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Intermediate Geometry : Triangles

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Example Question #8 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

173.47

189.62

198.52

174.55

Correct answer:

173.47

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(18 \times 9)=81$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 9^2= 81\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=81\pi-81=173.47$

Make sure to round to places after the decimal.

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Example Question #9 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

355.21

401.25

372.90

361.93

Correct answer:

361.93

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(26 \times 13)=169$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 13^2= 169\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=169\pi-169=361.93$

Make sure to round to places after the decimal.

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1036.53

1121.77

980.27

1092.14

Correct answer:

1036.53

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(44 \times 22)=484$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 22^2= 484\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=484\pi-484=1036.53$

Make sure to round to places after the decimal.

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Example Question #21 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1145.52

1190.02

1098.98

1132.90

Correct answer:

1132.90

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(46 \times 23)=529$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 23^2= 529\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=529\pi-529=1132.90$

Make sure to round to places after the decimal.

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Example Question #22 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

498.20

402.08

481.86

441.12

Correct answer:

481.86

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(30 \times 15)=225$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 15^2= 225\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=225\pi-225=481.86$

Make sure to round to places after the decimal.

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Example Question #23 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1190.21

1181.21

1447.72

1098.21

Correct answer:

1447.72

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(52 \times 26)=676$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 26^2= 676\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=676\pi-676=1447.72$

Make sure to round to places after the decimal.

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Example Question #24 : Triangles

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

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

Possible Answers:

1.25

0.75

1.5

Correct answer:

1.5

Explanation:

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.

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

$\text{Area of Parallelogram}=2\times 1 = 2$

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.

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

$\text{Height of Triangle}=\frac{1}{2}$

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

$\text{Area of Triangle}=\frac{1}{2}(2\times \frac{1}{2})=\frac{1}{2}=0.5$

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

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

$\text{Area of Shaded Region}=2-0.5=1.5$

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Example Question #25 : Triangles

Parallel 2

Refer to the above diagram. m ||n .

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

Possible Answers:

False

True

Correct answer:

True

Explanation:

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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Example Question #26 : Triangles

Parallel 2

Refer to the above diagram. $\angle ABV \cong \angle DCV$ .

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

$\angle AVB \cong \angle CVD$

$\angle ABV \cong \angle CDV$

$\angle BAV \cong \angle DCV$

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

$\angle ABV \cong \angle DCV$ , but this is not one of the statements we need to prove. Also, without further information - for example, whether and are parallel, which is not given to us - we have no way to prove either of the other two necessary statements.

The correct response is "false".

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Example Question #27 : Triangles

$\bigtriangleup ABC$ 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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

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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Intermediate Geometry : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #8 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #9 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #1 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #21 : Triangles

Example Question #22 : Triangles

Example Question #23 : Triangles

Example Question #24 : Triangles

Example Question #25 : Triangles

Example Question #26 : Triangles

Example Question #27 : Triangles

All Intermediate Geometry Resources

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