Right Triangles - GMAT Quantitative

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Question

What is the perimeter of isosceles triangle ABC?

(1)

(2)

Answer

(1) This statement gives us the length of one side of the triangle. This information is insufficient to solve for the perimeter.

(2) This statement gives us the length of one side of the triangle. This information is insufficient to solve for the perimeter.

Additionally, since we don't know which one of the sides ( or ) is one of the equal sides, it's impossible to determine the perimeter given the information provided.

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Question

Find the perimeter of right triangle $\Delta XLC$ .

I)

II) $CX=5\sqrt{2}$

Answer

If the two shorter sides of a right triangle are equal, that means our other two angles are 45 degrees. This means our triangle follows the ratios for a 45/45/90 triangle, so we can find the remaining sides from the length of the hypotenuse.

I) Tells us we have a 45/45/90 triangle. The ratio of side lengths for a 45/45/90 triangle is $x:x:x\sqrt2$ .

II) Tells us the length of the hypotenuse.

Together, we can find the remaining two sides and then the perimeter.

$CX=5\sqrt2\rightarrow XL=5, CL=5$

$P=5\sqrt2 +5+5$

$P=10+5\sqrt2$

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Question

Find the perimeter of the right triangle.

The product of the base and height measures .

The hypotenuse measures .

Answer

Statement 1: We need additional information.

$b \times h =48$

But this can mean our base and height measure 2 and 24, 4 and 12, or 6 and 8.

We cannot determine which one based solely on this statement.

Statement 2: We're given the length of the hypotenuse so we can narrow down the possible base and height values.

We have to see which pair of values makes the statement true.

The only pair that does is 6 and 8.

We can now find the perimeter of the right triangle:

or, if you're more familiar with the equation $perimeter=a+b+\sqrt{a^2+b^2}$ , then:

$6+8+\sqrt{6^2+8^2}=24$

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Question

Calculate the perimeter of the triangle.

The hypotenuse of the right triangle is .

The legs of the right triangle measure and .

Answer

Statement 1: In order to find the perimeter of a right triangle, we need to know the lengths of the legs, not the hypotenuse.

Statement 2: Since we have the values to both of the legs' lengths, we can just plug it into the equation for the perimeter:

$p=a+b+\sqrt{a^2+b^2}$

$p=3+4+\sqrt{9+16}=3+4+\sqrt{25}=12cm$

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Question

Two ships left New York at the same time. One ship has been moving due east the entire time at a speed of 50 nautical miles per hour. How far apart are the ships now?

The other ship has been moving due south the entire time.

The other ship has been moving at a rate of 60 nautical miles per hour the entire time.

Answer

In order to determine how far apart the ships are now, it is necessary to know how far each ship is from New York now. Even knowing both statements, however, you only know that the paths are at right angles, and that the ratio of the two distances is 6 to 5. Without knowing the time elapsed, which is not given, you cannot tell how far apart the ships are.

The answer is that both statements together are insufficent to answer the question.

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Question

Right triangle is similar to triangle .

Find the hypotenuse of triangle

I) Triangle has side lengths of .

II) The shortest side of is .

Answer

We need both statements here because we need to work with ratios.

If you notice that a 12/16/20 triangle follows the 3/4/5 triangle ratio, then it is even easier. Otherwise, you can set up the following proportion and solve to find the answer.

$\small \frac{12}{24}=\frac{20}x$

Where x is the hypotenuse of triangle HJK.

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Question

Find the hypotenuse of $\Delta HLC$ .

I) The shortest side is half of the second shortest side.

II) The middle side is fathoms long.

Answer

To find the hypotenuse of a right triangle we can use the classic Pythagorean Theorem. To do so, however, we need to know the other two sides.

I) Tells us how to relate the other two sides to eachother.

$SS=\frac{1}{2}MS$

II) Gives us the length of one side.

Use II) and I) to find the two short sides. Then use Pythagorean Theorem to find the perimeter.

$SS=\frac{1}{2}\cdot 13=\frac{13}{2}$

, where SS represents the shortest side, MS represents the middle side, and H represents the hypotenuse.

$\left(\frac{13}{2} \right )^2+13^2=211.25 \rightarrow H=14.5$

$P=\frac{13}{2}+13+14.5$

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Question

Find the length of the hypotenuse of the right triangle.

