# Area of an Isosceles Triangle

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• Oct 21st 2006, 11:17 AM
josh16
Area of an Isosceles Triangle
In grade 9 math i got homework about finding the area of an isosceles triangle and i don't know how to find the area.
Thanks,
Josh
• Oct 21st 2006, 11:22 AM
Ideasman
Quote:

Originally Posted by josh16
In grade 9 math i got homework about finding the area of an isosceles triangle and i don't know how to find the area.
Thanks,
Josh

The area of an isosceles triangle, and any triangle for that matter, is:

(1/2)*b*h, where b is the base and h is the height. Further, for an ISOSCELES triangle, height can be found by the following formula:

h = sqrt[(b^2 - (1/2)*a^2)]

The area of an isosceles triangle can also be found using trigonometry, but I am assuming you are not quite up to that level yet.

Another formula you can consider using is Heron's formula.

Given a triangle with sides of length a, b, c, the area of the triangle is:

A = sqrt(s*(s-a)*(s-b)*(s-c)),

where s = (a + b + c)/2.
• Oct 21st 2006, 11:30 AM
AfterShock
Quote:

Originally Posted by Ideasman
The area of an isosceles triangle, and any triangle for that matter, is:

(1/2)*b*h, where b is the base and h is the height. Further, for an ISOSCELES triangle, height can be found by the following formula:

h = sqrt[(b^2 - (1/2)*a^2)]

The area of an isosceles triangle can also be found using trigonometry, but I am assuming you are not quite up to that level yet.

Another formula you can consider using is Heron's formula.

Given a triangle with sides of length a, b, c, the area of the triangle is:

A = sqrt(s*(s-a)*(s-b)*(s-c)),

where s = (a + b + c)/2.

Small correction:

For: h = sqrt[(b^2 - (1/2)*a^2)], I think you mean:

h = sqrt[(b^2 - (1/4)*a^2)]