DSQ: Calculating whether right triangles are similar - GMAT Quantitative

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Question

You are given that $\small \Delta ABC$ and $\small \small \Delta DEF$ are right triangles with their right angles at and , respectively. Is it true that $\small \small \Delta ABC \sim \small \Delta DEF$ ?

$\small \small \angle A \cong \angle D$

and

Answer

All right angles are congruent, so $\small \small \small \angle B \cong \angle E$ .

Since Statement 1 tells us that $\small \small \angle A \cong \angle D$ , this sets up the conditions for the Angle-Angle Similarity Postulate, so $\small \small \Delta ABC \sim \small \Delta DEF$ .

Statement 2 alone only tells us their hypotenuses. Congruence between one pair of angles and the measures of one pair of sides is insufficient information to determine whether two triangles are similar (given one angle, at least two pairs of proportional sides are required).

Therefore, the answer is that Statement 1 alone, but not Statement 2, is sufficient.

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1: The ratio of the perimeter of $\triangle ABC$ to that of $\triangle XYZ$ is to .

Statement 2: $\frac{AB}{XY} = \frac{3}{2}$ .

Answer

Assume both statements are true.

While in two similar triangles, the ratio of the perimeters, given in Statement 1, is indeed equal to that of the ratios of the lengths of the hypotenuses, given in Statement 2, this is not a sufficient condition for similarity. For example:

Case 1:

Case 2:

In each case, the conditions of the main problem and both statements are met, since:

Both triangles are right - each Pythagorean triple is a multiple of Pythagorean triple 3-4-5;

The ratio of the perimeters is $\frac{9+12+15}{6+8+10}= \frac{36}{24} = \frac{3}{2}$ ; and,

$\frac{AC}{XY} = \frac{15}{10}$ .

But in Case 1,

$\frac{AC}{XZ} = \frac{BC}{YZ} = \frac{AB}{XY}$ , since $\frac{9}{6} = \frac{12}{8} = \frac{15}{10} = \frac{3}{2}$ , and the similarity follows by way of the Side-Side-Side Similarity Principle.

In Case 2,

$\frac{AC}{XZ} e \frac{AB}{XY}$ , since $\frac{9}{8} e \frac{15}{10}$ . This violates the conditions of similarity (note that in both cases, $\triangle ABC \sim \triangle YXZ$ , but this is a different statement).

The two statements together are inconclusive.

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1: $\angle A \cong \angle X$

Statement 2: $\angle B \cong \angle Y$

Answer

Assume Statement 1 alone is true. Then, since $\angle C \cong \angle Z$ , both being right angles, and $\angle A \cong \angle X$ from Statement 1, $\triangle ABC \sim \triangle XYZ$ follows by way of the Angle-Angle Similarity Postulate. A similar argument shows Statement 2 also provides sufficient information.

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1: $\angle A$ and $\angle Y$ are complimentary.

Statement 2: $\angle B$ and $\angle X$ are complimentary.

Answer

The acute angles of a right triangle are complementary, so $\angle A$ and $\angle B$ are a complementary pair, as are $\angle X$ and $\angle Y$ .

If Statement 1 is assumed—that is, if $\angle A$ and $\angle Y$ are a complementary pair—then, since two angles complementary to the same angle—here, $\angle Y$ —must be congruent, $\angle A \cong \angle X$ . Since right angles $\angle C \cong \angle Z$ , $\triangle ABC \sim \triangle XYZ$ follows by way of the Angle-Angle Similarity Postulate, and Statement 1 turns out to provide sufficient information. By a similar argument, Statement 2 is also sufficient.

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1: $3 \cdot AB = 4 \cdot XY$

Statement 2: $4 \cdot BC= 3 \cdot YZ$

Answer

Assume Statement 1 alone. The statement used to find a sidelength ratio:

$3 \cdot AB = 4 \cdot XY$

$\frac{3 \cdot AB}{3 \cdot XY} = \frac{4 \cdot XY}{3 \cdot XY}$

$\frac{ AB}{ XY} = \frac{4 }{3 }$

However, since we only know one sidelength ratio, similarity cannot be proved or disproved.

From Statement 2, another ratio can be found:

$4 \cdot BC= 3 \cdot YZ$

$\frac{ 4 \cdot BC}{4 \cdot YZ}=\frac{ 3 \cdot YZ }{4 \cdot YZ}$

$\frac{ BC}{ YZ}=\frac{ 3 }{4 }$

Again, since only one sidelength ratio is known, similarty can be neither proved nor disproved.

