Calculating the area of a rectangle - GMAT Quantitative

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Question

The area of a square that has sides with a length of 12 inches is equal to the area of a rectangle. If the rectangle has a width of 3 inches, what is the length of the rectangle?

Answer

If the area of the rectangle is equal to the area of the square, then it must have an area of $\dpi{100} \small 144in^{2}$ $\dpi{100} \small (12\times 12)$ . If the rectangle has an area of $\dpi{100} \small 144 in^{2}$ and a side with a lenth of 3 inches, then the equation to solve the problem would be $\dpi{100} \small 144=3x$ , where $\dpi{100} \small x$ is the length of the rectangle. The solution:

$\frac{144}{3} = 48$ .

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Question

Note: figure NOT drawn to scale

Give the area of the above rectangle.

Answer

The area of a rectangle is the product of its length and width;

$A = 3^{x+1} \cdot 3^{x} = 3 ^{x+1+x} = 3 ^{2x+1}$

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Question

A rectangle twice as long as it is wide has perimeter . Write its area in terms of .

Answer

Let be the width of the rectangle; then its length is , and its perimeter is

Set this equal to and solve for :

$\frac{6W}{6} =\frac{ 6N-12}{6}$

The width is and the length is $L = 2W = 2 \left (N - 2 \right ) = 2N - 4$ , so multiply these expressions to get the area:

$A = LW = (N-2) \cdot \left (2N-4 \right ) = 2N^{2} -8N + 8$

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Question

A rectangle has its vertices at . What part, in percent, of the rectangle is located in Quadrant III?

Answer

A rectangle with vertices has width and height , thereby having area $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant III is a rectangle with vertices

.

It has width and height , thereby having area $4 \times 3 = 12$ .

Therefore, $\frac{12}{50 }$ of the rectangle is in Quadrant III; this is equal to

$\frac{12}{50 } \times 100 = 24 %$

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Question

A rectangle has its vertices at . What percentage of the rectangle is located in Quadrant IV?

Answer

A rectangle with vertices has width and height ; it follows that its area is $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant IV has vertices . Its width is , and its height is , so its area is $1 \times 3 = 3$ .

Therefore, $\frac{3}{50 }$ , or $\frac{3}{50} \times 100 % = 6%$ , of this rectangle is in Quadrant IV.

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Question

What is the area of a rectangle given the length of and width of ?

Answer

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

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Question

What polynomial represents the area of a rectangle with length and width ?

Answer

The area of a rectangle is the product of the length and the width. The expression can be multplied by noting that this is the product of the sum and the difference of the same two terms; its product is the difference of the squares of the terms, or

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

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Question

The perimeter of a rectangle is $104\textup{ inches}$ and its length is times the width. What is the area?

Answer

The perimeter of a rectangle is the sum of all four sides, that is:

since , we can rewrite the equation as:

$w=13\textup{ in}$

We are being asked for the area so we still aren't done. The area of a rectangle is the product of the width and length. We know what the width is so we can find the length and then take their product.

$l=3w=3(13)=39\textup{ in}$

$A= l \cdot w = 39 \cdot 13$

$A = 507\textup{ in}^2$

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Question

Find the area of a rectangle whose side lengths are $7\textup{ and }9$ .

Answer

To calculate area, multiply width times height. Thus,

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Question

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

Answer

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

This problem asks us to find the area of a rectangle. We are given one side and asked to find the other. To find the other, we need to use the provided clues.

"...five less..."

"...three times the first side..." or

So put it together:

Next, find the area via the following formula:

So our answer is:

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Question

Find the area of a rectangle whose width is and length is .

Answer

To find area, simply multiply length times width. Thus

$2x\cdot2y=4xy$

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Question

The area of a square that has sides with a length of 12 inches is equal to the area of a rectangle. If the rectangle has a width of 3 inches, what is the length of the rectangle?

Answer

If the area of the rectangle is equal to the area of the square, then it must have an area of $\dpi{100} \small 144in^{2}$ $\dpi{100} \small (12\times 12)$ . If the rectangle has an area of $\dpi{100} \small 144 in^{2}$ and a side with a lenth of 3 inches, then the equation to solve the problem would be $\dpi{100} \small 144=3x$ , where $\dpi{100} \small x$ is the length of the rectangle. The solution:

$\frac{144}{3} = 48$ .

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Question

A rectangle twice as long as it is wide has perimeter . Write its area in terms of .

Answer

Let be the width of the rectangle; then its length is , and its perimeter is

Set this equal to and solve for :

$\frac{6W}{6} =\frac{ 6N-12}{6}$

The width is and the length is $L = 2W = 2 \left (N - 2 \right ) = 2N - 4$ , so multiply these expressions to get the area:

$A = LW = (N-2) \cdot \left (2N-4 \right ) = 2N^{2} -8N + 8$

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Question

A rectangle has its vertices at . What part, in percent, of the rectangle is located in Quadrant III?

Answer

A rectangle with vertices has width and height , thereby having area $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant III is a rectangle with vertices

.

It has width and height , thereby having area $4 \times 3 = 12$ .

Therefore, $\frac{12}{50 }$ of the rectangle is in Quadrant III; this is equal to

$\frac{12}{50 } \times 100 = 24 %$

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Question

A rectangle has its vertices at . What percentage of the rectangle is located in Quadrant IV?

Answer

A rectangle with vertices has width and height ; it follows that its area is $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant IV has vertices . Its width is , and its height is , so its area is $1 \times 3 = 3$ .

Therefore, $\frac{3}{50 }$ , or $\frac{3}{50} \times 100 % = 6%$ , of this rectangle is in Quadrant IV.

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Question

What is the area of a rectangle given the length of and width of ?

Answer

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

Compare your answer with the correct one above

Question

What polynomial represents the area of a rectangle with length and width ?

Answer

The area of a rectangle is the product of the length and the width. The expression can be multplied by noting that this is the product of the sum and the difference of the same two terms; its product is the difference of the squares of the terms, or

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

Compare your answer with the correct one above

Question

The perimeter of a rectangle is $104\textup{ inches}$ and its length is times the width. What is the area?

Answer

The perimeter of a rectangle is the sum of all four sides, that is:

since , we can rewrite the equation as:

$w=13\textup{ in}$

We are being asked for the area so we still aren't done. The area of a rectangle is the product of the width and length. We know what the width is so we can find the length and then take their product.

$l=3w=3(13)=39\textup{ in}$

$A= l \cdot w = 39 \cdot 13$

$A = 507\textup{ in}^2$

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Question

Find the area of a rectangle whose side lengths are $7\textup{ and }9$ .

Answer

To calculate area, multiply width times height. Thus,

Compare your answer with the correct one above

Question

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

Answer

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

This problem asks us to find the area of a rectangle. We are given one side and asked to find the other. To find the other, we need to use the provided clues.

"...five less..."

"...three times the first side..." or

So put it together:

Next, find the area via the following formula:

So our answer is:

Compare your answer with the correct one above

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