Triangles - Geometry

Card 0 of 1188

Question

Refer to the above diagram. .

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

Answer

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar.

$\angle AVB$ and $\angle CVD$ are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, $\angle AVB \cong \angle CVD$ .

$\angle ABV$ and $\angle CDV$ are alternating interior angles formed by two parallel lines and cut by a transversal . As a consequence, $\angle ABV \cong \angle CDV$ .

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

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Question

Two of the exterior angles of a triangle, taken at different vertices, measure $100^{\circ }$ . Is the triangle acute, right, or obtuse?

Answer

At a given vertex, an exterior angle and an interior angle of a triangle form a linear pair, making them supplementary - that is, their measures total $180^{\circ }$ . The measures of two interior angles can be calculated by subtracting the exterior angle measures from $180^{\circ }$ :

$180^{\circ } - 100 ^{\circ } = 80 ^{\circ }$

The triangle has two interior angles of measure $80 ^{\circ }$ .

The measures of the interior angles of a triangle add up to $180^{\circ }$ . If is the measure of the third angle, then

$t + 80^{\circ } + 80^{\circ } = 180^{\circ }$

$t +160^{\circ } = 180^{\circ }$

$t +160^{\circ }- 160 ^{\circ } = 180^{\circ } - 160 ^{\circ }$

$t = 20 ^{\circ }$

All three interior angles measure less than $90^{\circ }$ , making them acute. The triangle is, by definition, acute.

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Question

Triangle one and triangle two are similar triangles. Triangle one has two sides with lengths mm and mm. What are possible measurements for the corresponding sides in triangle two?

Answer

Since the two triangles are similar, each triangles three corresponding sides must have the same ratio.

The ratio of triangle one is:

If we look at the possible solutions we will see that ratio that is in triangle one is also seen in the triangle with side lengths as follows:

$23:35=(23\times 4):(35\times 4)=92:140$

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Question

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-fifth that of the parallelogram, find the area of the shaded region.

Answer

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=30 \times 25 = 750$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{5}$

$\text{Height of Triangle}=\frac{25}{5}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(30 \times 5)=75$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=750-75=675$

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Question

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Answer

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 30 =1020$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{30}{2}=15$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(34 \times 15)=255$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=1020-255=765$

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Question

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-ninth that of the parallelogram, find the area of the shaded region.

Answer

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=50 \times45 = 2250$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{9}$

$\text{Height of Triangle}=\frac{45}{9}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(50 \times 5)=125$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=2250-125=2125$

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Question

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Answer

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=52\times 44 = 2288$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{44}{2}=22$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(52\times 22)=572$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=2288-572=1716$

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Question

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Answer

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 20 = 680$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{20}{2}=10$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(340 \times 10)=170$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=680-170=510$

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Question

$\bigtriangleup ABC$ and $\bigtriangleup DEF$ are both isosceles triangles;

$m \angle B = 120 ^{\circ }$

$m \angle F = 30 ^{\circ }$

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

Answer

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

$\bigtriangleup DEF$ is an isosceles triangle, so it must have two congruent angles. $\angle F$ has measure $30^{\circ }$ , so either $\angle D$ has this measure, $\angle E$ has this measure, or $\angle D \cong \angle E$ . If we examine the second case, it immediately follows that $\angle B \ncong \angle E$ . One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that $\bigtriangleup ABC \sim \bigtriangleup DEF$ . This makes the correct response "false".

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Question

An isoceles triangle has a vertex angle that is twenty more than twice the base angle. What is the difference between the vertex and base angles?

Answer

A triangle has degrees. An isoceles triangle has one vertex angle and two congruent base angles.

Let = the base angle and $2x\ +\ 20$ = vertex angle

So the equation to solve becomes $x\ +\ x\ +\ 2x\ +\ 20 = 180$

or

$4x\ +\ 20=180$

so the base angle is and the vertex angle is and the difference is .

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Question

An ssosceles triangle has interior angles of degrees and degrees. Find the missing angle.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees.

Thus, the solution is:

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Question

The largest angle in an obtuse isosceles triangle is degrees. Find the measurement of one of the two equivalent interior angles.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles.

Thus, the solution is:

$82\div2=41$

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Question

The two equivalent interior angles of an obtuse isosceles triangle each have a measurement of degrees. Find the measurement of the obtuse angle.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees.

Thus, the solution is:

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Question

In an obtuse isosceles triangle the angle measurements are, $x^\circ$ , $x^\circ$ , and $(10x-2)=128^\circ$ . Find the measurement of one of the acute angles.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles.

The solution is:

However, degrees is the measurement of both of the acute angles combined.

Each individual angle is $26\div2=13$ .

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Question

In an acute isosceles triangle the two equivalent interior angles each have a measurement of degrees. Find the missing angle.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an acute isosceles triangle, all of the interior angles must be acute angles.

The solution is:

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Question

In an acute isosceles triangle the two equivalent interior angles are each degrees. Find the missing angle.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an acute isosceles triangle, all of the interior angles must be acute angles.

The solution is:

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Question

The largest angle in an obtuse isosceles triangle is degrees. Find the measurement of one of the equivalent interior angles.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an obtuse Isosceles triangle, the two missing angles must be acute angles.

Thus, the solution is:

$42\div2=21$

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Question

In an obtuse isosceles triangle the largest angle is degrees. Find the measurement of one of the acute angles.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles.

The solution is:

$62\div2=31$

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Question

In an acute isosceles triangle the measurement of the non-equivalent interior angle is degrees. Find the measurement of one of the equivalent interior angles.

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an acute isosceles triangle, all of the interior angles must be acute angles.

The solution is:

$94\div2=47$

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Question

In an obtuse isosceles triangle, the largest interior angle is degrees. What is the measurement of one of the equivalent interior angles?

Answer

Isosceles triangles always have two equivalent interior angles, and all three interior angles of any triangle always have a sum of degrees. Since this is an obtuse isosceles triangle, the two missing angles must be acute angles.

The solution is:

$45\div2=22.5$

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