Rectangles

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GMAT Quantitative › Rectangles

Questions 1 - 10

Rectangles

In the above diagram,

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

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

Explanation

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

$AE \cdot DE$

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

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

$= 60 \cdot 24$

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

$2N^{2} -8N + 8$

$N^{2} -4N + 4$

$4N^{2} -16N + 16$

$2N^{2} -6N + 4$

$N^{2} -3N + 2$

Explanation

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:

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

Find the length of the diagonal of a rectangle whose sides are lengths $4\textup{ and }7$ .

$\sqrt{65}$

$\sqrt{13}$

$\sqrt{11}$

Explanation

To find the diagonal, you must use the pythaorean theorem. Thus:

$c=\sqrt{65}$

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?

$\dpi{100} \small 48in.$

$\dpi{100} \small 24in.$

$\dpi{100} \small 36in.$

$\dpi{100} \small 40in.$

$\dpi{100} \small 12in.$

Explanation

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:

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

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

$A ^{2} - 16$

$A ^{2} + 16$

$A ^{2} -8A + 16$

Explanation

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

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

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

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

$\frac{CD}{YX}$

$\frac{BC}{WZ}$

$\frac{AC}{WY}$

The perimeter of $\textup{Rect }ABCD$ divided by the perimeter of $\textup{Rect }WXYZ$ .

All of the quantities in the other choices must be equal to .

Explanation

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.

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

Explanation

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

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

The area of a rectangle is 85; its length is . What is its width?

$\frac{85}{x-17 }$

$\frac{5}{x }$

$\frac{68}{x }$

$51 - \frac{1}{2}x$

Explanation

The product of the length and the width of a rectangle is its area, so divide area by length to get width. This is simply $\frac{85}{x-17 }$ .

The width of a rectangle is twice the length. Find an equation representing the perimeter.

$\textup{I. }l \cdot2l$

$\textup{II. }4l + 2l$

$\textup{III. }6l^2$

$\textup{II only}$

$\textup{I only}$

$\textup{III only}$

$\textup{II and III}$

$\textup{I and II}$

Explanation

The perimeter of a rectangle can be found by adding up all the sides:

We are told the width of the rectangle is twice its length so , inserting this into our equation for the perimeter leaves us with:

, which is II

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Rectangle has an area of and . What is the length of the diagonal ?

$13\sqrt{2}$

$5\sqrt{3}$

$11\sqrt{2}$

$7\sqrt{3}$

$3\sqrt{7}$

Explanation

Firstly, before we try to set up an equation for the length of the sides and the area, we should notice that the area is a perfect square. Indeed $13^{2}=169$ . Now let's try to see whether 13 could be the length of two consecutive sides of the rectangle, Indeed, we are told that , therefore ABDC is a square with side 13 and with diagonal $s\sqrt{2}$ , where is the length of the side of the square. Therefore, the final answer is $13\sqrt{2}$ .

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