Assume both statements to be true. Similarity, by definition, requires that

$\frac{ AB}{ XY} =\frac{ BC}{ YZ}$

From the two statements together, it can be seen that $\frac{ AB}{ XY} e \frac{ BC}{ YZ}$ , so $\triangle ABC sim \triangle XYZ$ .

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1: The ratio of the perimeter of $\triangle ABC$ to that of $\triangle XYZ$ is 7 to 6.

Statement 2: $AC \cdot YZ = BC \cdot XZ$

Answer

Assume Statement 1 alone. The ratio of the perimeters does not in and of itself establish similarity, since only one angle congruence is known.

Assume Statement 2 alone. The equation can be rewritten as a proportion statement:

$AC \cdot YZ = BC \cdot XZ$

$\frac{AC \cdot YZ }{BC \cdot YZ}= \frac{BC \cdot XZ}{BC \cdot YZ }$

$\frac{AC }{BC }= \frac{ XZ}{ YZ }$

This establishes that two pairs of corresponding sides are in proportion. Their included angles are both right angles, so $\angle C \cong \angle Z$ , and $\triangle ABC \sim \triangle XYZ$ follows from the Side-Angle-Side Similarity Theorem.

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Question

You are given two right triangles: $\triangle ABC$ with right angle , and $\triangle XYZ$ with right angle .

True or false: $\triangle ABC \sim \triangle XYZ$

Statement 1:

Statement 2:

Answer

Each statement alone only gives a relationship between two sides within one triangle, so neither alone answers the question of the similarity of the two triangles.

Assume both statements are true. Then, since , $\frac{1}{ XZ}> \frac{1}{YZ}$ .

By the multiplication property of inequality, since

and $\frac{1}{ XZ}> \frac{1}{YZ}$ ,

$AC \cdot \frac{1}{ XZ} > BC \cdot \frac{1}{YZ}$

$\frac{AC}{ XZ} > \frac{BC}{YZ}$

Since, by definition, $\triangle ABC \sim \triangle XYZ$ requires that $\frac{AC}{ XZ} = \frac{BC}{YZ}$ , $\triangle ABC sim \triangle XYZ$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure, which shows two right triangles circumscribed by circles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1: The area of the circle that circumscribes $\bigtriangleup ABC$ is four times the that of the circle that circumscribes $\bigtriangleup XYZ$ .

Statement 2: $AB = 2 \cdot DE$

Answer

Assume Statement 1 alone. The area of the circle that circumscribes $\bigtriangleup ABC$ is four times the that of the circle that circumscribes $\bigtriangleup XYZ$ , so the radius of the former is the square root of this, or two, times the radius of the latter; therefore, the ratio of the diameters of the circles is also two. Since in a right triangle inscribed inside a circle, the hypotenuse must be a diameter of that circle,

$\frac{BC}{EF} = 2$ or $BC = 2 \cdot EF$

However, no other information is given or can be determined about any other sides or angles, so the triangles cannot been proved or disproved to be similar.

Assume Statement 2 alone. $AB = 2 \cdot DE$ , or $\frac{AB}{DE} = 2$ , but again, no other information is given or can be determined about any other sides or angles, so the triangles have not been proved or disproved to be similar.

Now assume both statements. $\frac{AB}{DE} = 2 = \frac{BC}{EF}$ , setting up the conditions for the Hypotenuse-Leg Similarity Theorem; it follows that $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

True or false: $\bigtriangleup ABC \sim \bigtriangleup DEF$

Statement 1: Arcs $\widehat{AC}$ and $\widehat{DF}$ have the same degree measure.

Statement 2: Arcs $\widehat{ACB}$ and $\widehat{DFE}$ have the same degree measure.

