GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #2 : Calculating If Two Acute / Obtuse Triangles Are Similar

$\Delta ABC \sim \Delta D EF$

AB = 43, BC = 45, AC = 53 .

Order the angles of $\Delta D EF$ from least to greatest measure.

Possible Answers:

$\angle F, \angle D, \angle E$

The angles of $\Delta D EF$ cannot be ordered from the information given.

$\angle E, \angle F, \angle D$

$\angle E, \angle D, \angle F$

$\angle F, \angle E, \angle D$

Correct answer:

$\angle F, \angle D, \angle E$

Explanation:

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. AB < BC < AC , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

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Example Question #51 : Acute / Obtuse Triangles

The triangles are similar. What is the value of x?

Possible Answers:

$Not\ enough\ information$

Correct answer:

Explanation:

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\small \frac{8}{12}\ =\ \frac{10}{x}$ . $\small 8x\ =\ 120; x\ =\ 15$ .

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Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

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Triangle ABC has height . What is the length of , knowing that DC=4 and BC=5 ?

Possible Answers:

$2\sqrt{3}$

$\sqrt{6}$

$\sqrt{3}$

Correct answer:

Explanation:

To solve this equation, we need to calculate the length of the height with the Pythagorean Theorem.

We could also recognize that since DC=4 and BC=5 , the triangle is a Pythagorean Triple, in other words, its sides will be in ratio 3n,4n,5n where is a constant.

Here n=1 and therefore, the length of height BD must be , which is our final answer.

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Example Question #374 : Geometry

The largest angle of an obtuse isosceles triangle is $120^{\circ}$ . If two of the sides have an equal length of , what is the height of the triangle?

Possible Answers:

$9\sqrt{3}$

$\frac{9}{2}$

$18\sqrt{3}$

Correct answer:

Explanation:

If the largest angle of the obtuse isosceles triangle is $120^{\circ}$ , then this is the unique angle in between the two sides with an equal length of . We can imagine that the height of this isosceles triangle is simply the third side of a triangle formed by half of its base and the length of either equal side. That is, if we bisected the $120^{\circ}$ angle with a line perpendicular to the base of the obtuse isosceles triangle, this line would be the height of the triangle. If we bisected the $120^{\circ}$ angle, we would have two congruent triangles with angles of $60^{\circ}$ between the height and each side of equal length. This means the cosine of that angle will be equal to the length of the height over the length of either equal side, which gives us:

$\cos (60^{\circ})=\frac{h}{18}$

$h=18\cos(60^{\circ})=18\left(\frac{1}{2}\right)=9$

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Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

One angle of an obtuse isosceles triangle has a measure of $120^{\circ}$ . If the length of the two equivalent sides is , what is the height of the triangle?

Possible Answers:

$2\sqrt{30}$

$\sqrt{14}$

Correct answer:

Explanation:

If one measure of an obtuse isosceles triangle is $120^{\circ}$ , then this is obviously the unique angle that classifies the triangle as obtuse, which tells us that this is the angle between the two sides with an equivalent length of . The height of the triangle is given by a line that bisects this angle. This tells us that the angle between the height and the sides of equivalent length is $60^{\circ}$ , and because we know the length of the equivalent sides we can solve for the height as follows, where is the height of the triangle and is the length of the equivalent sides:

$\cos60^{\circ}=\frac{h}{l}$

$\frac{1}{2}=\frac{h}{14}$

$h=\frac{1}{2}(14)=7$

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Example Question #375 : Geometry

Given: $\bigtriangleup ABC$ with $m \angle A = m \angle C = 30 ^{\circ}$ and AC= 42 .

Construct the altitude of $\bigtriangleup ABC$ from to a point on $\overline{AC}$ . What is the length of $\overline{BM}$ ?

Possible Answers:

$7\sqrt{2}$

$14\sqrt{3}$

$7\sqrt{3}$

Correct answer:

$7\sqrt{3}$

Explanation:

$\bigtriangleup ABC$ is shown below, along with altitude $\overline{BM}$ .

Isosceles

By the Isosceles Triangle Theorem, since $\angle A \cong \angle C$ , $\bigtriangleup ABC$ is isosceles with $\overline{AB } \cong \overline {CB}$ . By the Hypotenuse-Leg Theorem, the altitude cuts $\bigtriangleup ABC$ into congruent triangles $\bigtriangleup AMB$ and $\bigtriangleup CMB$ , so $\overline{AM } \cong \overline {CM}$ ; this makes the midpoint of $\overline{AC}$ . $\overline{AC }$ has length 42, so $\overline{AM}$ measures half this, or 21.

