Isosceles Triangles - ACT Math

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Question

Triangle A and Triangle B are similar isosceles triangles. Triangle A's sides measure , , and . Two of the angles in Triangle A each measure $\small 83^o$ . Triangle B's sides measure , , and . What is the measure of the smallest angle in Triangle B?

Answer

Because the interior angles of a triangle add up to $\small 180^o$ , and two of Triangle A's interior angles measure $\small 83^o$ , we must simply add the two given angles and subtract from $\small 180^o$ to find the missing angle:

$\small 83^o+83^o=166^o$

$\small 180^o-166^o=14^o$

Therefore, the missing angle (and the smallest) from Triangle A measures $\small 14^o$ . If the two triangles are similar, their interior angles must be congruent, meaning that the smallest angle is Triangle B is also $\small 14^o$ .

The side measurements presented in the question are not needed to find the answer!

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Question

Triangle A and Triangle B are similar isosceles triangles. Triangle A has a base of and a height of . Triangle B has a base of . What is the length of Triangle B's two congruent sides?

Answer

We must first find the length of the congruent sides in Triangle A. We do this by setting up a right triangle with the base and the height, and using the Pythagorean Theorem to solve for the missing side ( $\small c$ ). Because the height line cuts the base in half, however, we must use for the length of the base's side in the equation instead of . This is illustrated in the figure below:

Using the base of $\small 3cm$ and the height of $\small 4cm$ , we use the Pythagorean Theorem to solve for $\small c$ :

$\small 3^2+4^2=c^2$

$\small 9+16=c^2$

$\small 25=c^2$

$\small c=5$

Therefore, the two congruent sides of Triangle A measure $\small 5cm$ ; however, the question asks for the two congruent sides of Triangle B. In similar triangles, the ratio of the corresponding sides must be equal. We know that the base of Triangle A is and the base of Triangle B is . We then set up a cross-multiplication using the ratio of the two bases and the ratio of $\small c$ to the side we're trying to find ( $\small x$ ), as follows:

$\small \frac{6}{12}=\frac{5}{x}$

$\small 6x=60$

$\small x=10$

Therefore, the length of the congruent sides of Triangle B is .

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Question

Isosceles triangles $\Delta ABC$ and $\Delta ABD$ share common side . $\Delta ABC$ is an obtuse triangle with sides . $\Delta ABD$ is also an obtuse isosceles triangle, where . What is the measure of $\angle AD$ ?

Answer

In order to prove triangle congruence, the triangles must have SAS, SSS, AAS, or ASA congruence. Here, we have one common side (S), and no other demonstrated congruence. Hence, we cannot guarantee that side is not one of the two congruent sides of $\Delta ABD$ , so we cannot state congruence with $\Delta ABC$ .

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Question

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

Answer

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

$6x=42$

$x=7$

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Question

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

Answer

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 $\small 12;cm$ .

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Question

A 44/45/90 triangle has a hypotenuse of $1\sqrt2$ . Find the length of one of its legs.

Answer

It's helpful to remember upon coming across a 45/45/90 triangle that it's a special right triangle. This means that its sides can easily be calculated by using a derived side ratio:

$s:s:s\sqrt2$

Here, represents the length of one of the legs of the 45/45/90 triangle, and $s\sqrt2$ represents the length of the triangle's hypotenuse. Two sides are denoted as congruent lengths () because this special triangle is actually an isosceles triangle. This goes back to the fact that two of its angles are congruent.

Therefore, using the side rules mentioned above, if $1\sqrt2=s\sqrt2$ , this problem can be resolved by solving for the value of :

$1\sqrt{2}=s\sqrt{2}$

$\frac{1\sqrt{2}}{\sqrt{2}}=s$

Therefore, the length of one of the legs is 1.

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Question

In a 45-45-90 triangle, if the hypothenuse is long, what is a possible side length?

Answer

If the hypotenuse of a 45-45-90 triangle is provided, its side length can only be one length, since the sides of all 45-45-90 triangles exist in a defined ratio of $1:1:\sqrt2$ , where represents the length of one of the triangle's legs and $\sqrt2$ represents the length of the triangle's hypotenuse. Using this method, you can set up a proportion and solve for the length of one of the triangle's sides:

$\frac{1}{\sqrt2}=\frac{x}{5:cm}$

Cross-multiply and solve for .

