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ISEE Upper Level Math : Rectangles

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Example Questions

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

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

What percent of the rectangle is located in Quadrant I?

Possible Answers:

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

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

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

$25 \%$

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

Correct answer:

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

Explanation:

The total area of the rectangle is

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

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

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

Therefore, the portion of the rectangle in Quadrant I is

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

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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 square centimeters; the length of the rectangle is centimeters. Give the width of the rectangle in centimeters.

Possible Answers:

$9\textrm{ cm}$

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

$18\textrm{ cm}$

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

$6\textrm{ cm}$

Correct answer:

$9\textrm{ cm}$

Explanation:

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

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

2L + 2W = P

$2 \cdot 21 + 2W = 60$

42 + 2W = 60

42 -42 + 2W = 60-42

2W = 18

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

W = 9

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Example Question #1 : Rectangles

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

Possible Answers:

Correct answer:

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 , 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:

$Area=x\cdot4x$

$Area=3\cdot4\cdot3$

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

96in^2

108in^2

96in

Correct answer:

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:

$A_{rectangle}=l*w$

Plug in our knowns and solve:

A=8in*12in=96in^2

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Example Question #302 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Possible Answers:

$125\text{in}^2$

$25\text{in}^2$

$75\text{in}^2$

$15\text{in}^2$

$50\text{in}^2$

Correct answer:

$75\text{in}^2$

Explanation:

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

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

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

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

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

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Example Question #303 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Possible Answers:

$148\text{cm}^2$

$22\text{cm}^2$

$56\text{cm}^2$

$28\text{cm}^2$

$196\text{cm}^2$

Correct answer:

$196\text{cm}^2$

Explanation:

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

$A = l \cdot w$

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

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

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

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Example Question #304 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Possible Answers:

$256\text{cm}^2$

$128\text{cm}^2$

$80\text{cm}^2$

$32\text{cm}^2$

$96\text{cm}^2$

Correct answer:

$256\text{cm}^2$

Explanation:

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

$A = l \cdot w$

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

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

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

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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:

9 ft^2

20 ft^2

13 ft^2

15 ft^2

Correct answer:

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.

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:

$81\text{cm}^2$

$324\text{cm}^2$

$216\text{cm}^2$

$243\text{cm}^2$

$27\text{cm}^2$

Correct answer:

$243\text{cm}^2$

Explanation:

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

$A = l \cdot w$

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.

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

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

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Example Question #5 : Rectangles

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

Possible Answers:

$56\text{in}^2$

$48\text{in}^2$

$24\text{in}^2$

$72\text{in}^2$

$36\text{in}^2$

Correct answer:

$48\text{in}^2$

Explanation:

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

$A = l \cdot w$

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

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

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

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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 : How To Find The Area Of A Rectangle

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

Example Question #1 : Rectangles

Example Question #1 : Rectangles

Example Question #302 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #303 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #304 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #2 : Rectangles

Example Question #71 : Quadrilaterals

Example Question #5 : Rectangles

All ISEE Upper Level Math Resources

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