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High School Math : Rectangles

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Example Question #1 : Rectangles

The front façade of a building is 100 feet tall and 40 feet wide. There are eight floors in the building, and each floor has four glass windows that are 8 feet wide and 6 feet tall along the front façade. What is the total area of the glass in the façade?

Possible Answers:

768 ft²

1536 ft²

192 ft²

1536 ft²

2464 ft²

Correct answer:

1536 ft²

Explanation:

Glass Area per Window = 8 ft x 6 ft = 48 ft²

Total Number of Windows = Windows per Floor * Number of Floors = 4 * 8 = 32 windows

Total Area of Glass = Area per Window * Total Number of Windows = 48 * 32 = 1536 ft²

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Example Question #71 : Quadrilaterals

Mark is making a plan to build a rectangular garden. He has 160 feet of fence to form the outside border of the garden. He wants the dimensions to look like the plan outlined below:

Screen_shot_2013-03-19_at_9.17.30_pm

What is the area of the garden, rounded to the nearest square foot?

Possible Answers:

160

486

126

1472

Correct answer:

1472

Explanation:

Perimeter: Sum of the sides:

4x + 4x + 2x+8 +2x+8 = 160

12x + 6 = 160

12x = 154

x = $\frac{77}{6}$

Therefore, the short side of the rectangle is going to be:

$\dpi{100} \dpi{100} 2\cdot \frac{77}{6}+3=\frac{172}{6}=\frac{86}{3}$

And the long side is going to be:

$\dpi{100} 4\cdot \frac{77}{6}=\frac{308}{6}=\frac{154}{3}$

The area of the rectangle is going to be as follows:

Area = lw

$\dpi{100} Area = \frac{86}{3}\cdot \frac{154}{3}=\frac{13,244}{9}=1471.55\approx 1472\ ft^{2}$

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Example Question #331 : Plane Geometry

Two circles of a radius of {2}'' each sit inside a square with a side length of {8}'' . If the circles do not overlap, what is the area outside of the circles, but within the square?

Possible Answers:

$16+8\pi$

$36-6\pi$

$36-4\pi$

$36-8\pi$

$64-8\pi$

Correct answer:

$64-8\pi$

Explanation:

The area of a square = $\dpi{100} \small side^{2}$

The area of a circle is $\dpi{100} \small \pi r^{2}$

Area = Area of Square $\dpi{100} \small -$ 2(Area of Circle) =

$8^{2}-2\cdot \pi (2)^{2}$

$64-8\pi$

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Example Question #24 : Quadrilaterals

If the area Rectangle A is $25\%$ larger than Rectangle B and the sides of Rectangle A are and , what is the area of Rectangle B?

Possible Answers:

Correct answer:

Explanation:

$A = \text{length}\: \times \:\text{width} = 10 \times 6 = 60$

A = 1.25 B

60 = 1.25B

$B =\frac{60}{1.25} = 48$

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Example Question #122 : Quadrilaterals

Erin is getting ready to plant her tulip garden. She wants to plant two tulips per square foot of garden. If her rectangular garden is enclosed by 24 feet of fencing, and the length of the fence is twice as long as its width, how many tulips will Erin plant?

Possible Answers:

Correct answer:

Explanation:

We know that the following represents the formula for the perimeter of a rectangle:

P = 2l + 2w

In this particular case, we are told that the length of the fence is twice as long as the width. We can write this as the following expression:

l = 2w

Use this information to substitute in a variable for the length that matches the variable for width in our perimeter equation.

P = 2(2w) + 2w

24=6w

w=4

We also know that the length is two times the width; therefore, we can write the following:

l = 2w

$l = 2\times4$

l =8

The area of a rectangle is found by using this formula:

A = lw

$A=8\times4$

A=32

The area of the garden is 32 square feet. Erin will plant two tulips per square foot; thus, she will plant 64 tulips.

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High School Math : Rectangles

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Rectangles

Example Question #71 : Quadrilaterals

Example Question #331 : Plane Geometry

Example Question #24 : Quadrilaterals

Example Question #122 : Quadrilaterals

All High School Math Resources

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