ACT Math - Isosceles Triangles

Find the hypotenuse of an isosceles right triangle given side length of 3.

$3\sqrt2$

$9\sqrt2$

Explanation

To solve, simply use the Pythagorean Theorem.

Recall that an isosceles right triangle has two leg lengths that are equal.

Therefore, to solve for the hypotenuse let and in the Pythagorean Theorem.

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{2*3^2}=3\sqrt{2}$

Find the hypotenuse of an isosceles right triangle given side length of 3.

$3\sqrt2$

$9\sqrt2$

Explanation

To solve, simply use the Pythagorean Theorem.

Recall that an isosceles right triangle has two leg lengths that are equal.

Therefore, to solve for the hypotenuse let and in the Pythagorean Theorem.

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{2*3^2}=3\sqrt{2}$

Find the hypotenuse of an isosceles right triangle given side length of 3.

$3\sqrt2$

$9\sqrt2$

Explanation

To solve, simply use the Pythagorean Theorem.

Recall that an isosceles right triangle has two leg lengths that are equal.

Therefore, to solve for the hypotenuse let and in the Pythagorean Theorem.

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{2*3^2}=3\sqrt{2}$

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 $\small 12;cm$ .

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 $\small 12;cm$ .

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 $\small 12;cm$ .

The base angle of an isosceles triangle is thirteen more than three times the vertex angle. What is the difference between the vertex angle and the base angle?

Explanation

Every triangle has . An isosceles triangle has one vertex ange, and two congruent base angles.

Let be the vertex angle and be the base angle.

The equation to solve becomes , since the base angle occurs twice.

Now we can solve for the vertex angle.

The difference between the vertex angle and the base angle is .

The base angle of an isosceles triangle is thirteen more than three times the vertex angle. What is the difference between the vertex angle and the base angle?

Explanation

Every triangle has . An isosceles triangle has one vertex ange, and two congruent base angles.

Let be the vertex angle and be the base angle.

The equation to solve becomes , since the base angle occurs twice.

Now we can solve for the vertex angle.

The difference between the vertex angle and the base angle is .

The base angle of an isosceles triangle is thirteen more than three times the vertex angle. What is the difference between the vertex angle and the base angle?