Also, since $m \angle A = 30^{\circ }$ , and $\overline{BM}$ , by definition, is perpendicular to $\overline{AC}$ , $\bigtriangleup AMB$ is a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, $\overline{BM}$ , as the shorter leg of $\bigtriangleup AMB$ , has length equal to that of longer leg $\overline{AM}$ divided by $\sqrt{3}$ ; that is,

$BM = \frac{AM}{\sqrt{3}} = \frac{21}{\sqrt{3}} = \frac{21\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}= \frac{21\cdot \sqrt{3}}{3} = 7 \sqrt{3}$

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Example Question #611 : Problem Solving Questions

Given: $\bigtriangleup ABC$ with $m \angle A = 45^{\circ }$ , $m \angle C = 60 ^{\circ}$ , AC= 24 .

Construct the altitude of $\bigtriangleup ABC$ from to a point on $\overline{AC}$ . What is the length of $\overline{BM}$ ?

Possible Answers:

$48- 24\sqrt{2}$

$24\sqrt{2} - 12\sqrt{3}$

$36- 12\sqrt{3}$

$8\sqrt{2}$

Correct answer:

$36- 12\sqrt{3}$

Explanation:

$\bigtriangleup ABC$ is shown below, along with altitude $\overline{BM}$ .

Triangle_1

Since $\overline{BM}$ is, by definition, perpendicular to $\overline{AC}$ , it divides the triangle into 45-45-90 triangle $\bigtriangleup AMB$ and 30-60-90 triangle $\bigtriangleup BMC$ .

Let be the length of $\overline{BM}$ . By the 45-45-90 Theorem, $\overline{BM}$ and $\overline{AM}$ , the legs of $\bigtriangleup AMB$ , are congruent, so AM = BM = h ; by the 30-60-90 Theorem, short leg $\overline{MC}$ of $\bigtriangleup BMC$ has as its length that of $\overline{BM}$ divided by $\sqrt{3}$ , or $\frac{h}{ \sqrt{3}}= \frac{h \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}} = \frac{h \sqrt{3}}{ 3}$ . Therefore, the length of $\overline{AC}$ is:

$AC = AM + MC =h+\frac{h \sqrt{3} }{3}= h\left ( 1+ \frac{\sqrt{3}}{3} \right ) = h\left ( \frac{3+ \sqrt{3}}{3} \right )$

We are given that AC= 24 , so

$h\left ( \frac{3+ \sqrt{3}}{3} \right ) = 24$

$h\left ( \frac{3+ \sqrt{3}}{3} \right ) \left ( \frac{3}{3+ \sqrt{3}} \right )= 24 \left ( \frac{3}{3+ \sqrt{3}} \right )$

$h = \frac{72}{3+ \sqrt{3}}$

We can simplify this by multiplying both numerator and denominator by $3- \sqrt{3}$ , thereby rationalizing the denominator:

$h = \frac{72\left (3- \sqrt{3} \right )}{\left (3+ \sqrt{3} \right )\left (3- \sqrt{3} \right )}$

$= \frac{72\left (3- \sqrt{3} \right )}{3^{2}- \left (\sqrt{3} \right )^{2}}$

$= \frac{72\left (3- \sqrt{3} \right )}{9-3}$

$= \frac{72\left (3- \sqrt{3} \right )}{6}$

$= 12\left (3- \sqrt{3} \right )$

$=36- 12\sqrt{3}$

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Example Question #6 : Calculating The Height Of An Acute / Obtuse Triangle

Given: $\bigtriangleup ABC$ with AB = BC = 8, AC=10

Construct the altitude of $\bigtriangleup ABC$ from to a point on $\overline{BC}$ . Between which two consecutive integers does the length of $\overline{AN}$ fall?

Possible Answers:

Between 7 and 8

Between 6 and 7

Between 8 and 9

Between 5 and 6

Between 9 and 10

Correct answer:

Between 7 and 8

Explanation:

Construct two altitudes of the triangle, one from to a point on $\overline{AC}$ , and the one stated in the question.

Isosceles_4

$\bigtriangleup ABC$ is isosceles, so the median $\overline{BM}$ cuts it into two congruent triangles; is the midpoint, so (as marked above) $\overline{AM}$ has length half that of $\overline{AC}$ , or half of 10, which is 5. By the Pythagorean Theorem,

$BM = \sqrt{\left (AB \right )^{2} - \left ( AM\right )^{2} }$

$= \sqrt{8^{2} - 5^{2} }$

$= \sqrt{64- 25 }$

$= \sqrt{39 }$

The area of a triangle is one half the product of the length of any base and its corresponding height; this is $\frac{1}{2} \cdot AC \cdot BM$ , but it is also $\frac{1}{2} \cdot BC \cdot AN$ . Since we know all three sidelengths other than that of $\overline{AN}$ , we can find the length of the altitude $\overline{AN}$ by setting the two expressions equal to each other and solving for :

$\frac{1}{2} \cdot BC \cdot AN = \frac{1}{2} \cdot AC \cdot BM$

$\frac{1}{2} \cdot 8 \cdot AN = \frac{1}{2} \cdot 10 \cdot \sqrt{39}$

$4 \cdot AN = 5 \sqrt{39}$

$\frac{4 \cdot AN}{4} = \frac{5 \sqrt{39}}{4}$

$AN= \frac{5 \sqrt{39}}{4}$

To find out what two integers this falls between, square it:

$(AN)^{2}=\left ( \frac{5 \sqrt{39}}{4} \right )^{2}= \frac{25 \cdot 39}{16} = 60 \frac{15}{16}$

Since $49 < (AN)^{2}< 64$ , it follows that 7 < AN< 8 .

