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Basic Geometry : Plane Geometry

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9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #38 : Triangles

If the width of the rectangle is half the length of the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

5.29

5.85

5.43

5.10

Correct answer:

5.43

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{5^2+5^2}$

$\text{Hypotenuse}=5\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{5\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=2\times\frac{5\sqrt2}{2}$

$\text{Area of Rectangle}=5\sqrt2$

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

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

Substitute in the given base and height to find the area.

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

$\text{Area of Triangle}=\frac{25}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{25}{2}-5\sqrt2$

Solve.

$\text{Area of Shaded Region}=5.43$

Report an Error

Example Question #39 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

9.57

9.51

9.92

9.22

Correct answer:

9.51

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{6^2+6^2}=6\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{6\sqrt2}{2}$

$\text{Width}=3\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=2\times3\sqrt2$

$\text{Area of Rectangle}=6\sqrt2$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{6\times 6}{2}$

$\text{Area of Triangle}=18$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=18-6\sqrt2$

Solve.

$\text{Area of Shaded Region}=9.51$

Report an Error

Example Question #40 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

9.29

9.34

9.18

9.65

Correct answer:

9.65

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{7^2+7^2}$

$\text{Hypotenuse}=7\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{7\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=3\times\frac{7\sqrt2}{2}$

$\text{Area of Rectangle}=\frac{21\sqrt2}{2}$

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

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

Substitute in the given base and height to find the area.

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

$\text{Area of Triangle}=\frac{49}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{49}{2}-\frac{21\sqrt2}{2}$

Solve.

$\text{Area of Shaded Region}=9.65$

Report an Error

Example Question #41 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

15.41

15.22

15.03

16.72

Correct answer:

15.03

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{8^2+8^2}$

$\text{Hypotenuse}=8\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{8\sqrt2}{2}$

$\text{Width}=4\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=3\times4\sqrt2$

$\text{Area of Rectangle}=12\sqrt2$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{8\times 8}{2}$

$\text{Area of Triangle}=32$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=32-12\sqrt2$

Solve.

$\text{Area of Shaded Region}=15.03$

Report an Error

Example Question #42 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

8.91

8.77

8.68

12.21

Correct answer:

8.68

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{9^2+9^2}$

$\text{Hypotenuse}=9\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{9\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=5\times\frac{9\sqrt2}{2}$

$\text{Area of Rectangle}=\frac{45\sqrt2}{2}$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{9\times 9}{2}$

$\text{Area of Triangle}=\frac{81}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{81}{2}-\frac{45\sqrt2}{2}$

Solve.

$\text{Area of Shaded Region}=8.68$

Report an Error

Example Question #43 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

7.89

7.57

8.12

7.99

Correct answer:

7.57

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{10^2+10^2}$

$\text{Hypotenuse}=10\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{10\sqrt2}{2}$

$\text{Width}=5\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=6\times5\sqrt2$

$\text{Area of Rectangle}=30\sqrt2$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{10 \times 10}{2}$

$\text{Area of Triangle}=50$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=50-30\sqrt2$

Solve.

$\text{Area of Shaded Region}=7.57$

Report an Error

Example Question #44 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

12.11

13.83

15.58

14.22

Correct answer:

13.83

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{11^2+11^2}$

$\text{Hypotenuse}=11\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{11\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=6\times\frac{11\sqrt2}{2}$

$\text{Area of Rectangle}=33\sqrt2$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{11 \times 11}{2}$

$\text{Area of Triangle}=\frac{121}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{121}{2}-33\sqrt2$

Solve.

$\text{Area of Shaded Region}=13.83$

Report an Error

Example Question #45 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

23.79

24.88

28.02

27.65

Correct answer:

27.65

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{15^2+15^2}$

$\text{Hypotenuse}=15\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{15\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=8\times\frac{15\sqrt2}{2}$

$\text{Area of Rectangle}=60\sqrt2$

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

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

Substitute in the given base and height to find the area.

$\text{Area of Triangle}=\frac{15 \times 15}{2}$

$\text{Area of Triangle}=\frac{225}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{225}{2}-60\sqrt2$

Solve.

$\text{Area of Shaded Region}=27.65$

Report an Error

Example Question #46 : Triangles

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

59.24

57.11

56.69

58.40

Correct answer:

58.40

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

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

Now, in order to find the length of the rectangle, we will first need to find the length of the hypotenuse of the triangle. The question provides us with the following relationship between the lengths of the rectangle and hypotenuse:

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

Recall how to find the hypotenuse of a right triangle:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

Simplify.

$\text{Hypotenuse}^2=2(\text{side})^2$

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=14\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{14\sqrt2}{2}$

$\text{length}=7\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=7\sqrt2 \times 4$

$\text{Area of Rectangle}=28\sqrt2$

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

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

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{14 \times 14}{2}$

$\text{Area of Triangle}=98$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=98-28\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=58.40$

Report an Error

Example Question #1021 : Basic Geometry

A rectangle is inscribed in a right isosceles triangle. The length of the rectangle is half the hypotenuse of the triangle and is shown in the provided figure. What is the area of the shaded region?

Possible Answers:

50.50

49.12

44.10

48.50

Correct answer:

48.50

Explanation:

In order to find the area of the shaded region, you will need to first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle.

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

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

Recall how to find the hypotenuse of a right triangle:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

Simplify.

$\text{Hypotenuse}^2=2(\text{side})^2$

Find the square root of each side.

$\text{Hypotenuse}=\sqrt{2(\text{side})^2}$

$\text{Hypotenuse}=\text{side}\sqrt2$

Substitute in the value of the side of the triangle to find the length of the hypotenuse.

$\text{Hypotenuse}=14\sqrt2$

Now, find the length of the rectangle:

$\text{length}=\frac{14\sqrt2}{2}$

$\text{length}=7\sqrt2$

Next, find the area of the rectangle:

$\text{Area of Rectangle}=7\sqrt2 \times 5$

$\text{Area of Rectangle}=35\sqrt2$

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

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

Since we are given a right isosceles triangle, the base and the height are the same length. Substitute in the given side length to find the area of the triangle.

$\text{Area of Triangle}=\frac{14 \times 14}{2}$

$\text{Area of Triangle}=98$

Now that we have the areas of the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=98-35\sqrt2$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=48.50$

Report an Error

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Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #38 : Triangles

Example Question #39 : Triangles

Example Question #40 : Triangles

Example Question #41 : Triangles

Example Question #42 : Triangles

Example Question #43 : Triangles

Example Question #44 : Triangles

Example Question #45 : Triangles

Example Question #46 : Triangles

Example Question #1021 : Basic Geometry

All Basic Geometry Resources

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