The area of the triangle is .

The legs are measured to be and .

Answer

Statement 1: We'll need more information to answer the question.

$A=\frac{b\times h}{2}=6$

$b\times h=12$

Which means the base and height can either be 4 and 3, or 6 and 2. We cannot find the length of the hypotenuse unless we know which one.

Statement 2: We're given the lengths of the base and height so we can just plug them into the Pythagorean Theorem.

$c=\sqrt{a^2+b^2}=\sqrt{3^2+4^2}=\sqrt{25}=5in$

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Question

Calculate the length of the hypotenuse of a right triangle.

The area of the right triangle is .

One of the angles is $35^{\circ}$ .

Answer

Statement 1:In order to find the hypotenuse, we need to find the values for the base and length. Although we're given the area, we need additional information.

$A=\frac{1}{2}b\times h=18$

$36= b\times h$

but the base and height can be 2 and 18, 3 and 12, 4 and 9, or 6 and 6.

Statement 2: We don't need this information to find the length of the hypotenuse.

Statements 1 and 2 are not sufficient, and additional data is needed to answer the question.

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Question

Calculate the length of the hypotenuse of the right triangle.

The area of the right triangle is .

The difference between the height and base is .

Answer

Statement 1: We're given the area but require additional information.

$A=\frac{1}{2}b \times h=15$

$b \times h=30$

The base and height, however, can be 2 and 15, 3 and 10, or 5 and 6.

Statement 2: We're provided the information that narrows our possible values. The difference between the base and height is so our values must be 5 and 6.

Using BOTH statements, we can find the hypotenuse of the right triangle via the Pythagorean Theorem:

$c\approx7.8cm$

Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.

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Question

$\bigtriangleup ABC$ has right angle $\angle B$ ; $\bigtriangleup DEF$ has right angle $\angle E$ . Which, if either, is longer, $\overline{AC}$ or $\overline{DF}$ ?

Statement 1: $m \angle A = 30^{\circ }$

Statement 2: $m \angle D = 45^{\circ }$

Answer

The two statements together only give information about the angle measures of the two triangles. Without any information about the relative or absolute lengths of the sides, no comparison can be drawn between their hypotenuses.

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Question

Data sufficiency question- do not actually solve the question

The lengths of two sides of a triangle are 6 and 8. What is the length of the third side?

1. The length of the longest side is 8.

2. The triangle contains a right angle.

Answer

Knowing that the triangle has a right angle indicates the question can be solved using the Pythagorean Theorem, however it is unclear which side is the hypotenuse. For example, if 8 is not the hypotenuse, the length of the third side is 10. If 8 is the hypotenuse, the length of the third side is 5.3.

Additionally, if you only know that 8 is the longest side, the length of the third side could be anything greater than 2 and less than 8. Therefore, having both pieces of data will allow you to solve the problem.

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Question

Which of the three sides of $\Delta ABC$ has the greatest measure?

Statement 1: $\angle A$ and $\angle B$ are complementary angles.

Statement 2: $\angle C$ is not an acute angle.

Answer

The side opposite the angle of greatest measure is the longest of the three, so if we can determine which angle is the longest, we can answer this question.

It follows from Statement 1 by definition that $m\angle A + m\angle B = 90^{\circ }$ , so, since the measures of the three angles total $180^{\circ }$ , $m\angle C = 90^{\circ }$ , making $\angle C$ right and the other two acute. This proves $\angle C$ has the greatest measure of the three.

It follows from Statement 2 that $\angle C$ is either right or obtuse; therefore, $m\angle C \geq 90^{\circ }$ . Subsequently, the other two angles are acute, so again, $\angle C$ has the greatest measure of the three.

From either statement alone, we can therefore identify $\overline{AB}$ as the side of greatest measure.

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Question

What is the base length of the right triangle?

The width is four times the length.

The area of the right triangle is .

Answer

Statement 1: All we're given is the equation for finding the width, , which we'll use in the next statement.

Statement 2: Using the information from statement 1, we can set up an equation and solve for the length.

$A=\frac{1}{2}l\times w= \frac{1}{2} (l)(4l)=\frac{1}{2}4l^2$

$18=\frac{1}{2}\times 4 l^2$

Statement 2 alone would not have provided sufficient information because we would have ended up with

$36=l \times w$

and would not have been able to determine what the the values were.

