# Area of an Isosceles Triangle

• Aug 29th 2008, 01:01 PM
magentarita
Area of an Isosceles Triangle
Show that the area A of an isosceles triangle whose equal sides are of length s and theta is the angle between them is (1/2)(s^2)(sintheta)

HINT GIVEN BY TEACHER:

The height h bisects the angle theta and is the perpendicular bisector of the base.

• Aug 29th 2008, 03:17 PM
mr fantastic
Quote:

Originally Posted by magentarita
Show that the area A of an isosceles triangle whose equal sides are of length s and theta is the angle between them is (1/2)(s^2)(sintheta)

HINT GIVEN BY TEACHER:

The height h bisects the angle theta and is the perpendicular bisector of the base.

Let the height of the isosceles triangle be h and its base length be 2x.

Draw in the height. Then calculate x and h from the right angle triangle:

$\cos \left( \frac{\theta}{2} \right) = \frac{h}{s} \Rightarrow h = ....$

$\sin \left( \frac{\theta}{2} \right) = \frac{x}{s} \Rightarrow x = ....$

Then Area = xh = ......

I've left several details for you to complete. I'm confident you can do this (or at least, mostly do this) with a bit of effort. This is important. It helps you to develop your understanding.

Contrary to what some students might think, getting a complete solution to a problem is not necessarily beneficial for them if they want to develop understanding ..... Maths is a doing subject, not a spectator sport. The best way of learning and developing understanding is to get a few hints and then try the problem again. You can always ask for more help if you get stuck.

Edit: I've just seen Jhevon's post (beaten to the punch doh!). My comments were made without having seen his post. They are not directed towards him, his post or any of his other posts. I felt the comments were appropriate given the stage the OP appears to currently be at.
• Aug 29th 2008, 07:31 PM
magentarita
Thanks
Quote:

Originally Posted by mr fantastic
Let the height of the isosceles triangle be h and its base length be 2x.

Draw in the height. Then calculate x and h from the right angle triangle:

$\cos \left( \frac{\theta}{2} \right) = \frac{h}{s} \Rightarrow h = ....$

$\sin \left( \frac{\theta}{2} \right) = \frac{x}{s} \Rightarrow x = ....$

Then Area = xh = ......

I've left several details for you to complete. I'm confident you can do this (or at least, mostly do this) with a bit of effort. This is important. It helps you to develop your understanding.

Contrary to what some students might think, getting a complete solution to a problem is not necessarily beneficial for them if they want to develop understanding ..... Maths is a doing subject, not a spectator sport. The best way of learning and developing understanding is to get a few hints and then try the problem again. You can always ask for more help if you get stuck.

Edit: I've just seen Jhevon's post (beaten to the punch doh!). My comments were made without having seen his post. They are not directed towards him, his post or any of his other posts. I felt the comments were appropriate given the stage the OP appears to currently be at.

I thank you very much.