GMAT Math : Geometry

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Example Questions

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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Example Question #621 : Gmat Quantitative Reasoning

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

Construct two altitudes of $\bigtriangleup ABC$ : one from to a point on $\overline{AC}$ , and the other from to a point on $\overleftrightarrow{BC}$ . Give the ratio of the length of $\overline{AN}$ to that of $\overline{BM}$ .

Possible Answers:

3:1

$\sqrt{2}:1$

$\sqrt{3}:1$

1:1

2:1

Correct answer:

$\sqrt{3}:1$

Explanation:

$\bigtriangleup ABC$ is shown below, along with altitudes $\overline{AN}$ and $\overline{BM}$ ; note that $\overline{CB}$ has been extended to a ray $\overrightarrow{CB}$ to facilitate the location of the point .

Isosceles_3

For the sake of simplicity, we will call the measure of $\overline{BM}$ 1; the ratio is the same regarless of the actual measure, and the measure of $\overline{AN}$ willl give us the desired ratio.

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 Theorem, hypotenuse $\overline{AB}$ of $\bigtriangleup AMB$ has length twice that of short leg $\overline{BM}$ , so AB = 2 .

Since an exterior angle of a triangle has as its measure the sum of those of its remote interior angles,

$m \angle ABN = m \angle BAC + m \angle C = 30^{\circ }+ 30 ^{\circ } = 60^{\circ }$ .

By defintiion of an altitude, $\overline{AN}$ is perpendicular to $\overrightarrow{CB}$ , making $\bigtriangleup ANB$ a 30-60-90 triangle. By the 30-60-90 Theorem, shorter leg $\overline{BN}$ of $\bigtriangleup ANB$ has half the length of hypotenuse $\overline{AB}$ , so BN = 1 ; also, longer leg $\overline{AN}$ has length $\sqrt{3}$ times this, or $\sqrt{3}$ .

The correct choice is therefore that the ratio of the lengths is $\sqrt{3}:1$ .

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Example Question #10 : Calculating The Height Of An Acute / Obtuse Triangle

Given: $\bigtriangleup ABC$ with $m \angle A = m \angle C = 30 ^{\circ}$ and AB = 48 .

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

Possible Answers:

$24 \sqrt{3}$

$48 \sqrt{3}$

$24 \sqrt{2}$

$48 \sqrt{2}$

Correct answer:

$24 \sqrt{3}$

Explanation:

$\bigtriangleup ABC$ is shown below, along with altitude $\overline{AN}$ ; note that $\overline{CB}$ has been extended to a ray $\overrightarrow{CB}$ to facilitate the location of the point .

Isosceles_2

Since an exterior angle of a triangle has as its measure the sum of those of its remote interior angles,

$m \angle ABN = m \angle BAC + m \angle C = 30^{\circ }+ 30 ^{\circ } = 60^{\circ }$

By definition of an altitude, $\overline{AN}$ is perpendicular to $\overrightarrow{CB}$ , making $\angle ANB$ a right triangle and $\bigtriangleup ANB$ a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, shorter leg $\overline{BN}$ of $\bigtriangleup ANB$ has half the length of hypotenuse $\overline{AB}$ —that is, half of 48, or 24; longer leg $\overline{AN}$ has length $\sqrt{3}$ times this, or $24 \sqrt{3}$ , which is the correct choice.

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Example Question #11 : Calculating The Height Of An Acute / Obtuse Triangle

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

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

Possible Answers:

$10 \sqrt{2}$

$10 \sqrt{3}$

$15 \sqrt{3}$

Correct answer:

Explanation:

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

Isosceles

Since $m \angle A = 30$ , 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 half the length of hypotenuse $\overline{AB}$ ; this is half of 30, or 15.

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Example Question #12 : Calculating The Height Of An Acute / Obtuse Triangle

Given: $\bigtriangleup ABC$ with AB = 8, BC = 12, AC = 16 , construct two altitudes of $\bigtriangleup ABC$ : one from to a point on $\overleftrightarrow{BC}$ , and another from to a point on $\overleftrightarrow{AC}$ . Which of the following is true of the relationship of the lengths of $\overline{AM}$ and $\overline{BN}$ ?

Possible Answers:

The length of $\overline{BN}$ is nine-sixteenths that of $\overline{AM}$ .

The length of $\overline{BN}$ is two-thirds that of $\overline{AM}$ .

The length of $\overline{BN}$ is four-ninths that of $\overline{AM}$ .

The length of $\overline{BN}$ is twice that of $\overline{AM}$ .

The length of $\overline{BN}$ is three-fourths that of $\overline{AM}$ .

Correct answer:

The length of $\overline{BN}$ is three-fourths that of $\overline{AM}$ .

Explanation:

The area of a triangle is one half the product of the length of any base and its corresponding height; this is $\frac{1}{2} \cdot AC \cdot BN$ , but it is also $\frac{1}{2} \cdot BC \cdot AM$ . Set these equal, and note the following:

$\frac{1}{2} \cdot AC \cdot BN= \frac{1}{2} \cdot BC \cdot AM$

$\frac{1}{2} \cdot 16 \cdot BN= \frac{1}{2} \cdot 12 \cdot AM$

$8 \cdot BN= 6 \cdot AM$

$\frac{8 \cdot BN}{8}= \frac{6 \cdot AM}{8}$

$BN = \frac{3}{4} \cdot AM$

That is, the length of $\overline{BN}$ is three fourths that of that of $\overline{AM}$ .

