Right Triangles - Geometry

Card 0 of 1412

Question

Figures and are triangles.

$\overline{AB}=\overline{A'B'}=\overline{BC}=\overline{B'C'}=1, \overline{AC}=\overline{A'C'}=\sqrt{2}$

Are $\Delta ABC$ and $\Delta A'B'C'$ congruent?

Answer

We know that congruent triangles have equal corresponding angles and equal corresponding sides. We are given that the corresponding sides are equal and are in the ratio of $1:1:\sqrt{2}$ . A triangle whose sides are in this ratio is a $45^{\circ}-45^{\circ}-90^{\circ}$ , where the shorter sides lies opposite the $45^{\circ}$ angles, and the longer side is the hypotenuse and lies opposite the right angle. So we know the corresponding angles are equal. Therefore, the triangles are congruent.

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Question

Figures and are triangles.

$\overline{AB}=\overline{A'B'}=\overline{BC}=\overline{B'C'}=\sqrt{2}, \overline{AC}=\overline{A'C'}=2$

Are $\Delta ABC$ and $\Delta A'B'C'$ congruent?

Answer

We know that congruent triangles have equal corresponding angles and equal corresponding sides. We are given that the corresponding sides are equal and are in the ratio of $\sqrt{2}:\sqrt{2}:2$ .

Simplify the ratio by dividing by $\sqrt{2}$

$\frac{\sqrt{2}}{\sqrt{2}}:\frac{\sqrt{2}}{\sqrt{2}}:\frac{2}{\sqrt{2}}$

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

$\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

Thus, the corresponding sides are in the ratio $1:1:\sqrt{2}$ and we know both triangles are $45^{\circ}-45^{\circ}-90^{\circ}$ triangles. Since the corresponding angles and the corresponding sides are equal, the triangles are congruent.

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Question

Are these right triangles congruent?

Answer

Right now we can't directly compare these triangles because we do not know all three side lengths. However, we can use Pythagorean Theorem to determine both missing sides. The left triangle is missing the hypotenuse:

$\small 48^2 + 55^2 = c^2$

$\small 2,304 + 3,025 = c^2$

$\small 5,329 = c$

$\small c = \sqrt{5,329}= 73$

The right triangle is missing one of the legs:

$\small x^2 + 48^2 = 78^2$

$\small x^2 + 2,304 = 5,329$ subtract 2,304 from both sides

$\small x^2 = 3,025$

$\small x = \sqrt{3,025} = 55$

This means that the two triangles both have side lengths 48, 55, 73, so they must be congruent.

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Question

The hypotenuse and acute angle are given for several triangles. Which if any are congruent? Triangle A- Hypotenuse=15; acute angle=56 degrees. Triangle B- Hypotenuse=18; acute angle=56 degrees. Triangle C-Hypotenuse=18; acute angle= 45 degrees.

Answer

The correct answer is none of these. There are several pairs of angles and sides or sides and angles that must be the same in order for two triangles to be congruent.

In our case, we need the acute angle and the hypotenuse to both be equal. No two triangles above have this relationship and therefore no two are congruent.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ .

$\angle B$ and $\angle E$ are both right angles.

$\overline{AB } \cong \overline{DE}$

$\overline{AC} \cong \overline{DF}$

True or false: From the above information, it follows that $\bigtriangleup ABC \cong \bigtriangleup DEF$ .

Answer

If we seek to prove that $\bigtriangleup ABC \cong \bigtriangleup DEF$ , then , , and correspond to , , and , respectively.

By the Hypotenuse-Leg Theorem (HL), if the hypotenuse and one leg of a triangle are congruent to those of another, the triangles are congruent.

$\angle B$ and $\angle E$ are both right angles, so $\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both right triangles. $\overline{AB }$ and $\overline{DE}$ are congruent corresponding sides, and moreover, since, each includes the right-angle vertex as an endpoint, they are congruent corresponding legs. $\overline{AC}$ and $\overline{DF}$ are opposite the right angles, making them congruent corresponding hypotenuses.

The conditions of HL are satisfied, so $\bigtriangleup ABC \cong \bigtriangleup DEF$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ .

$\angle B$ and $\angle E$ are both right angles.

$\overline{AB} \cong \overline{BC}$

$\overline{DE} \cong \overline{EF}$

True or false: From the given information, it follows that $\bigtriangleup ABC \cong \bigtriangleup DEF$ .

Answer

The congruence of $\bigtriangleup ABC$ and $\bigtriangleup DEF$ cannot be proved from the given information alone. Examine the two triangles below:

$\overline{AB} \cong \overline{BC}$ , $\overline{DE} \cong \overline{EF}$ , and $\angle B$ and $\angle E$ are both right angles, so the conditions of the problem are met; however, since the sides are not congruent between triangles - for example, $\overline{AB} \ncong \overline{DE}$ - the triangles are not congruent either.

