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Intermediate Geometry : Intermediate Geometry

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 #29 : How To Find The Area Of A Parallelogram

In the figure, the height of the parallelogram is half the height of the equilateral triangle. Find the area of the shaded region.

Possible Answers:

82.00

90.07

95.26

85.63

Correct answer:

90.07

Explanation:

First, we will need to find the height of the equilateral triangle.

Recall that the height of an equilateral triangle will cut the equilateral triangle into two congruent $30^\circ-60^\circ-90^\circ$ triangles that have side lengths in the the following ratio:

$1:\sqrt3:2$

Use the given side length of the equilateral triangle in order to find the length of the height.

$\frac{\text{side length}}{2}=\frac{\text{height}}{\sqrt3}$

$\text{height}=(\sqrt3)\times \frac{\text{side length}}{2}$

$\text{height}=\frac{26(\sqrt3)}{2}$

$\text{height}=13\sqrt3$

Now, find the area of the equilateral triangle.

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

$\text{Area of Equilateral Triangle}=\frac{1}{2}(26\times13\sqrt3)$

$\text{Area of Equilateral Triangle}=169\sqrt3$

Now, use the height of the equilateral triangle to find the height of the parallelogram.

$\text{height of parallelogram}=\frac{\text{height of triangle}}{2}$

$\text{height of parallelogram}=\frac{13\sqrt3}{2}$

Next, find the area of the parallelogram.

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

$\text{Area of Parallelogram}=18\times \frac{13\sqrt3}{2}$

$\text{Area of Parallelogram}=117\sqrt3$

Now, from the figure, we can see that we will need to subtract the area of the parallelogram from the area of the triangle in order to find the area of the shaded region.

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

$\text{Area of Shaded Region}=169\sqrt3-117\sqrt3$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=90.07$

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Example Question #30 : How To Find The Area Of A Parallelogram

In the figure, the height of the parallelogram is half the height of the equilateral triangle. Find the area of the shaded region.

Possible Answers:

174.06

162.39

173.21

181.29

Correct answer:

173.21

Explanation:

First, we will need to find the height of the equilateral triangle.

Recall that the height of an equilateral triangle will cut the equilateral triangle into two congruent $30^\circ-60^\circ-90^\circ$ triangles that have side lengths in the the following ratio:

$1:\sqrt3:2$

Use the given side length of the equilateral triangle in order to find the length of the height.

$\frac{\text{side length}}{2}=\frac{\text{height}}{\sqrt3}$

$\text{height}=(\sqrt3)\times \frac{\text{side length}}{2}$

$\text{height}=\frac{40(\sqrt3)}{2}$

$\text{height}=20\sqrt3$

Now, find the area of the equilateral triangle.

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

$\text{Area of Equilateral Triangle}=\frac{1}{2}(40\times20\sqrt3)$

$\text{Area of Equilateral Triangle}=400\sqrt3$

Now, use the height of the equilateral triangle to find the height of the parallelogram.

$\text{height of parallelogram}=\frac{\text{height of triangle}}{2}$

$\text{height of parallelogram}=\frac{20\sqrt3}{2}$

$\text{height of parallelogram}=10\sqrt3$

Next, find the area of the parallelogram.

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

$\text{Area of Parallelogram}=30\times 10\sqrt3$

$\text{Area of Parallelogram}=300\sqrt3$

Now, from the figure, we can see that we will need to subtract the area of the parallelogram from the area of the triangle in order to find the area of the shaded region.

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

$\text{Area of Shaded Region}=400\sqrt3-300\sqrt3$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=173.21$

Report an Error

Example Question #71 : Parallelograms

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Cannot be determined

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+1)=6

Expand this equation.

x^2+x=6

x^2+x-6=0

Now factor this equation.

(x-2)(x+3)=6

Solve for .

x=2, x=-3

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

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Example Question #32 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 100 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-48)=100

Expand this equation.

x^2-48x=100

x^2-48x-100=0

Now factor this equation.

(x-50)(x+2)=0

Solve for .

x=50, x=-2

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

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Example Question #31 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 120 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-14)=120

Expand this equation.

x^2-14x=120

x^2-14x-120=0

Now factor this equation.

(x-20)(x+6)=0

Solve for .

x=20, x=-6

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

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Example Question #31 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 240 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+34)=240

Expand this equation.

x^2+34x=240

x^2+34x-240=0

Now factor this equation.

(x-6)(x+40)=0

Solve for .

x=6, x=-40

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #411 : Intermediate Geometry

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+11)=80

Expand this equation.

x^2+11x=80

x^2+11x-80=0

Now factor this equation.

(x-5)(x+16)=0

Solve for .

x=5, x=-16

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #412 : Intermediate Geometry

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+13)=90

Expand this equation.

x^2+13x=90

x^2+13x-90=0

Now factor this equation.

(x-5)(x+18)=0

Solve for .

x=5, x=-18

Since the values of lengths of shapes can only be positive, we know that the base of the parallelogram must be .

Report an Error

Example Question #71 : Parallelograms

In the figure, the area of the parallelogram is . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-5)=84

Expand this equation.

x^2-5x=84

x^2-5x-84=0

Now factor this equation.

(x-12)(x+7)=0

Solve for .

x=12, x=-7

Since lengths of bases and heights can only be positive, x=12 .

Notice that the length of the base is given by the expression x-5 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x-5$

$\text{Length of Base}=12-5$

$\text{Length of Base}=7$

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Example Question #81 : Parallelograms

In the figure, the area of the parallelogram is 150 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+5)=150

Expand this equation.

x^2+5x=150

x^2+5x-150=0

Now factor this equation.

(x-10)(x+15)=0

Solve for .

x=10, x=-5

Since lengths of bases and heights can only be positive, x=10 .

Notice that the length of the base is given by the expression x+5 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+5$

$\text{Length of Base}=10+5$

$\text{Length of Base}=15$

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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #29 : How To Find The Area Of A Parallelogram

Example Question #30 : How To Find The Area Of A Parallelogram

Example Question #71 : Parallelograms

Example Question #32 : How To Find The Area Of A Parallelogram

Example Question #31 : How To Find The Area Of A Parallelogram

Example Question #31 : How To Find The Area Of A Parallelogram

Example Question #411 : Intermediate Geometry

Example Question #412 : Intermediate Geometry

Example Question #71 : Parallelograms

Example Question #81 : Parallelograms

All Intermediate Geometry Resources

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