Answer

Assume Statement 1 alone. $\widehat{AC}$ and $\widehat{DF}$ have the same degree measure, so the inscribed angles that intercept these arcs must also have the same degree measure - that is, $\angle B \cong \angle E$ . Since $\angle A \cong \angle D$ , both being right angles, this sets up the conditions of the Angle-Angle Postulate, so it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Assume Statement 2 alone. Major arc $m \widehat{ACB} = 360^{\circ} - m \widehat{AB}$ and major arc $m \widehat{DFE} = 360^{\circ} - m \widehat{DE}$ . $m \widehat{ACB} = m \widehat{DFE}$ by Statement 2, so

$360^{\circ} - m \widehat{AB} = 360^{\circ} - m \widehat{DE}$

$- m \widehat{AB} = - m \widehat{DE}$

$m \widehat{AB} = m \widehat{DE}$

Again, the inscribed angles that intercept these arcs must also be congruent - that is, $\angle C \cong \angle F$ . Again, this, along with $\angle A \cong \angle D$ , prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ by way of the Angle-Angle Postulate.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ , where $\angle B$ and $\angle Y$ are right angles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1:

Statement 2: $\angle A cong \angle X$

Answer

Similarity of two triangles does not require that corresponding sides be congruent - only that they be in proportion. Therefore, the inequality of one pair of corresponding sides, as given in Statement 1 alone, is insufficient to prove or disprove that $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

However, similarity of two triangles requires that all three pairs of corresponding angles be congruent. Therefore, Statement 2 alone, which establishes that one such pair is noncongruent, proves that $\bigtriangleup ABC sim \bigtriangleup XYZ$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ , where $\angle B$ and $\angle Y$ are right angles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1: Both triangles are isosceles.

Statement 2: The triangles have the same perimeter.

Answer

Assume Statement 1 alone. By the 45-45-90 Theorem, an isosceles right triangle has two acute angles that each measure $45^{\circ }$ ; specifically, $m \angle A = m \angle X = 45^{\circ }$ and $m \angle C = m \angle Z = 45^{\circ }$ . By the Angle-Angle Similarity Postulate, $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

Assume Statement 2 alone. The equality of the perimeters of two right triangles does not imply their similarity or nonsimilarity. For example:

Let

and

Since $\overline{AB } \cong \overline{XY }$ and $\overline{AC} \cong \overline{XZ}$ , it follows by the Hypotenuse-Leg Theorem that $\bigtriangleup ABC \cong \bigtriangleup XYZ$ , and, since congruent triangles are similar, it further follows that $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

Now, let

and .

The sides are not in proportion, as $\frac{AB}{XY} = \frac{3}{4} e \frac{4}{3} = \frac{BC}{YZ}$ , so $\bigtriangleup ABC sim \bigtriangleup XYZ$ .

In both scenarios, since $3^{2}+4^{2}= 5^{2}$ , by the converse of the Pythagorean Theorem, the triangles are right, and the triangles both have perimeter , thereby satisfying the main premise and Statement 2. Statement 2 alone is therefore insufficient.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ , where $\angle B$ and $\angle Y$ are right angles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1: $m \angle A = 45^{\circ }$

Statement 2: $\bigtriangleup XYZ$ is an isosceles triangle.

Answer

Each of Statement 1 and Statement 2 gives information about only one of the triangles, so neither statement alone is sufficient.

Assume both statements are true. From Statement 1, $m \angle A = 45^{\circ }$ and $\angle B$ is right and measures $90^{\circ }$ .

From Statement 2 alone, $\bigtriangleup XYZ$ is isosceles; the acute angles of an isosceles right triangle must both measure $45^{\circ }$ , so, in particular, $m \angle X = 45^{\circ }$ . Also, it is given that $\angle Y$ is right.

$\angle A \cong \angle X$ and $\angle B \cong \angle Y$ (both of the latter being right angles), and by the Angle-Angle Postulate, $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ , where $\angle B$ and $\angle Y$ are right angles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1: Both triangles are scalene.

Statement 2: The triangles have the same perimeter.

Answer

Assume both statements are true. Consider these two scenarios:

Let

and

Since $\overline{AB } \cong \overline{XY }$ and $\overline{AC} \cong \overline{XZ}$ , then by the Hypotenuse-Leg Theorem, $\bigtriangleup ABC \cong \bigtriangleup XYZ$ , and it follows that $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

Now, let

and .

The sides are not in proportion, as $\frac{AB}{XY} = \frac{3}{4} e \frac{4}{3} = \frac{BC}{YZ}$ , so $\bigtriangleup ABC sim \bigtriangleup XYZ$ .

