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

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

$2,816 \textrm{ in}^{2}$

$7,128 \textrm{ in}^{2}$

$1,386 \textrm{ in}^{2}$

$2,475 \textrm{ in}^{2}$

$2,772 \textrm{ in}^{2}$

Explanation

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

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.

It is impossible to determine the area from the given information.

$40,560 \textrm{ in}^{2}$

$29,040 \textrm{ in}^{2}$

$48,400\textrm{ in}^{2}$

$67,600\textrm{ in}^{2}$

Explanation

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

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

$2.2L \div 2.2 = 260$

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

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?

Explanation

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.

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

What is the area of your book cover?

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 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

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

(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 $6 \times 2 = 12$ .

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

(A) is greater

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Explanation

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.

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

$256\text{cm}^2$

$128\text{cm}^2$

$32\text{cm}^2$

$96\text{cm}^2$

$80\text{cm}^2$

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

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

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

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?

Explanation

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

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

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

$=42L \div 5$

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?

$83 \frac{1}{3} %$

$87 \frac{1}{2} %$

Explanation

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.

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?

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.

How to find the area of a rectangle

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