$\sqrt2\cdot x=5:cm$

$x=\frac{5}{\sqrt2}:cm$

Rationalize the denominator.

$\frac{5}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{5\sqrt2}{2}:cm$

You can also solve this problem using the Pythagorean Theorem.

In a 45-45-90 triangle, the side legs will be equal, so . Substitute for and rewrite the formula.

Substitute the provided length of the hypothenuse and solve for .

$a^2 = \frac{25}{2}$

$a=\sqrt{\frac{25}{2}}:cm$

While the answer looks a little different from the result of our first method of solving this problem, the two represent the same value, just written in different ways.

$\sqrt{}\frac{25}{2}=\frac{\sqrt{5\cdot 5}}{\sqrt2}=\frac{5}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{5\sqrt2}{2}$

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Question

In a triangle, if the length of the hypotenuse is $5\sqrt{2}$ , what is the perimeter?

Answer

1. Remember that this is a special right triangle where the ratio of the sides is:

$x:x:x\sqrt{2}$

In this case that makes it:

$5:5:5\sqrt{2}$

2. Find the perimeter by adding the side lengths together:

$5+5+5\sqrt{2}=10+5\sqrt{2}=17.07$

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Question

The height of a triangle is . What is the length of the hypotenuse?

Answer

Remember that this is a special right triangle where the ratio of the sides is:

$x:x:x\sqrt{2}$

In this case that makes it:

$3:3:3\sqrt{2}$

Where $3\sqrt{2}$ is the length of the hypotenuse.

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Question

What is the area of an isosceles right triangle with a hypotenuse of ?

Answer

Now, this is really your standard triangle. Since it is a right triangle, you know that you have at least one -degree angle. The other two angles must each be degrees, because the triangle is isosceles.

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

This is derived from your reference triangle for the triangle:

For our triangle, we could call one of the legs . We know, then:

$\frac{x}{26}=\frac{1}{\sqrt{2}}$

Thus, $x=\frac{26}{\sqrt{2}}$ .

The area of your triangle is:

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

For your data, this is:

$\frac{1}{2}\frac{26}{\sqrt{2}}\frac{26}{\sqrt{2}}=\frac{26^2}{4}=169:cm^2$

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Question

What is the area of an isosceles right triangle with a hypotenuse of $9\sqrt{2}:cm$ ?

Answer

Now, this is really your standard triangle. Since it is a right triangle, you know that you have at least one -degree angle. The other two angles must each be degrees because the triangle is isosceles.

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

This is derived from your reference triangle for the triangle:

For our triangle, we could call one of the legs . We know, then:

$\frac{x}{1}=\frac{9\sqrt{2}}{\sqrt{2}}$

Thus, .

The area of your triangle is:

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

For your data, this is:

$\frac{1}{2}99=40.5:cm^2$

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Question

What is the area of an isosceles right triangle with a hypotenuse of $7\sqrt{2}:cm$ ?

Answer

Now, this is really your standard triangle. Since it is a right triangle, you know that you have at least one -degree angle. The other two angles must each be degrees because the triangle is isosceles.

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

This is derived from your reference triangle for the triangle:

For our triangle, we could call one of the legs . We know, then:

$\frac{x}{1}=\frac{7\sqrt{2}}{\sqrt{2}}$

Thus, .

The area of your triangle is:

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

For your data, this is:

$\frac{1}{2}77=24.5:cm^2$

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Question

$\Delta PQR$ is a right isosceles triangle with hypotenuse . What is the area of $\Delta PQR$ ?

Answer

Right isosceles triangles (also called "45-45-90 right triangles") are special shapes. In a plane, they are exactly half of a square, and their sides can therefore be expressed as a ratio equal to the sides of a square and the square's diagonal:

$x: x: x\sqrt{2}$ , where $x\sqrt{2}$ is the hypotenuse.

