DSQ: Calculating whether rectangles are similar - GMAT Quantitative Reasoning

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure, which shows a rectangle divided into two smaller rectangles.

True or false: $\textup{Rect } ABCF \sim \textup{Rect } CDEF$ .

Statement 1: $CF = \sqrt{AF \cdot FE}$

Statement 2: $AF \cdot FE = AB \cdot DE$

Answer

The statements are actually equivalent. Since , Statement 2 can be rewritten as $AF \cdot FE = CF \cdot CF$ , or $\left (CF \right )^{2} =AF \cdot FE$ - and, since all quantities are positive, $CF = \sqrt{AF \cdot FE}$ . The question is therefore whether either statement alone answers the question or both together do not.

Each statement is equivalent to

$AF \cdot FE = CF \cdot CF$ .

Divide both sides by $FE \cdot CF$ to yield a proportion statement:

$\frac{AF \cdot FE}{FE \cdot CF} = \frac{CF \cdot CF}{FE \cdot CF}$

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

The sides of the rectangles are in proportion; subsequently, the rectangles are similar.

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Question

You are given two rectangles, $\textup{Rect } ABCD$ and $\textup{Rect } EFGH$

True or false: $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

Statement 1: $\frac{AC}{EG} = \frac{BD}{FH}$

Statement 2: $\frac{AB}{EF} = \frac{BC}{FG}$

Answer

Statement 1 compares the lengths of the diagonals of the two rectangles. Since the diagonals of any rectangle are congruent, , , and, as a consequence, $\frac{AC}{EG} = \frac{BD}{FH}$ regardless of whether the rectangles are similar or not. Statement 1 is a superfluous statement and is therefore unhelpful.

Statement 2 asserts that sides of the two rectangles are in proportion. This is a necessary and sufficient condition for the rectangles to be similar.

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Question

You are given two rectangles, $\textup{Rect } ABCD$ and $\textup{Rect } EFGH$

True or false: $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

Statement 1: $AB \cdot FG= BC \cdot EF$

Statement 2: $AB \cdot GH= CD \cdot EF$

Answer

Assume Statement 1 alone. Divide both sides by $EF \cdot FG$ ,

$AB \cdot FG= BC \cdot EF$

$\frac{AB \cdot FG}{EF \cdot FG}= \frac{BC \cdot EF}{EF \cdot FG}$

$\frac{AB}{EF} = \frac{BC}{FG}$ .

This proportion statement asserts that sides of the two rectangles are in proportion. This is a necessary and sufficient condition for the rectangles to be similar.

Now examine Statement 2.

By the congruence of opposite sides of a rectangle,

, ,

and, regardless of whether the rectangles are similar or not,

$AB \cdot GH= CD \cdot EF$ .

Therefore, Statement 2 provides superfluous and unhelpful information.

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Question

You are given two rectangles, $\textup{Rect } ABCD$ and $\textup{Rect } EFGH$ .

Let the perimeter of $\textup{Rect } ABCD$ be , and let the perimeter of $\textup{Rect } EFGH$ be .

True or false: $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

Statement 1: $\frac{AB}{EF} = \frac{p}{q}$

Statement 2: $\frac{AD}{EH} = \frac{p}{q}$

Answer

The perimeter of $\textup{Rect } ABCD$ is the sum of its sidelengths, and opposite sides are congruent, so

$p = 2 \cdot AB + 2\cdot BC$

Similarly,

$q = 2 \cdot EF + 2 \cdot FG$

Therefore,

$\frac{p }{q}= \frac{2 \cdot AB + 2\cdot BC}{2 \cdot EF + 2 \cdot FG}$ ,

and, reducing,

$\frac{p }{q}= \frac{ AB + BC}{ EF + FG}$

Assume Statement 1 alone. Then

$\frac{p}{q} = \frac{AB}{EF}$ , or $\frac{p}{q} = \frac{-AB}{-EF}$ , and by a property of proportions,

$\frac{p }{q}= \frac{\left ( AB + BC \right )+(-AB)}{ \left (EF + FG \right )+(-EF)} = \frac{BC}{EF}$

Therefore,

$\frac{AB}{EF}= \frac{BC}{EF}$ ,

thereby proving the sides of the rectangles to be in proportion. As a consequence, $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

By a similar argument, Statement 2 also proves $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

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Question

You are given two rectangles, $\textup{Rect } ABCD$ and $\textup{Rect } EFGH$ .

