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Intermediate Geometry : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

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8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : Acute / Obtuse Triangles

Find the area of the triangle. Round to the nearest tenths.

Possible Answers:

$35.4\:units^2$

$22.9\:units^2$

$42.5\:units^2$

$65.1\:units^2$

Correct answer:

$35.4\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(41^{\circ})=\frac{height}{9}$

Now, solve for the height.

$\text{Height}=9\sin(41^{\circ})=5.90$

Now you can find the area.

$\text{Area}=\frac{12\times 5.90}{2}=35.4\:units^2$

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Example Question #72 : Acute / Obtuse Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$24.8\:units^2$

$8.2\:units^2$

$14.7\:units^2$

$16.5\:units^2$

Correct answer:

$16.5\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(36^{\circ})=\frac{height}{4}$

Now, solve for the height.

$\text{Height}=4\sin(36^{\circ})=2.35$

Now you can find the area.

$\text{Area}=\frac{14\times 2.35}{2}=16.5\:units^2$

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Example Question #133 : Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$145.2\:units^2$

$97.9\:units^2$

$80.3\:units^2$

$79.8\:units^2$

Correct answer:

$80.3\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(42^{\circ})=\frac{height}{8}$

Now, solve for the height.

$\text{Height}=8\sin(42^{\circ})=5.35$

Now you can find the area.

$\text{Area}=\frac{30\times 5.35}{2}=80.3\:units^2$

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Example Question #73 : Acute / Obtuse Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$116.0\:units^2$

$111.4\:units^2$

$104.4\:units^2$

$92.8\:units^2$

Correct answer:

$116.0\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(48^{\circ})=\frac{height}{12}$

Now, solve for the height.

$\text{Height}=12\sin(48^{\circ})=8.92$

Now you can find the area.

$\text{Area}=\frac{26\times 8.92}{2}=116.0\:units^2$

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Example Question #131 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=6 \times 12$

$\text{Area of Rectangle}= 72$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{6}{2}$

$\text{height}=3$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 3$

$\text{Area of Triangle}=18$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=72-18$

Solve.

$\text{Area of Shaded Region}=54$

Report an Error

Example Question #131 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=8 \times 12$

$\text{Area of Rectangle}= 96$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{8}{2}$

$\text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 4$

$\text{Area of Triangle}=24$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=96-24$

Solve.

$\text{Area of Shaded Region}=72$

Report an Error

Example Question #17 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

168

154

132

170

Correct answer:

168

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=14 \times 16$

$\text{Area of Rectangle}= 224$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{14}{2}$

$\text{height}=7$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 16\times 7$

$\text{Area of Triangle}=56$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=224-56$

Solve.

$\text{Area of Shaded Region}=168$

Report an Error

Example Question #18 : How To Find The Area Of An Acute / Obtuse Triangle

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{5}{2}$

$\frac{7}{2}$

Correct answer:

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=1 \times 4$

$\text{Area of Rectangle}= 4$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{1}{2}$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 4\times \frac{1}{2}$

$\text{Area of Triangle}=1$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=4-1$

Solve.

$\text{Area of Shaded Region}=3$

Report an Error

Example Question #131 : Triangles

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{19}{2}$

$\frac{17}{2}$

$\frac{23}{2}$

$\frac{21}{2}$

Correct answer:

$\frac{21}{2}$

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=2 \times 7$

$\text{Area of Rectangle}= 14$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{2}{2}$

$\text{height}=1$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 7\times 1$

$\text{Area of Triangle}=\frac{7}{2}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=14-\frac{7}{2}$

Solve.

$\text{Area of Shaded Region}=\frac{21}{2}$

Report an Error

Example Question #581 : Plane Geometry

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Possible Answers:

$\frac{11}{2}$

$\frac{13}{2}$

$\frac{15}{2}$

$\frac{17}{2}$

Correct answer:

$\frac{15}{2}$

Explanation:

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=2 \times 5$

$\text{Area of Rectangle}= 10$

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

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

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{2}{2}$

$\text{height}=1$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 5\times 1$

$\text{Area of Triangle}=\frac{5}{2}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

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

$\text{Area of Shaded Region}=10-\frac{5}{2}$

Solve.

$\text{Area of Shaded Region}=\frac{15}{2}$

Report an Error

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Intermediate Geometry : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #71 : Acute / Obtuse Triangles

Example Question #72 : Acute / Obtuse Triangles

Example Question #133 : Triangles

Example Question #73 : Acute / Obtuse Triangles

Example Question #131 : Triangles

Example Question #131 : Triangles

Example Question #17 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #18 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #131 : Triangles

Example Question #581 : Plane Geometry

All Intermediate Geometry Resources

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