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ISEE Upper Level Math : Rectangles

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6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Rectangles

A rectangle on the coordinate plane has its vertices at the points $\displaystyle (3,4), (-6,4), (-6,-2), (3,-2)$.

What percent of the rectangle is located in Quadrant I?

Possible Answers:

$25 \%$ $\displaystyle 25 \%$

$44 \frac{4}{9} \%$ $\displaystyle 44 \frac{4}{9} \%$

$22 \frac{2}{9} \%$ $\displaystyle 22 \frac{2}{9} \%$

$11 \frac{1}{9} \%$ $\displaystyle 11 \frac{1}{9} \%$

$33 \frac{1}{3} \%$ $\displaystyle 33 \frac{1}{3} \%$

Correct answer:

$22 \frac{2}{9} \%$ $\displaystyle 22 \frac{2}{9} \%$

Explanation:

The total area of the rectangle is

$\left [ 3- (-6) \right ] \times\left [ 4- (-2) \right ] = 9 \times 6 = 54$ $\displaystyle \left [ 3- (-6) \right ] \times\left [ 4- (-2) \right ] = 9 \times 6 = 54$.

The area of the portion of the rectangle in Quadrant I is

$\left ( 3- 0 \right ) \times\left ( 4- 0 \right) = 3 \times 4 = 12$ $\displaystyle \left ( 3- 0 \right ) \times\left ( 4- 0 \right) = 3 \times 4 = 12$.

Therefore, the portion of the rectangle in Quadrant I is

$\frac{12}{54} = \frac{2}{9} = \frac{2}{9} \times 100 \% = 22 \frac{2}{9} \%$ $\displaystyle \frac{12}{54} = \frac{2}{9} = \frac{2}{9} \times 100 \% = 22 \frac{2}{9} \%$.

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Example Question #1 : How To Find The Area Of A Rectangle

A rectangle and a square have the same perimeter. The area of the square is 225 $\displaystyle 225$ square centimeters; the length of the rectangle is $\displaystyle 21$ centimeters. Give the width of the rectangle in centimeters.

Possible Answers:

$6\textrm{ cm}$ $\displaystyle 6\textrm{ cm}$

$9\textrm{ cm}$ $\displaystyle 9\textrm{ cm}$

$10 \frac{5}{7}\textrm{ cm}$ $\displaystyle 10 \frac{5}{7}\textrm{ cm}$

$21 \frac{3}{7}\textrm{ cm}$ $\displaystyle 21 \frac{3}{7}\textrm{ cm}$

$18\textrm{ cm}$ $\displaystyle 18\textrm{ cm}$

Correct answer:

$9\textrm{ cm}$ $\displaystyle 9\textrm{ cm}$

Explanation:

The sidelength of a square with area 225 $\displaystyle 225$ square centimeters is $\sqrt{225} = 15$ $\displaystyle \sqrt{225} = 15$ centimeters; its perimeter, as well as that of the rectangle, is therefore $15 \times 4 = 60$ $\displaystyle 15 \times 4 = 60$ centimeters.

Using the formula for the perimeter of a rectangle, substitute L = 21 $\displaystyle L = 21$ and solve for $\displaystyle W$ as follows:

$\displaystyle 2L + 2W = P$

$2 \cdot 21 + 2W = 60$ $\displaystyle 2 \cdot 21 + 2W = 60$

$\displaystyle 42 + 2W = 60$

$\displaystyle 42 -42 + 2W = 60-42$

2W = 18 $\displaystyle 2W = 18$

$2W \div 2 = 18\div 2$ $\displaystyle 2W \div 2 = 18\div 2$

W = 9 $\displaystyle W = 9$

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Example Question #1 : Rectangles

In a rectangle, the width is $\displaystyle x$ while the length is $\displaystyle 4x$. If 3x=9 $\displaystyle 3x=9$, what is the area of the rectangle?

