How to find the area of a rectangle - ISEE Upper Level Quantitative Reasoning

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Question

Which is the greater quantity?

(a) The area of the rectangle on the coordinate plane with vertices

(b) The area of the rectangle on the coordinate plane with vertices

Answer

(a) The first rectangle has width and height ; its area is $A = x \cdot 4y = 4xy$ .

(b) The second rectangle has width and height ; its area is $A = 2x \cdot 2y = 4xy$ .

The areas of the rectangle are the same.

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Question

. Which is the greater quantity?

(a) The area of the square on the coordinate plane with vertices

(b) The area of the rectangle on the coordinate plane with vertices

Answer

(a) The square has sidelength , and therefore has area $A ^{2}$ .

(b) The rectangle has width and height , and therefore has area $(A + 1)(A - 1) = A^{2} - 1$ .

Since $A^{2} > A^{2} - 1$ , the square in (a) has the greater area.

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Question

A rectangle on the coordinate plane has its vertices at the points .

Which is the greater quantity?

(a) The area of the portion of the rectangle in Quadrant I

(b) The area of the portion of the rectangle in Quadrant III

Answer

(a) The portion of the rectangle in Quadrant I is a rectangle with vertices , so its area is $3 \times 4 = 12$ .

(a) The portion of the rectangle in Quadrant III is a rectangle with vertices , so its area is $6 \times 2 = 12$ .

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Question

The perimeter of a rectangle is one yard. The rectangle is three times as long as it is wide. Which is the greater quantity?

(a) The area of the rectangle

(b) 60 square inches

Answer

Let be the width of the rectangle. Then its length is , and its perimeter is

Since the perimeter is one yard, or 36 inches,

$8W\div 8 = 36 \div 8$

$W = 4 \frac{1}{2}$ inches is the wiidth, and $3W = 3\cdot 4 \frac{1}{2}= 13 \frac{1}{2}$ inches is the length, so the area is

$4 \frac{1}{2} \cdot 13 \frac{1}{2} = 60 \frac{3}{4}$ square inches. (a) is the greater quantity.

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Question

If one rectangular park measures $45ft\times 20ft$ and another rectangular park measures $30ft\times 90ft$ , how many times greater is the area of the second park than the area of the first park?

Answer

First, you must compute the area of both parks. The area of a rectangle is length times width. Therefore, the area of park one is $45\cdot 20$ , which is . The area of park two is $30\cdot 90$ , which is Then, divide the area of the second park by the area of the first park ( $2700\div 900$ ). This yields 3 as the answer.

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Question

A rectangle is two feet shorter than twice its width; its perimeter is six yards. Give its area in square inches.

Answer

The length of the rectangle is two feet, or 24 inches, shorter than twice the width, so, if is the width in inches, the length in inches is

Six yards, the perimeter of the rectangle, is equal to $36 \times 6 = 216$ inches. The perimeter, in terms of length and width, is , so we can set up the equation:

$6W \div 6 = 264 \div 6$

The length and width are 64 inches and 44 inches; the area is their product, which is

$44 \times 64 = 2,816$ square inches

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Question

The perimeter of a rectangle is 210 inches. The width of the rectangle is 40% of its length. What is the area of the rectangle?

Answer

If the width of the rectangle is 40% of the length, then

.

The perimeter of the rectangle is:

The perimeter is 210 inches, so we can solve for the length:

$2.8L \div 2.8 = 210 \div 2.8$

$W = 0.40 \times 75 = 30$

The length and width of the rectangle are 75 and 30 inches; the area is their product, or

$75\times 30 = 2,250$ square inches.

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Question

The sum of the lengths of three sides of a rectangle is 572 inches; the width of the rectangle is 60% of its length. Give its area in square inches.

Answer

Since the width of the rectangle is 60% of its length, we can write .

However, it is not clear from the problem which three sides - two lengths and a width or two widths and a length - we are choosing to have sum 572 inches. Depending on the three sides chosen, we can either set up

$2.6L \div 2.6 = 572 \div 2.6$

or

$2.2L \div 2.2 = 260$

Since the length cannot be determined with certainty, neither can the width, and, subsequently, neither can the area.

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Question

Five rectangles each have the same length, which we will call . The widths of the five rectangles are 7, 5, 8, 10, and 12. Which of the following expressions is equal to the mean of their areas?

Answer

The area of a rectangle is the product of its width and its length. The areas of the five rectangles, therefore, are . The mean of these five areas is their sum divided by 5, or

$\left (7L+5L+ 8L+10L+ 12L \right ) \div 5$

$= \left (7+5+ 8+10+ 12 \right ) L \div 5$

$=42L \div 5$

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Question

Two rectangles, A and B, each have perimeter 32 feet. Rectangle A has length 12 feet; Rectangle B has length 10 feet. The area of Rectangle A is what percent of the area of Rectangle B?

Answer

The perimeter of a rectangle can be given by the formula

Since for both rectangles, 30 is the perimeter, this becomes

, and subsequently

.

Rectangle A has length 12 feet and, subsequently. width 4 feet, making its area

$12 \times 4 = 48$ square feet

Rectangle B has length 10 feet and, subsequently. width 6 feet, making its area

$10 \times 6 = 60$ square feet

The area of Rectangle A is

$\frac{48}{60} \times 100 = 80 %$

of the area of Rectangle B.

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Question

Give the area of the above rectangle in terms of .

Answer

The area of a rectangle is equal to the product of its length and height, which here are 5 and . This product is .

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Question

is a positive integer.

Rectangle A has length and width ; Rectangle B has length and length . Which is the greater quantity?

(A) The area of Rectangle A

(B) The area of Rectangle B

Answer

This might be easier to solve if you set .

Then the dimensions of Rectangle A are and . The area of Rectangle A is their product:

$(u+2) (u - 2) = u ^{2} -4$

The dimensions of Rectangle B are and . The area of Rectangle B is their product:

$(u+4) (u - 4) = u ^{2} -16$

$u ^{2} - 4 > u ^{2} -16$ regardless of the value of (or, subsequently, ), so Rectangle A has the greater area.

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Question

In Rectangle , $RE = 2 ^{x}$ and $EC = 4 ^{x}$ . Give the area of this rectangle in terms of .

Answer

The area of a rectangle is the product of its length and its width, which here are $2 ^{x}$ and $4^{x}$ . Mulitply:

$A = 2 ^{x} \cdot 4^{x} = (2 \cdot 4)^{x} = 8^{x}$

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Question

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

Answer

The sidelength of a square with area 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 and solve for as follows:

$2 \cdot 21 + 2W = 60$

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

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Question

A rectangle on the coordinate plane has its vertices at the points .

What percent of the rectangle is located in Quadrant I?

Answer

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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Question

In a rectangle, the width is while the length is . If , what is the area of the rectangle?

Answer

In a rectangle in which the width is x while the length is 4x, the first step is to solve for x. If , 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:

$Area=x\cdot4x$

$Area=3\cdot4\cdot3$

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Question

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

What is the area of your book cover?

Answer

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:

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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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