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $\angle F$ is a right angle; $\overline{AB}=20$ ; $\overline{AC}=10$ ; $\overline{DE}=30$

Find $\overline{EF}$ .

Answer

Since $\bigtriangleup ABC \sim \bigtriangleup DEF$ and $\angle F$ is a right angle, $\angle C$ is also a right angle.

$\overline{AB}$ is the hypotenuse of the first triangle; since one of its legs $\overline{AC}$ is half the length of that hypotenuse, $\bigtriangleup ABC$ is 30-60-90 with $\overline{AC}$ the shorter leg and $\overline{BC}$ the longer.

Because the two are similar triangles, $\overline{DE}$ is the hypotenuse of the second triangle, and $\overline{EF}$ is its longer leg.

Therefore, $EF = \frac{\sqrt{3}}{2} \cdot DE = \frac{\sqrt{3}}{2} \cdot 30 = 15 \sqrt{3}$ .

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Question

Which of the following is sufficient to say that two right triangles are similar?

Answer

If all three angles of a triangle are congruent but the sides are not, then one of the triangles is a scaled up version of the other. When this happens the proportions between the sides still remains unchanged which is the criteria for similarity.

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Question

Which of the following statements is true regarding the two triangles?

Answer

Though we must do a little work, we can show these triangles are similar. First, right triangles are not necessarily always similar. They must meet the necessary criteria like any other triangles; furthermore, there is no Hypotenuse-Leg Theorem for similarity, only for congruence; therefore, we can eliminate two answer choices.

However, we can use the Pythagorean Theorem with the smaller triangle to find the missing leg. Doing so gives us a length of 48. Comparing the ratio of the shorter legs in each trangle to the ratio of the longer legs we get

$\frac{20}{10}=\frac{2}{1}$

$\frac{48}{24}=\frac{2}{1}$

In both cases, the leg of the larger triangle is twice as long as the corresponding leg in the smaller triangle. Given that the angle between the two legs is a right angle in each triangle, these angles are congruent. We now have enough evidence to conclude similarity by Side-Angle-Side.

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Question

Two triangles, $\Delta A$ and $\Delta B$ , are similar when:

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. Either condition alone is not sufficient. If two figures have both equal corresponding angles and equal corresponding lengths then they are congruent, not similar.

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Question

$\Delta ABC$ and $\Delta DEF$ are triangles.

$m\angle B=90^{\circ}$

$m\angle E=90^{\circ}$

Are $\Delta ABC$ and $\Delta DEF$ similar?

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. In other words, we need to know both the measures of the corresponding angles and the lengths of the corresponding sides. In this case, we know only the measures of $\angle B$ and $\angle E$ . We don't know the measures of any of the other angles or the lengths of any of the sides, so we cannot answer the question -- they might be similar, or they might not be.

It's not enough to know that both figures are right triangles, nor can we assume that angles are the same measurement because they appear to be.

Similar triangles do not have to be the same size.

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Question

$\Delta ABC$ and $\Delta DEF$ are similar triangles.

$m\angle A=30^{\circ}, m\angle B=90^{\circ}, m\angle C=60^{\circ}$

$\overline{AB}=\sqrt{3}, \overline{BC}=1, \overline{AC}=2$

$m\angle D=30^{\circ}, m\angle E=90^{\circ}, m\angle F=60^{\circ}$

$\overline{EF}=2$

What is the length of $\overline{DE}$ ?

Answer

Since $\Delta ABC$ and $\Delta DEF$ are similar triangles, we know that they have proportional corresponding lengths. We must determine which sides correspond. Here, we know $\overline{EF}$ corresponds to $\overline {BC}$ because both line segments lie opposite $30^{\circ}$ angles and between $90^{\circ}$ and $60^{\circ}$ angles. Likewise, we know $\overline{DE}$ corresponds to $\overline {AB}$ because both line segments lie opposite $60^{\circ}$ angles and between $30^{\circ}$ and $90^{\circ}$ angles. We can use this information to set up a proportion and solve for the length of $\overline{DE}$ .

$\frac{\overline{EF}}{\overline{BC}}=\frac{\overline{DE}}{\overline{AB}}$

Substitute the known values.

$\frac{2}{1}=\frac{\overline{DE}}{\sqrt{3}}$

Cross-multiply and simplify.

$2\sqrt{3}=\overline{DE}$

and result from setting up an incorrect proportion. $\sqrt{6}$ results from incorrectly multiplying and $\sqrt{3}$ .

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Question

Are these triangles similar? If so, list the scale factor.

