Rectangles - 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

In the above diagram,

$\textup{Rect }ABCF \sim \textup{Rect }CDEF$ .

and . Give the area of $\textup{Rect }ABDE$ .

Answer

$\textup{Rect }ABCF \sim \textup{Rect }CDEF$ , so

$\frac{AF}{CF} = \frac{CF}{EF}$

$\frac{AF}{24} = \frac{24}{12}$

$AF = \frac{24}{12} \cdot 24 = 48$

The area of the rectangle is

$AE \cdot DE$

$= \left (AF + EF \right ) \cdot CF$

$= \left (48+12 \right ) \cdot 24$

$= 60 \cdot 24$

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Question

$\textup{Rect }ABCD \sim \textup{Rect }WXYZ$ , and $\frac{AB}{WX} = 2$ .

All the following quantities MUST be equal to 2 except for .

Answer

The two rectangles are similar with similarity ratio 2.

Corresponding sides of similar rectangles are in proportion, so

$\frac{BC}{XY} = \frac{AB}{WX} = 2$ .

Since opposite sides of a rectangle are congruent, , so

$\frac{BC}{WZ}= 2$ .

$\overline{AC}$ and $\overline{WY}$ are diagonals of the rectangle. If they are constructed, then, since $\frac{AB}{WX} = \frac{BC}{XY}$ and $\angle B \cong \angle X$ (both are right angles), by the Side-Angle-Side Similarity Theorem, $\bigtriangleup ABC \cong \bigtriangleup WXY$ . By similarity, $\frac{AC}{WY} = \frac{AB}{WX} =2$ .

The ratio of the perimeters of the rectangles is

$\frac{2 \cdot AB + 2 \cdot BC }{2\cdot WX + 2 \cdot XY}$

$= \frac{ AB + BC }{ WX + XY}$ ,

It follows from $\frac{AB}{WX} =\frac{BC}{XY} = 2$ and a property of proportions that this ratio is equal to .

However,

$\frac{CD}{YX}$

is not a ratio of corresponding sides of the rectangle, so it does not have any restrictions on it. This is the correct choice.

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Question

$\textup{Rect }ABCD \sim \textup{Rect }WXYZ$ ;

$\frac{AB}{WX} = 3$ .

Which of the following must be equal to $\frac{1}{3}$ ?

Answer

$\frac{AB}{WX} = 3$ , so $\frac{WX}{AB} = \frac{1}{3}$ ; this is the similarity ratio of $\textup{Rect }WXYZ$ to $\textup{Rect }ABCD$ .

$\overline{WZ}$ and $\overline{AB}$ are not corresponding sides, nor are they congruent to corresponding sides, so it may or may not be true that $\frac{WZ}{AB} = \frac{1}{3}$ .

The ratio of the perimeters of similar rectangles is the same as their actual similarity ratio, but the choice gives the rectangles in the original order; the quotient of the perimeters as given is 3, not $\frac{1}{3}$ .

The ratio of the areas of similar rectangles is the square of the similarity ratio, so the quotient in the given choice is $\left (\frac{1}{3} \right )^{2} = \frac{1}{9}$ .

However, by similarity, and the fact that opposite sides $\overline{BD }$ and $\overline{AC}$ are congruent,

$\frac{WY}{BD} = \frac{WY}{AC} = \frac{WX}{AB} = \frac{1}{3}$ .

The correct choice is $\frac{WY}{BD}$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. $\textup{Rect }ABCD \sim \textup{Rect }WXYZ$ .

$\frac{AD}{WZ} = \frac{7}{5}$ .

Give the ratio of the area of the shaded region to the area of $\textup{Rect }WXYZ$ .

Answer

The ratio of the areas of similar rectangles is the square of the similarity ratio, so the ratio of the area of $\textup{Rect }ABCD$ to that of $\textup{Rect }WXYZ$ is

$\left ( \frac{7}{5} \right )^{2} = \frac{49}{25}$ .

So if the area of $\textup{Rect }WXYZ$ is , the area of $\textup{Rect }ABCD$ is $\frac{49}{25} N$

The area of the shaded region is the difference between the areas of the rectangles, making this area

$\frac{49}{25} N - N = \frac{24}{25} N$ .

The desired ratio is 24 to 25.

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Question

A certain rectangle has a length of and a width of . Which of the following dimensions would another rectangle need in order for it to be similar?

Answer

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can first check the ratio of the length to the width for the given rectangle, and then see which option has the same ratio, which will tell us whether or not the rectangle is similar:

$Given:\frac{L}{W}=\frac{15}{6}=\frac{5}{2}$

So in order for a rectangle to be similar, the ratio of its length to its width must be the same. All we must do then is check the answer options, in no particular order, for the rectangle with the same ratio:

$Option1:\frac{L}{W}=\frac{18}{7}$

$Option2:\frac{L}{W}=\frac{12}{5}$

$Option3:\frac{L}{W}=\frac{24}{10}=\frac{12}{5}$

$Option4:\frac{L}{W}=\frac{32}{12}=\frac{8}{3}$

$Option5:\frac{L}{W}=\frac{20}{8}=\frac{5}{2}$

A rectangle with a length of and a width of has the same ratio of dimensions as a rectangle with a length of and a width of , so these two rectangles are similar.

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Question

An engineer is making a scale model of a building. The real building needs to have a width of and a length of . If the engineer's scale model has a width of , what does the length of the model need to be?

Answer

An engineer is making a scale model of a building. The real building needs to have a width of and a length of . If the engineer's scale model has a width of , what does the length of the model need to be?

To begin, we need to know what a scale model is. A scale model is a smaller version of something that is "to scale." In other words, it is similar but not congruent.

So, we find a length that will make the model accurate, we need a ratio. Try the following:

$\frac{0.05}{54}=\frac{l_m}{164}$

$l_m=\frac{.05}{54}*164=0.15m$

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Question

A rectangle has a length of and width of . Solve for the perimeter.

Answer

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Question

A farmer decides to build a fence two feet around his rectangular field. The field is feet long and feet wide. How long should the fencing be in order to build the fence around the field?

Answer

The fence is 2 feet around the field. Therefore the area actually enclosed by the fence is 4 feet longer and 4 feet wider than the field.

The length of fencing needed is the perimeter of the area enclosed which is calculated as follows:

$2\times((21+4)+(16+4))=2\times(25+20)=2\times45=90$

The fencing should be 90 feet.

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Question

If a rectangle has an area of and a length of , what is its perimeter?

Answer

To find the perimeter, we need the length and the width. We are only given the length, so first we must find the width using the given area:

$63=9W\rightarrow W=7$

The perimeter is simply two times the length plus two times the width, so we can now use the known length and width to calculate the perimeter:

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