Possible Answers:

$\displaystyle 36$

$\displaystyle 32$

$\displaystyle 40$

$\displaystyle 44$

Correct answer:

$\displaystyle 36$

Explanation:

In a rectangle in which the width is x while the length is 4x, the first step is to solve for x. If 3x=9 $\displaystyle 3x=9$, the value of x can be found by dividing each side of this equation by 3.

Doing so gives us the information that x is equal to 3.

Thus, the area is equal to:

$Area=x\cdot4x$ $\displaystyle Area=x\cdot4x$

$Area=3\cdot4\cdot3$ $\displaystyle Area=3\cdot4\cdot3$

Area=36 $\displaystyle Area=36$

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Example Question #1 : Rectangles

Your geometry book has a rectangular front cover which is 12 inches by 8 inches.

What is the area of your book cover?

Possible Answers:

40in^2 $\displaystyle 40in^2$

96in $\displaystyle 96in$

108in^2 $\displaystyle 108in^2$

96in^2 $\displaystyle 96in^2$

Correct answer:

96in^2 $\displaystyle 96in^2$

Explanation:

Your geometry book has a rectangular front cover which is 12 inches by 8 inches.

What is the area of your book cover?

To find the area of a rectangle, use the following formula:

$A_{rectangle}=l*w$ $\displaystyle A_{rectangle}=l*w$

Plug in our knowns and solve:

$\displaystyle A=8in*12in=96in^2$

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Example Question #302 : Plane Geometry

Find the area of a rectangle with a width of 5in and a length that is three times the width.

Possible Answers:

$15\text{in}^2$ $\displaystyle 15\text{in}^2$

$125\text{in}^2$ $\displaystyle 125\text{in}^2$

$50\text{in}^2$ $\displaystyle 50\text{in}^2$

$75\text{in}^2$ $\displaystyle 75\text{in}^2$

$25\text{in}^2$ $\displaystyle 25\text{in}^2$

Correct answer:

$75\text{in}^2$ $\displaystyle 75\text{in}^2$

Explanation:

To find the area of a rectangle, we will use the following formula:

$\text{area of rectangle} = l \cdot w$ $\displaystyle \text{area of rectangle} = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 5in. We also know the length is three times the width. Therefore, the length is 15in.

Knowing this, we can substitute into the formula. We get

$\text{area of rectangle} = 15\text{in} \cdot 5\text{in}$ $\displaystyle \text{area of rectangle} = 15\text{in} \cdot 5\text{in}$

$\text{area of rectangle} = 75\text{in}^2$ $\displaystyle \text{area of rectangle} = 75\text{in}^2$

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Example Question #1 : Rectangles

Find the area of a rectangle with a width of 7cm and a length that is four times the width.

Possible Answers:

$28\text{cm}^2$ $\displaystyle 28\text{cm}^2$

$196\text{cm}^2$ $\displaystyle 196\text{cm}^2$

$22\text{cm}^2$ $\displaystyle 22\text{cm}^2$

$56\text{cm}^2$ $\displaystyle 56\text{cm}^2$

$148\text{cm}^2$ $\displaystyle 148\text{cm}^2$

Correct answer:

$196\text{cm}^2$ $\displaystyle 196\text{cm}^2$

Explanation:

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$ $\displaystyle A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 7cm. We also know the length is four times the width. Therefore, the length is 28cm.

Knowing this, we will substitute into the formula. We get

$A = 28\text{cm} \cdot 7\text{cm}$ $\displaystyle A = 28\text{cm} \cdot 7\text{cm}$

$A = 196\text{cm}^2$ $\displaystyle A = 196\text{cm}^2$

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Example Question #303 : Plane Geometry

Find the area of a rectangle with a width of 8cm and a length that is four times the width.

Possible Answers:

$256\text{cm}^2$ $\displaystyle 256\text{cm}^2$

$80\text{cm}^2$ $\displaystyle 80\text{cm}^2$

$96\text{cm}^2$ $\displaystyle 96\text{cm}^2$

$128\text{cm}^2$ $\displaystyle 128\text{cm}^2$

$32\text{cm}^2$ $\displaystyle 32\text{cm}^2$

Correct answer:

$256\text{cm}^2$ $\displaystyle 256\text{cm}^2$

Explanation:

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$ $\displaystyle A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 8cm. We also know the length of the rectangle is four times the width. Therefore, the length of the rectangle is 32cm.