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Question

$\bigtriangleup ABC$ is a right triangle with right angle $\angle B$ . Evaluate .

Statement 1: and $m \angle A = 30^{\circ }$ .

Statement 2: and $m \angle C = 60^{\circ }$ .

Answer

Either statement alone is sufficient.

From either statement alone, it can be determined that $m \angle A = 30^{\circ }$ and $m \angle C = 60^{\circ }$ ; each statement gives one angle measure, and the other can be calculated by subtracting the first from $90^{\circ }$ , since the acute angles of a right triangle are complementary.

Also, since $\angle B$ is the right angle, $\overline{AC}$ is the hypotenuse, and $\overline{BC}$ , opposite the $30 ^{\circ }$ angle, the shorter leg of a 30-60-90 triangle. From either statement alone, the 30-60-90 Theorem can be used to find the length of longer leg $\overline{AB}$ . From Statement 1 alone, $\overline{AB}$ has length $\frac{\sqrt{3}}{2}$ times that of the hypotenuse, or $\frac{\sqrt{3}}{2} \cdot 22 = 11\sqrt{3}$ . From Statement 2 alone, $\overline{AB}$ has length $\sqrt{3}$ of the shorter leg, or $11\sqrt{3}$ .

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Question

$\bigtriangleup ABC$ is a right triangle with right angle $\angle B$ . Evaluate .

Statement 1: $\bigtriangleup ABC$ can be inscribed in a circle with circumference $20 \pi$ .

Statement 2: $\bigtriangleup ABC$ can be inscribed in a cricle with area $100 \pi$ .

Answer

From Statement 1 alone, the circumscribed circle has as its diameter the circumference $20 \pi$ divided by $\pi$ , or 20. From Statement 2 alone, the circle has area $100 \pi$ , so its radius can be found using the area formula;

$\pi r^{2}= A$

$\pi r^{2}= 100 \pi$

$r^{2}= 100$

The diameter is the radius doubled, which here is 20.

The hypotenuse of a right triangle is a diameter of the circle that circumscribes it, so the diameter of the circle gives us the length of the hypotenuse. However, we are looking for the length of a leg, $\overline{AB}$ . Either statement alone gives us only the length of the hypotenuse, which, without other information, does not give us any further information about the right triangle.

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Question

$\bigtriangleup ABC$ is a right triangle with right angle $\angle B$ . Evaluate .

Statement 1: $\bigtriangleup ABC$ has area 24.

Statement 2: $\bigtriangleup ABC$ can be circumscribed by a circle with area $25 \pi$ .

Answer

Since $\angle B$ is given as the right angle of the triangle $\bigtriangleup ABC$ , we are being asked to evaluate the length of hypotenuse $\overline{AC}$ .

Statement 1 alone gives insufficient information. We note that the area of a right triangle is half the product of the lengths of its legs, and we examine two scenarios:

Case 1:

The area is $\frac{1}{2} \cdot 6 \cdot 8 = 24$

By the Pythagorean Theorem, hypotenuse $\overline{AC}$ has length

$\sqrt{6^{2}+8^{2}} = \sqrt{36+64} = \sqrt{100}= 10$

Case 2:

The area is $\frac{1}{2} \cdot 4 \cdot 12 = 24$

By the Pythagorean Theorem, hypotenuse $\overline{AC}$ has length

$\sqrt{4^{2}+12^{2}} = \sqrt{16+144} = \sqrt{160}= \sqrt{16} \cdot \sqrt{10} = 4\sqrt{10}$

Both triangles have area 24 but the hypotenuses have different lengths.

Assume Statement 2 alone. A circle that circumscribes a right triangle has the hypotenuse of the triangle as one of its diameters, so the length of the hypotenuse is the diameter - or, twice the radius - of the circle. Since the area of the circumsctibed circle is $25 \pi$ , its radius can be determined using the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 25 \pi$

$r^{2}= 25$

The diameter - and the length of hypotenuse $\overline{AC}$ - is twice this, or 10.

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Question

$\bigtriangleup ABC$ is a right triangle. Evaluate .

Statement 1: and

Statement 2: $\bigtriangleup ABC$ is not a 30-60-90 triangle.