Answer

The two triangles are similar, but we can't be sure of that until we can compare all three corresponding pairs of sides and make sure the ratios are the same. In order to do that, we first have to solve for the missing sides using the Pythagorean Theorem.

$\small 9^2 + 12^2 = c^2$

$\small 81 + 144 = c^2$

$\small 225 = c^2$

$\small \sqrt{225 } = 15 = c$

The smaller triangle is missing not the hypotenuse, c, but one of the legs, so we'll use the formula slightly differently.

$\small x^2 + 6^2 = 10^2$

$\small x^2 + 36 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x = \sqrt{64} = 8$

Now we can compare all three ratios of corresponding sides:

$\small \frac{6}{9 } =? \frac{8}{12} =? \frac{10}{15}$ one way of comparing these ratios is to simplify them.

We can simplify the leftmost ratio by dividing top and bottom by 3 and getting $\small \frac{2}{3}$ .

We can simplify the middle ratio by dividing top and bottom by 4 and getting $\small \frac{2}{3}$ .

Finally, we can simplify the ratio on the right by dividing top and bottom by 5 and getting $\small \frac{2}{3}$ .

This means that the triangles are definitely similar, and $\small \frac{2}{3}$ is the scale factor.

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Question

Are these right triangles similar? If so, state the scale factor.

Answer

In order to compare these triangles and determine if they are similar, we need to know all three side lengths in both triangles. To get the missing ones, we can use Pythagorean Theorem:

$\small 15^2 + 8^2 = c^2$

$\small 225 + 64 = c^2$

$\small 289 = c^2$ take the square root

$\small \sqrt{289} = 17 = c$

The other triangle is missing one of the legs rather than the hypotenuse, so we'll adjust accordingly:

$\small 6^2 + x^2 = 10^2$

$\small 36 + x^2 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x= 8$

Now we can compare ratios of corresponding sides:

$\small \frac{8}{6} =? \frac{15}{8 } =? \frac{17}{10}$

The first ratio simplifies to $\small \frac{4}{3}$ , but we can't simplify the others any more than they already are. The three ratios clearly do not match, so these are not similar triangles.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ .

$\angle B$ and $\angle E$ are both right angles.

$\overline{AB} \cong \overline{BC}$

$\overline{DE} \cong \overline{EF}$

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

If we seek to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond to , , and , respectively.

By the Side-Angle-Side Similarity Theorem (SASS), if two sides of a triangle are in proportion with the corresponding sides of another triangle, and the included angles are congruent, then the triangles are similar.

and , so by the Division Property of Equality, $\frac{AB}{DE} = \frac{BC}{EF}$ . Also, $\angle B$ and $\angle E$ , their respective included angles, are both right angles, so $\angle B \cong \angle E$ . The conditions of SASS are met, so $\bigtriangleup ABC \sim \bigtriangleup DEF$

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Question

The ratio for the side lengths of a right triangle is 3:4:5. If the perimeter is 48, what is the area of the triangle?

Answer

We can model the side lengths of the triangle as 3x, 4x, and 5x. We know that perimeter is 3x+4x+5x=48, which implies that x=4. This tells us that the legs of the right triangle are 3x=12 and 4x=16, therefore the area is A=1/2 bh=(1/2)(12)(16)=96.

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Question

A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?

Answer

The area of a triangle is denoted by the equation 1/2 b x h.

b stands for the length of the base, and h stands for the height.

Here we are told that the perimeter (total length of all three sides) is 12, and the hypotenuse (the side that is neither the height nor the base) is 5 units long.

So, 12-5 = 7 for the total perimeter of the base and height.

7 does not divide cleanly by two, but it does break down into 3 and 4,

and 1/2 (3x4) yields 6.

Another way to solve this would be if you recall your rules for right triangles, one of the very basic ones is the 3,4,5 triangle, which is exactly what we have here

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Question

The length of one leg of an equilateral triangle is 6. What is the area of the triangle?

Answer

$A=\frac{1}{2}(B)(H)$

The base is equal to 6.

The height of an quilateral triangle is equal to $\frac{B\sqrt{3}}{2}$ , where is the length of the base.

$A=\frac{1}{2}(6)(\frac{6\sqrt{3}}{2})=\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}$

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Question

Given:

A = 3 cm

B = 4 cm

What is the area of the right triangle ABC?

Answer

The area of a triangle is given by the equation:

$A_{\Delta } = (1/2)(base)(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$A_{\Delta }=(1/2)34 = 6 cm^{2}$

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Question

Given:

A = 4 cm

B = 6 cm

What is the area of the right triangle ABC?

Answer

The area of a triangle is given by the equation:

$A_{\Delta } = (1/2)(base)(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$A_{\Delta }=(1/2)46 = 12 cm^{2}$

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