In this case, maps to $x\sqrt{2}$ , so to find the length of a side (so we can use the triangle area formula), just divide the hypotenuse by $\sqrt2$ :

$\frac{46}{\sqrt2} = \frac{46}{\sqrt2} \cdot (\frac{\sqrt2}{\sqrt2}) = \frac{46\sqrt{2}}{2} = 23\sqrt{2}$

So, each side of the triangle is $23\sqrt2$ long. Now, just follow your formula for area of a triangle:

$A\Delta = \frac{lh}{2} = \frac{(23\sqrt{2})(23\sqrt{2})}{2} = \frac{1058}{2} = 529$

Thus, the triangle has an area of $529\textup{ units}^{2}$ .

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Question

Square has a side length of . What is the length of its diagonal?

Answer

The answer can be found two different ways. The first step is to realize that this is really a triangle question, even though it starts with a square. By drawing the square out and adding the diagonal, you can see that you form two right triangles. Furthermore, the diagonal bisects two ninety-degree angles, thereby making the resulting triangles a triangle.

From here you can go one of two ways: using the Pythagorean Theorem to find the diagonal, or recognizing the triangle as a triangle.

Using the Pythagorean Theorem

Once you recognize the right triangle in this question, you can begin to use the Pythagorean Theorem. Remember the formula: $a^{2}+b^{2}=c^{2}}$ , where and are the lengths of the legs of the triangle, and is the length of the triangle's hypotenuse.

In this case, . We can substitute these values into the equation and then solve for , the hypotenuse of the triangle and the diagonal of the square:

$3^{2}+3^{2}=c^{2}}$

$c=\sqrt{18}$

$c=\sqrt{92}$

$c=\sqrt{9}{\sqrt{2}}$

**$c=3*\sqrt{2}$

$c=3\sqrt2$**

The length of the diagonal is $3\sqrt2$ .

Using Properties of Triangles

The second approach relies on recognizing a triangle. Although one could solve this rather easily with Pythagorean Theorem, the following method could be faster.

triangles have side length ratios of $x:x:x\sqrt2$ , where represents the side lengths of the triangle's legs and $x\sqrt2$ represents the length of the hypotenuse.

In this case, because it is the side length of our square and the triangles formed by the square's diagonal.

Therefore, using the triangle ratios, we have $3\sqrt{2}$ for the hypotenuse of our triangle, which is also the diagonal of our square.

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Question

What is the length of the hypotenuse of an isosceles right triangle with an area of ?

Answer

Recall that an isosceles right triangle is also a triangle. It has sides that appear as follows:

Therefore, the area of the triangle is:

$A=\frac{1}{2}x^2$ , since the base and the height are the same.

For our data, this means:

$10.125=\frac{1}{2}x^2$

Solving for , you get:

So, your triangle looks like this:

Now, you can solve this with a ratio and easily find that it is $4\sqrt{2}$ . You also can use the Pythagorean Theorem. To do the latter, it is:

$h = \sqrt{4.5^2+4.5^2}$

Now, just do your math carefully:

$h = \sqrt{4.5^2(1+1)}$

That is a weird kind of factoring, but it makes sense if you distribute back into the group. This means you can simplify:

$h = \sqrt{4.5^2(2)} = \sqrt{4.5^2}\sqrt{2}=4.5\sqrt{2}:in$

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Question

When the sun shines on a pole, it leaves a shadow on the ground that is also . What is the distance from the top of the pole to the end of its shadow?

Answer

The pole and its shadow make a right angle. Because they are the same length, they form an isosceles right triangle (45/45/90). We can use the Pythagorean Theorem to find the hypotenuse. . In this case, . Therefore, we do . So $c=\sqrt{32}\approx5.66$ .

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Question

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

Answer

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,

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

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Question

There are two obtuse triangles. The obtuse angle of triangle one is $101^{\circ}$ . The sum of two angles in the second triangle is $80^{\circ}$ . When are these two triangles congruent?

Answer

In order for two obtuse triangles to be congruent, the sum of the two smaller angles must equal the sum of the two smaller angles of the second triangle. That is, excluding the obtuse angle.

The first triangle has an obtuse angle of $101^{\circ}$ . That means the sum of the other two angles is $79^{\circ}$ . The sum of the corresponding angles in triangle 2 is $80^{\circ}$ . Therefore, because $79^{\circ}$ is not equal to $80^{\circ}$ , the two obtuse triangles cannot be congruent.

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Question

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

Answer

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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Question

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

Answer

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

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:

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 $\frac{1}{2}24\sqrt{3}=4\sqrt{3}:cm^2$

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