Answer

Statement 1 alone gives insufficient information. and , but it is not clear which of the three sides is the hypotenuse of $\bigtriangleup ABC$ . $\overline {BC}$ is not the longest side, so we know that $\overline {AB}$ or $\overline {AC}$ is the hypotenuse, and the other is the second leg. We explore the two possibilities:

If $\overline {AB}$ is the hypotenuse, then the legs are $\overline {AC}$ and $\overline {BC}$ ; since the lengths of the legs are 12 and 24, by the Pythagorean Theorem, $\overline {AB}$ has length

$\sqrt{24^{2}+12^{2}} = \sqrt{576+144}= \sqrt{720} = \sqrt{144} \cdot \sqrt{5} = 12 \sqrt{5}$ .

If $\overline {AB}$ is a leg, then the hypotenuse, being the longest side, is $\overline {AC}$ , and $\overline {BC}$ is the other leg; by the Pythagorean Theorem, $\overline {AB}$ has length

$\sqrt{24^{2}-12^{2}} = \sqrt{576-144}= \sqrt{432} = \sqrt{144} \cdot \sqrt{3} = 12 \sqrt{3}$ .

Statement 2 alone gives insufficient information in that it only gives information about the angles, not the sides.

Assume both statements are true. If $\overline {AC}$ is the hypotenuse and $\overline {BC}$ is a leg, then, since the hypotenuse measures twice the length of a leg from Statement 1, the triangle is 30-60-90, contradicting Statement 2. Therefore, by elimination, $\overline {AB}$ is the hypotenuse, and, consequently, $AB= 12 \sqrt{5}$ .

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Question

$\bigtriangleup ABC$ is a right triangle. Evaluate .

Statement 1: and

Statement 2: $\bigtriangleup ABC$ has a $60^{\circ }$ angle.

Answer

Assume Statement 1 alone. Since we do not know whether $\overline {AB}$ is the hypotenuse or a leg of $\bigtriangleup ABC$ , we can show that can take one of two different values.

Case 1: If $\overline {AB}$ is the hypotenuse, then the legs are $\overline {AC}$ and $\overline {BC}$ ; since their lengths are 10 and 20, by the Pythagorean Theorem, $\overline {AB}$ has length

$\sqrt{20^{2}+10^{2}} = \sqrt{400+100}= \sqrt{500} = \sqrt{100} \cdot \sqrt{5} = 10 \sqrt{5}$ .

Case 2: If $\overline {AB}$ is a leg, then the hypotenuse, being the longest side, is $\overline {AC}$ , and $\overline {BC}$ is the other leg; by the Pythagorean Theorem, $\overline {AB}$ has length

$\sqrt{20^{2}-10^{2}} = \sqrt{400-100}= \sqrt{300}= \sqrt{100} \cdot \sqrt{3} = 10 \sqrt{3}$ .

Statement 2 gives insufficient information, since it only clues us in to the measures of the angles, not the sides.

Now assume both statements. Since one of the angles of the right triangle has measure $60^{\circ }$ , the other has measure $90^{\circ } - 60^{\circ }= 30^{\circ }$ ; the triangle is a 30-60-90 triangle, and therefore, its hypotenuse has twice the length of its shorter leg. Since, from Statement 1, $BC \cdot 2 = AC$ , $\overline {AC}$ is the hypotenuse, and $\overline {AB}$ is the longer leg, the length of which, by the 30-60-90 Theorem, is $\sqrt{3}$ times that of shorter leg $\overline {BC}$ , or $10\sqrt{3}$ .

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Question

$\bigtriangleup ABC$ is a right triangle. Evaluate .

Statement 1:

Statement 2:

Answer

Neither statement alone gives enough information to find , as each alone gives only one sidelength.

Assume both statements are true. While neither side is indicated to be a leg or the hypotenuse, the hypotenuse of a right triangle is longer than either leg; therefore, since $\overline {AC}$ and $\overline {BC}$ are of equal length, they are the legs. $\overline {AB}$ is the hypotenuse of an isosceles right triangle with legs of length 10, and by the 45-45-90 Theorem, the length of $\overline{AB}$ is $\sqrt{2}$ times that of a leg, or $10 \sqrt{2}$ .

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