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Example Question #7 : Calculating The Height Of An Acute / Obtuse Triangle

Given: $\bigtriangleup ABC$ with $m \angle A = 45^{\circ }$ , $m \angle C = 30 ^{\circ}$ , AC= 24 .

Construct the altitude of $\bigtriangleup ABC$ from to a point on $\overline{AC}$ . What is the length of $\overline{BM}$ ?

Possible Answers:

$24\sqrt{3}- 24\sqrt{2}$

$24\sqrt{2}- 24$

$12 \sqrt{3}-12$

Correct answer:

$12 \sqrt{3}-12$

Explanation:

$\bigtriangleup ABC$ is shown below, along with altitude $\overline{BM}$ .

Triangle_1

Since $\overline{BM}$ is, by definition, perpendicular to $\overline{AC}$ , it divides the triangle into 45-45-90 triangle $\bigtriangleup AMB$ and the 30-60-90 triangle $\bigtriangleup BMC$ .

Let be the length of $\overline{BM}$ . By the 45-45-90 Theorem, $\overline{BM}$ , and $\overline{AM}$ , the legs of $\bigtriangleup AMB$ , are congruent, so AM = BM = h ; by the 30-60-90 Theorem, long leg $\overline{MC}$ of $\bigtriangleup BMC$ has length $\sqrt{3}$ times that of $\overline{BM}$ , or $h \sqrt{3}$ . Therefore, the length of $\overline{AC}$ is:

$AC = AM + MC =h+h \sqrt{3} = h (\sqrt{3}+ 1)$

We are given that AC= 24 , so

$s(\sqrt{3}+ 1) = 24$

and

$s = \frac{24}{\sqrt{3}+1}$

We can simplify this by multiplying both numerator and denominator by $\sqrt{3} - 1$ , thereby rationalizing the denominator:

$s = \frac{24\left (\sqrt{3}-1 \right )}{\left (\sqrt{3}+1 \right )\left (\sqrt{3}-1 \right )}$

$= \frac{24\left (\sqrt{3}-1 \right )}{ \left (\sqrt{3} \right )^{2}-\left (1 \right )^{2}}$

$= \frac{24\left (\sqrt{3}-1 \right )}{ 3-1}$

$= \frac{24\left (\sqrt{3}-1 \right )}{2}$

$= 12 \left (\sqrt{3}-1 \right )$

$= 12 \sqrt{3}-12$

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Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

Given: $\bigtriangleup ABC$ with AB = 12, BC = 8, AC = 17 , construct three altitudes of $\bigtriangleup ABC$ - one from to a point on $\overleftrightarrow{BC}$ , another from to a point on $\overleftrightarrow{AC}$ , and a third from to a point on $\overleftrightarrow{AB}$ . Order the altitudes, $\overline{AM}$ , $\overline{BN}$ , and $\overline{CO}$ from shortest to longest.

Possible Answers:

$\overline{CO},\overline{BN},\overline{AM}$

$\overline{CO},\overline{AM},\overline{BN}$

$\overline{BN}, \overline{CO},\overline{AM}$

$\overline{AM},\overline{BN},\overline{CO}$

$\overline{AM},\overline{CO},\overline{BN}$

Correct answer:

$\overline{BN}, \overline{CO},\overline{AM}$

Explanation:

The area of a triangle is half the product of the lengths of a base and that of its corresponding altitude. If we let and (height) stand for those lengths, respectively, the formula is

$a = \frac{1}{2} bh$ ,

which can be restated as:

$\frac{1}{2} bh = a$

$2 \cdot \frac{1}{2} bh =2a$

bh =2a

It follows that in the same triangle, the length of an altitude is inversely proportional to the length of the corresponding base, so the longest base will correspond to the shortest altitude, and vice versa.

Since, in descending order by length, the sides of the triangle are

$\overline{AC } , \overline{AB } , \overline{BC }$ ,

their corresponding altitudes are, in ascending order by length,

$\overline{BN}, \overline{CO},\overline{AM}$ .

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #2 : Calculating If Two Acute / Obtuse Triangles Are Similar

Example Question #51 : Acute / Obtuse Triangles

Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #374 : Geometry

Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #375 : Geometry

Example Question #611 : Problem Solving Questions

Example Question #6 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #7 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #1 : Calculating The Height Of An Acute / Obtuse Triangle

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