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

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

Questions 1 - 2

1

What is the height of an isosceles triangle which has a base of and an area of ?

Explanation

The area of a triangle is given by the equation:

$\small A=\frac{1}{2}bh$

In this case, we are given the area ( $\small A$ ) and the base ( $\small b$ ) and are asked to solve for height ( $\small h$ ).

To do this, we must plug in the given values for $\small A$ and $\small b$ , which gives the following:

$\small 24;cm^2=\frac{1}{2}(4; cm)h$

We then must multiply the right side, and then divide the entire equation by 2, in order to solve for $\small h$ :

$\small 24;cm^2=2;cm(h)$

$\small \frac{24; cm^2}{2; cm}=\frac{2; cm(h)}{2; cm}$

$\small 12;cm=h$

$\small h=12;cm$

Therefore, the height of the triangle is $\small 12;cm$ .

2

An isosceles triangle has a base of $12\ cm$ and an area of $42\ cm^{2}$ . What must be the height of this triangle?

$7\ cm$

$6\ cm$

$8\ cm$

$9\ cm$

$10\ cm$

Explanation

$A=\frac{1}{2}bh$ .

$6x=42$

$x=7$

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