Let the perimeter of $\textup{Rect } ABCD$ be , and let the perimeter of $\textup{Rect } EFGH$ be .

Let the area of $\textup{Rect } ABCD$ be and the area $\textup{Rect } EFGH$ be .

True or false: $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

Statement 1: $\frac{p^{2}}{q^{2}} = \frac{r}{s}$

Statement 2: $\frac{AB}{EF} = \frac{p}{q}$

Answer

Assume Statement 1 alone.

Examine these three rectangles. The one of the left is $\textup{Rect } ABCD$ ; the other two have the same dimensions, and both are called $\textup{Rect } EFGH$ , except that the names of the vertices are differently arranged:

Regardless of which $\textup{Rect } EFGH$ is chosen, the ratio of the perimeters is $\frac{14}{28} = \frac{1}{2}$ and the ratio of the areas is $\frac{12}{48} = \frac{1}{4} =\left ( \frac{1 }{2} \right )^{2}$ . The conditions of the problem are met for both pairings, but in one case, $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ and in the other, $\textup{Rect } ABCD sim \textup{Rect } EFGH$ (that is, the rectangles are similar, but the given similarity statement may or may not be true).

Assume Statement 2 alone.

The perimeter of $\textup{Rect } ABCD$ is the sum of its sidelengths, and opposite sides are congruent, so

$p = 2 \cdot AB + 2\cdot BC$

Similarly,

$q = 2 \cdot EF + 2 \cdot FG$

Therefore,

$\frac{p }{q}= \frac{2 \cdot AB + 2\cdot BC}{2 \cdot EF + 2 \cdot FG}$ ,

and, reducing,

$\frac{p }{q}= \frac{ AB + BC}{ EF + FG}$

From Statement 2,

$\frac{p}{q} = \frac{AB}{EF}$ , or $\frac{p}{q} = \frac{-AB}{-EF}$ , and by a property of proportions,

$\frac{p }{q}= \frac{\left ( AB + BC \right )+(-AB)}{ \left (EF + FG \right )+(-EF)} = \frac{BC}{EF}$

Therefore,

$\frac{AB}{EF}= \frac{BC}{EF}$ ,

thereby proving the sides of the rectangles to be in proportion. As a consequence, $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

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Question

You are given two rectangles, $\textup{Rect } ABCD$ and $\textup{Rect } EFGH$ .

Let the perimeter of $\textup{Rect } ABCD$ be , and let the perimeter of $\textup{Rect } EFGH$ be .

Let the area of $\textup{Rect } ABCD$ be and the area $\textup{Rect } EFGH$ be .

True or false: $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ .

Statement 1: $\left (\frac{p }{q} \right ) ^{2}= \frac{r}{s}$

Statement 2: $\frac{AC}{EG} = \frac{p}{q}$

Answer

Assume both statements to be true.

Examine these three rectangles. The one of the left is $\textup{Rect } ABCD$ ; the other two have the same dimensions, and both are called $\textup{Rect } EFGH$ , except that the names of the vertices are differently arranged:

Regardless of which $\textup{Rect } EFGH$ is chosen:

The ratio of the perimeters is $\frac{14}{28} = \frac{1}{2}$ ;

The ratio of the areas is $\frac{12}{48} = \frac{1}{4} =\left ( \frac{1 }{2} \right )^{2}$ ; and,

The ratio $\frac{AC}{EG} = \frac{5}{10} = \frac{1}{2}$ .

The conditions of the problem are met for both pairings, but in one case, $\textup{Rect } ABCD \sim \textup{Rect } EFGH$ and in the other, $\textup{Rect } ABCD sim \textup{Rect } EFGH$ (that is, the rectangles are similar but the similarity statement given may be true or false).

The two statements together provide insufficient information.

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