Knowing this, we will substitute into the formula. We get

$A = 32\text{cm} \cdot 8\text{cm}$ $\displaystyle A = 32\text{cm} \cdot 8\text{cm}$

$A = 256\text{cm}^2$ $\displaystyle A = 256\text{cm}^2$

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Example Question #2 : Rectangles

You are designing a poster to put on the front of your refrigerator. If the refrigerator door is 2 feet wide by 4.5 feet tall, what is the area of the largest poster you could fit on the door?

Possible Answers:

13 ft^2 $\displaystyle 13 ft^2$

15 ft^2 $\displaystyle 15 ft^2$

20 ft^2 $\displaystyle 20 ft^2$

9 ft^2 $\displaystyle 9 ft^2$

Correct answer:

9 ft^2 $\displaystyle 9 ft^2$

Explanation:

You are designing a poster to put on the front of your refrigerator. If the refrigerator door is 2 feet wide by 4.5 feet tall, what is the area of the largest poster you could fit on the door?

We need to find the area of a shape. Given the context of a refrigerator door and a poster, we can assume that the poster will be a rectangle. To find the area of a rectangle, we need to multiply length and width.

$\displaystyle A=l*w=2ft*4.5ft=9ft^2$

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Example Question #71 : Quadrilaterals

A rectangle has a width of 9cm and a length that is three times the width. Find the area.

Possible Answers:

$81\text{cm}^2$ $\displaystyle 81\text{cm}^2$

$324\text{cm}^2$ $\displaystyle 324\text{cm}^2$

$216\text{cm}^2$ $\displaystyle 216\text{cm}^2$

$243\text{cm}^2$ $\displaystyle 243\text{cm}^2$

$27\text{cm}^2$ $\displaystyle 27\text{cm}^2$

Correct answer:

$243\text{cm}^2$ $\displaystyle 243\text{cm}^2$

Explanation:

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$ $\displaystyle A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 9cm. We also know the length of the rectangle is three times the width. Therefore, the length is 27cm. So, we can substitute.

$A = 27\text{cm} \cdot 9\text{cm}$ $\displaystyle A = 27\text{cm} \cdot 9\text{cm}$

$A = 243\text{cm}^2$ $\displaystyle A = 243\text{cm}^2$

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Example Question #3 : How To Find The Area Of A Rectangle

Find the area of a rectangle with a length of 12in and a width that is a third of the length.

Possible Answers:

$72\text{in}^2$ $\displaystyle 72\text{in}^2$

$36\text{in}^2$ $\displaystyle 36\text{in}^2$

$56\text{in}^2$ $\displaystyle 56\text{in}^2$

$48\text{in}^2$ $\displaystyle 48\text{in}^2$

$24\text{in}^2$ $\displaystyle 24\text{in}^2$

Correct answer:

$48\text{in}^2$ $\displaystyle 48\text{in}^2$

Explanation:

To find the area of a rectangle, we will use the following formula:

$A = l \cdot w$ $\displaystyle A = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the length of the rectangle is 12in. We also know the width of the rectangle is a third of the length. Therefore, the width is 4in.

So, we get

$A = 12\text{in} \cdot 4\text{in}$ $\displaystyle A = 12\text{in} \cdot 4\text{in}$

$A = 48\text{in}^2$ $\displaystyle A = 48\text{in}^2$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Rectangles

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : Rectangles

Example Question #1 : How To Find The Area Of A Rectangle

Example Question #1 : Rectangles

Example Question #1 : Rectangles

Example Question #302 : Plane Geometry

Example Question #1 : Rectangles

Example Question #303 : Plane Geometry

Example Question #2 : Rectangles

Example Question #71 : Quadrilaterals

Example Question #3 : How To Find The Area Of A Rectangle

All ISEE Upper Level Math Resources

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