In both scenarios, since $3^{2}+4^{2}= 5^{2}$ , the triangles are right by the converse of the Pythagorean Theorem, satisfying the main premise; the triangles are scalene, having sides of three different lengths, satisfying Statement 1; and the triangles both have perimeter , satisfying Statement 2. The two statements together are insufficient.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. True or false: $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Statement 1: $m \angle CDV + 1^{\circ } = m \angle ABV$

Statement 2:

Answer

Assume Statement 1 alone. If $\bigtriangleup AVB \sim \bigtriangleup CVD$ , then corresponding angles are congruent - specifically, $m \angle CDV = m \angle ABV$ . Statement 1 directly contradicts this, so $\bigtriangleup AVB sim \bigtriangleup CVD$ .

Statement 2 alone does not prove or disprove similarity, since information is only given about one set of corresponding sides. No information is given about corresponding angles, or any other set of corresponding sides.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. True or false: $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Statement 1: Quadrilateral is a trapezoid.

Statement 2: $\frac{AC}{CV} = \frac{BD}{DV}$

Answer

Assume Statement 1 alone. Since Quadrilateral is a trapezoid, it follows that $\overline{AB} ||\overline{CD}$ ; by the Corresponding Angles Principle, $\angle ABV \cong \angle CDV$ . By reflexivity, $\angle AVB \cong \angle CVD$ . By the Angle-Angle Postulate, $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

Assume Statement 2 alone.

$\frac{AC}{CV} = \frac{BD}{DV}$

$\frac{AC}{CV} + 1 = \frac{BD}{DV} + 1$

$\frac{AC}{CV} + \frac{CV}{CV} = \frac{BD}{DV} + \frac{DV}{DV}$

$\frac{AC+CV}{CV} = \frac{BD+DV}{DV}$

$\frac{A V}{CV} = \frac{B V}{DV}$

Also, by reflexivity, $\angle V \cong \angle V$ . By the Side-Angle-Side Similarity Theorem, it follows that $\bigtriangleup AVB \sim \bigtriangleup CVD$ .

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Question

Given: Rectangles and with diagonals $\overline{AC}$ and $\overline{EG}$ , respectively.

True or false: $\bigtriangleup ABC \sim \bigtriangleup EFG$

Statement 1: Of the two rectangles, only Rectangle is a rhombus.

Statement 2:

Answer

Assume Statement 1 alone. Rectangle is not a rhombus, so or . Rectangle is a rhombus, so . Therefore, either

$\frac{EF}{AB} < \frac{FG}{BC}$

or

$\frac{EF}{AB} > \frac{FG}{BC}$

Either way, the corresponding sides are not in proportion; it follows that $\bigtriangleup ABC sim \bigtriangleup EFG$ .

Assume Statement 2 alone. No information is given about , so the quantities $\frac{EF}{AB}$ and $\frac{FG}{BC}$ cannot be proved to be equal or unequal. If they are unequal, since the sides are not in proportion, it follows that $\bigtriangleup ABC sim \bigtriangleup EFG$ . If they are equal, then, since $m \angle B = m \angle F = 90^{\circ }$ , it follows by the Side-Angle-Side similarity Theorem that $\bigtriangleup ABC \sim \bigtriangleup EFG$ .

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Question

Given: Rectangles and with diagonals $\overline{AC}$ and $\overline{EG}$ , respectively.

True or false: $\bigtriangleup ABC \sim \bigtriangleup EFG$

Statement 1: Rectangles and are both rhombuses.

Statement 2: Rectangles and are both squares.

Answer

Both rectangles, by definition, have four right angles.

A rhombus is a quadrilateral with four sides of equal length. A square is a quadrilateral with four right angles and four sides of equal length. Therefore, the two statements are actually equivalent in this context, so either they together provide insufficient information, or either alone does. We show that the latter is the case.

If both statements are true, and the rectangles are rhombuses and squares, then

and .

Consequently,

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

$m \angle B = m \angle F = 90^{\circ }$ ,

so by the Side-Angle-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup EFG$ .

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Question

Given: Rectangles and with diagonals $\overline{AC}$ and $\overline{EG}$ , respectively.

True or false: $\bigtriangleup ABC \sim \bigtriangleup EFG$

Statement 1: $\bigtriangleup ADC \sim \bigtriangleup EHG$

Statement 2: $\angle BAC \cong \angle FEG$

Answer

Refer to the figure below, which gives the two rectangles with their diagonals as described.

