GMAT Quantitative - Calculating the height of an acute / obtuse triangle

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

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

$7\sqrt{3}$

$7\sqrt{2}$

$14\sqrt{3}$

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,

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

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

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

$12 \sqrt{3}-12$

$24\sqrt{2}- 24$

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

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 ; 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 , 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:

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

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

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

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

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

$= 12 \sqrt{3}-12$

Given: $\bigtriangleup ABC$ with

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?

Between 7 and 8

Between 8 and 9

Between 6 and 7

Between 9 and 10

Between 5 and 6

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 .

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

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

$24 \sqrt{3}$

$48 \sqrt{3}$

$24 \sqrt{2}$

$48 \sqrt{2}$

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.

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

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

$36- 12\sqrt{3}$

$48- 24\sqrt{2}$

$8\sqrt{2}$

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

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 ; 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 , 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:

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

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

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

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

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

$=36- 12\sqrt{3}$

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 $\overline{BM}$ .

$\sqrt{3}:1$

$\sqrt{2}:1$

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 .

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 ; 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 $\sqrt{3}:1$ .

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

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

$10 \sqrt{3}$

$10 \sqrt{2}$

$15 \sqrt{3}$

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.

Given: $\bigtriangleup ABC$ with , 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 $\overline{BN}$ ?

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

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

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

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

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

Given: $\bigtriangleup ABC$ with , 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.

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

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

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

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

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

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:

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

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

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,

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

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?

$\frac{9}{2}$

$9\sqrt{3}$

$18\sqrt{3}$

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: