cdot \sin{\theta}$

3. Re: Area of an Isoscles Triangle

Thank you so much for replying!! Still confused as to why you made those first steps, having h = s(cos theta/2) and b/2 = s(si theta/2). Where did that come from? Why are the thetas over 2? And why cos/sin for the respective parts?

4. Re: Area of an Isoscles Triangle

Recall what you learned in geometry ... the altitude $h$ bisects angle $\theta$ and is perpendicular to the base, forming two congruent right triangles.

Recall right triangle trig ... does SOHCAHTOA ring a bell?

$\sin\left(\dfrac{\theta}{2}\right) = \dfrac{opposite}{hypotenuse} = \dfrac{(b/2)}{s} \implies \dfrac{b}{2} = s \sin\left(\dfrac{\theta}{2}\right)$

$\cos\left(\dfrac{\theta}{2}\right) = \dfrac{adjacent}{hypotenuse} = \dfrac{h}{s} \implies h = s \sin\left(\dfrac{\theta}{2}\right)$

5. Re: Area of an Isoscles Triangle

Also, what happens to the cos theta/2 at the end? And wouldn't it be sin 2theta? Very confused, but again thank you so much for taking the time to help it means a lot

6. Re: Area of an Isoscles Triangle

$2\sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right) = \sin\bigg[2 \cdot \left(\dfrac{\theta}{2}\right) \bigg] = \sin{\theta}$