Assume Statement 1 alone. The diagonals of a rectangle bisect the rectangle into congruent triangles, so $\bigtriangleup ABC \cong \bigtriangleup CDA$ and $\bigtriangleup EFG \cong \bigtriangleup GHE$ . Congruent triangles are also similar, so it follows that $\bigtriangleup ABC \sim \bigtriangleup CDA$ and $\bigtriangleup EFG \sim \bigtriangleup GHE$ . Since, by Statement 1, $\bigtriangleup ADC \sim \bigtriangleup EHG$ - or, stated differently, $\bigtriangleup CDA \sim \bigtriangleup GHE$ - by transitivity of similarity,

$\bigtriangleup ABC \sim \bigtriangleup CDA \sim \bigtriangleup GHE \sim \bigtriangleup EFG$ , and

$\bigtriangleup ABC \sim \bigtriangleup EFG$ .

Assume Statement 2 alone. The quadrilaterals are rectangles, so $\angle ABC \cong \angle EFG$ , both being right angles. From Statement 2, $\angle BAC \cong \angle FEG$ , setting up the conditions of the Angle-Angle Postulate; therefore, $\bigtriangleup ABC \sim \bigtriangleup EFG$ .

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Question

Given: Rectangles and with diagonals $\overline{AC}$ and $\overline{EG}$ , respectively.

True or false: $\bigtriangleup ABC \sim \bigtriangleup EFG$

Statement 1: The perimeter of Rectangle is three times that of Rectangle .

Statement 2: The area of Rectangle is nine times that of Rectangle .

Answer

Assume both statements are true. We show the two statements together are insufficent to prove or disprove that $\bigtriangleup ABC \sim \bigtriangleup EFG$

Suppose

.

Then Rectangle has perimeter

and area

$AB \cdot AD = 9 \cdot 3 = 27$

Now set up two cases with different dimensions for Rectangle .

Case 1:

The perimeter of Rectangle is

,

one-third that of Rectangle .

The area of Rectangle is

$EF \cdot EH = 3 \cdot 1 = 3$ ,

one-ninth that of Rectangle .

$\frac{AB }{EF} = \frac{9}{3} = 3$

$\frac{BC}{FG} = \frac{3}{1} = 3$

$m \angle B = m \angle F = 90^{\circ }$ ,

so by the Side-Angle-Side Similarity Theorem, $\bigtriangleup ABC \sim \bigtriangleup EFG$ .

Case 2:

The perimeter of Rectangle is

,

one-third that of Rectangle .

The area of Rectangle is

$EF \cdot EH = 1 \cdot 3= 3$ ,

one-ninth that of Rectangle .

$\frac{AB }{EF} = \frac{9}{1} = 1$

$\frac{BC}{FG} = \frac{3}{3} = 1$

Sides are not in proportion, making the statement $\bigtriangleup ABC \sim \bigtriangleup EFG$ false.

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup XYZ$ , where $\angle B$ and $\angle Y$ are right angles.

True or false: $\bigtriangleup ABC \sim \bigtriangleup XYZ$

Statement 1: Both triangles are isosceles.

Statement 2: The triangles have the same perimeter.

Answer

Assume Statement 1 alone. By the 45-45-90 Theorem, an isosceles right triangle has two acute angles that each measure $45^{\circ }$ ; specifically, $m \angle A = m \angle X = 45^{\circ }$ and $m \angle C = m \angle Z = 45^{\circ }$ . By the Angle-Angle Similarity Postulate, $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

Assume Statement 2 alone. The equality of the perimeters of two right triangles does not imply their similarity or nonsimilarity. For example:

Let

and

Since $\overline{AB } \cong \overline{XY }$ and $\overline{AC} \cong \overline{XZ}$ , it follows by the Hypotenuse-Leg Theorem that $\bigtriangleup ABC \cong \bigtriangleup XYZ$ , and, since congruent triangles are similar, it further follows that $\bigtriangleup ABC \sim \bigtriangleup XYZ$ .

Now, let

and .

The sides are not in proportion, as $\frac{AB}{XY} = \frac{3}{4} e \frac{4}{3} = \frac{BC}{YZ}$ , so $\bigtriangleup ABC sim \bigtriangleup XYZ$ .

In both scenarios, since $3^{2}+4^{2}= 5^{2}$ , by the converse of the Pythagorean Theorem, the triangles are right, and the triangles both have perimeter , thereby satisfying the main premise and Statement 2. Statement 2 alone is therefore insufficient.

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