Calculating whether rectangles are similar - GMAT Quantitative

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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?

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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

In the above diagram,

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

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

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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 .

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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 $\frac{1}{3}$ ?

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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 $\textup{Rect }WXYZ$ .

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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

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?

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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

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 $\textup{Rect }WXYZ$ .

Tap to reveal answer

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

In the above diagram,

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

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

Tap to reveal answer

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 .

Tap to reveal answer

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 $\frac{1}{3}$ ?

Tap to reveal answer

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

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?

Tap to reveal answer

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?

Tap to reveal answer

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:

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

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

← Didn't Know|Knew It →

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 $\textup{Rect }WXYZ$ .

Tap to reveal answer

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

In the above diagram,

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

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

Tap to reveal answer

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 .

Tap to reveal answer

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 $\frac{1}{3}$ ?

Tap to reveal answer

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

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?

Tap to reveal answer

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?

Tap to reveal answer

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:

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

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

← Didn't Know|Knew It →

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 $\textup{Rect }WXYZ$ .

Tap to reveal answer

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

In the above diagram,

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

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

Tap to reveal answer

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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