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Example Question #381 : Geometry

Find the equation of the line that is perpendicular to the line connecting the points $\dpi{100} \small (0,-4)\ and\ (-1,-7)$ .

Possible Answers:

$\dpi{100} \small y=3x-1$

the line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$

$\dpi{100} \small y=\frac{x}{3}+1$

$\dpi{100} \small y=-4x+8$

the line between points $\dpi{100} \small (0,0)\ and\ (2,2)$

Correct answer:

the line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$

Explanation:

Lines are perpendicular if their slopes are negative reciprocals of each other. First we need to find the slope of the line in the question stem.

$slope = \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-7 + 4}{-1 - 0} = \frac{-3}{-1} = 3$

The negative reciprocal of 3 is $\dpi{100} \small -\frac{1}{3}$ , so our answer will have a slope of $\dpi{100} \small -\frac{1}{3}$ . Let's go through the answer choices and see.

$\dpi{100} \small y=3x-1$ : This line is of the form $\dpi{100} \small y=mx+b$ , where $\dpi{100} \small m$ is the slope. The slope is 3, so this line is parallel, not perpendicular, to our line in question.

$\dpi{100} \small y=-4x+8$ : The slope here is $\dpi{100} \small -4$ , also wrong.

$\dpi{100} \small y=\frac{x}{3}+1$ : The slope of this line is $\dpi{100} \small \frac{1}{3}$ . This is the reciprocal, but not the negative reciprocal, so this is also incorrect.

The line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$ : $\dpi{100} \small slope = \frac{2}{(-3-3)}=\frac{2}{-6}=-\frac{1}{3}$ .

This is the correct answer! Let's check the last answer choice as well.

The line between points $\dpi{100} \small (0,0)\ and\ (2,2)$ : $\dpi{100} \small slope = \frac{2}{2}=1$ , which is incorrect.

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Example Question #382 : Geometry

Determine whether the lines with equations 3y+x=2 and 4y-2x=3 are perpendicular.

Possible Answers:

They are not perpendicular

They are perpendicular

There is not enough information to determine the answer

Correct answer:

They are not perpendicular

Explanation:

If two equations are perpendicular, then they will have inverse negative slopes of each other. So if we compare the slopes of the two equations, then we can find the answer. For the first equation we have $3y+x=2\Rightarrow y=\frac{-1}{3}x+\frac{2}{3}$

so the slope is $-\frac{1}{3}$ .

So for the equations to be perpendicular, the other equation needs to have a slope of 3. For the second equation, we have

$4y-2x=3\Rightarrow y=\frac{1}{2}x+\frac{3}{4}$

so the slope is $\frac{1}{2}$ .

Since the slope of the second equation is not equal to 3, then the lines are not perpendicular.

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Example Question #2 : Lines

Transversal

Refer to the above figure. l || m . True or false: $m \perp t$

Statement 1: $\angle 3 \cong \angle 5$

Statement 2: $\angle 2$ and $\angle 6$ are supplementary.

Possible Answers:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Correct answer:

EITHER statement ALONE is sufficient to answer the question.

Explanation:

If transversal crosses two parallel lines and , then same-side interior angles are supplementary, so $\angle 3$ and $\angle 5$ are supplementary angles. Also, corresponding angles are congruent, so $\angle 2 \cong \angle 6$ .

By Statement 1 alone, angles $\angle 3$ and $\angle 5$ are congruent as well as supplementary; by Statement 2 alone, $\angle 2$ and $\angle 6$ are also supplementary as well as congruent. Two angles that are both supplementary and congruent are both right angles, so from either statement alone, and intersect at right angles, so, consequently, $m \perp t$ .

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Example Question #383 : Geometry

Transversal

Figure NOT drawn to scale.

Refer to the above figure.

True or false: $m \perp t$

Statement 1: $\angle 4$ is a right angle.

Statement 2: $\angle 3$ and $\angle 5$ are supplementary.

Possible Answers:

BOTH statements TOGETHER are insufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Explanation:

Statement 1 alone establishes by definition that $l\perp t$ , but does not establish any relationship between and .

By Statement 2 alone, since same-side interior angles are supplementary, l || m , but no conclusion can be drawn about the relationship of , since the actual measures of the angles are not given.

Assume both statements are true. If two lines are parallel, then any line in their plane perpendicular to one must be perpendicular to the other. l || m and $l\perp t$ , so it can be established that $m \perp t$ .

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

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

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

Example Question #621 : Gmat Quantitative Reasoning

Example Question #10 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #11 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #12 : Calculating The Height Of An Acute / Obtuse Triangle

Example Question #381 : Geometry

Example Question #382 : Geometry

Example Question #2 : Lines

Example Question #383 : Geometry

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