How to find the area of an acute / obtuse isosceles triangle

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ACT Math › How to find the area of an acute / obtuse isosceles triangle

Questions 1 - 3

1

An isosceles triangle has a base length of and a height that is twice its base length. What is the area of this triangle?

Explanation

1. Find the height of the triangle:

$height=2\cdot base$

2. Use the formula for area of a triangle:

$A=\frac{1}{2}(b)(h)$

$A=\frac{1}{2}(10)(20)=100$

2

What is the area of an isosceles triangle with a vertex of degrees and two sides equal to ?

$4\sqrt{3}:cm^2$

$20\sqrt{3}:cm^2$

$12\sqrt{3}:cm^2$

$15\sqrt{2}:cm^2$

Explanation

Based on the description of your triangle, you can draw the following figure:

_tri71

You can do this because you know:

The two equivalent sides are given.
Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.

Now, based on the properties of an isosceles triangle, you can draw the following as well:

_tri72

Based on your standard reference triangle, you know:

$\frac{h}{4}=\frac{1}{2}$

Therefore, is .

This means that is $2\sqrt{3}$ and the total base of the triangle is $4\sqrt{3}$ .

Now, the area of the triangle is:

$\frac{1}{2}bh$ or $\frac{1}{2}*2*4\sqrt{3}=4\sqrt{3}:cm^2$

3

An isosceles triangle has a height of and a base of . What is its area?

Explanation

Use the formula for area of a triangle:

$A=\frac{1}{2}(base)(height)$

$A=\frac{1}{2}(4)(7)=14$

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