GMAT Math : Right Triangles

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Example Question #1 : Calculating The Perimeter Of A Right Triangle

What is the perimeter of a 3-4-5 traingle?

Possible Answers:

Correct answer:

Explanation:

A 3-4-5 triangle is a right triangle. To find the perimeter, we must add up all the sides of the triangle.

3+4+5=12

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Example Question #2 : Calculating The Perimeter Of A Right Triangle

A right triangle has a hypotenuse of length 13 yards and a leg of length 5 yards. Give the perimeter of this triangle in inches.

Possible Answers:

$960\textrm{ in}$

$1,040\textrm{ in}$

$1,080\textrm{ in}$

$1,200\textrm{ in}$

$1,400\textrm{ in}$

Correct answer:

$1,080\textrm{ in}$

Explanation:

We can use the Pythagorean Theorem to calculate the length of the second leg in yards of the triangle by setting c = 13, a = 5 in this formula:

$b ^{2} = c ^{2 } - a ^{2}$

$b ^{2} = 13 ^{2 } - 5 ^{2} = 169 - 25 = 144$

$b = \sqrt{ 144} = 12$

The perimeter of the triangle is yards is

P = a + b + c = 5 + 12 + 13 = 30 yards. Multiply this by 36 to covert to inches:

$30 \times 36 = 1,080$ inches.

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Example Question #21 : Triangles

A right triangle has legs of length feet and and $6 \frac{2}{3}$ feet. Give the perimeter of this triangle in yards.

Possible Answers:

$6 \frac{2}{3} \textrm{ yd}$

$7 \textrm{ yd}$

$6 \textrm{ yd}$

$5 \frac{2}{3} \textrm{ yd}$

$7 \frac{1}{3} \textrm{ yd}$

Correct answer:

$6 \frac{2}{3} \textrm{ yd}$

Explanation:

We can use the Pythagorean Theorem to calculate the hypotenuse of the triangle by setting $a = 5, b = 6 \frac{2}{3}= \frac{20}{3}$ in this formula:

$c^{2} = a ^{2}+ b^{2}$

$c^{2} = 5 ^{2}+ \left ( \frac{20}{3} \right ) ^{2}$

$c^{2} = 25 + \frac{400}{9} = \frac{225}{9} + \frac{400}{9} = \frac{625}{9}$

$c = \sqrt{\frac{625}{9} } = \frac{25}{3} = 8 \frac{1}{3}$

The perimeter is $P = a + b + c = 5 + 6\frac{2}{3} + 8 \frac{1}{3} = 20$ feet.

Divide by 3 to convert to yards:

$20 \div 3 = 6 \frac{2}{3}$ yards

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Example Question #1 : Calculating The Perimeter Of A Right Triangle

What is the perimeter of a 30-60-90 triangle?

1) One of the sides measures 10 inches.

2) One of the sides measures 20 inches.

Possible Answers:

Statement 1 ALONE is sufficient, but Statement 2 alone is not sufficient.

Statement 2 ALONE is sufficient, but Statement 1 alone is not sufficient.

BOTH statements TOGETHER are sufficient, but neither statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements 1 and 2 TOGETHER are not sufficient.

Correct answer:

BOTH statements TOGETHER are sufficient, but neither statement ALONE is sufficient.

Explanation:

The hypotenuse of a 30-60-90 triangle measures twice its shorter leg; its longer leg measures $\small \sqrt{3}$ times its shorter leg. Given one side length alone, there is no indication which of the three sides it measures; but given two, one of which is twice the other, as is the case here (10 and 20), 10 must be the shorter leg and 20 the hypotenuse. The longer leg is therefore $10\sqrt{3}$ , and the perimeter is

$P=10+20+10\sqrt{3}=30+10\sqrt{3}$ .

Therefore, both statements together are sufficent but neither alone is sufficient.

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Example Question #5 : Calculating The Perimeter Of A Right Triangle

A right triangle has a base of 4 and a height of 3. What is the perimeter of the triangle?

Possible Answers:

Correct answer:

Explanation:

We are given two sides of the right triangle, so in order to calculate the perimeter we must first find the length of the third side, the hypotenuse, using the Pythagorean theorem:

a^2+b^2=c^2

3^2+4^2=c^2

$c^2=25\rightarrow c=5$

Now that we know the length of the third side, we can add the lengths of the three sides to calculate the perimeter of the right triangle:

P=3+4+5=12

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Example Question #6 : Calculating The Perimeter Of A Right Triangle

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$\Delta BCD$ is a right triangle and CD=4 and DB=3 . What is half the circle's circumference added to the triangle's perimeter?

Possible Answers:

$\frac{5}{2}\pi+12$

$\pi+12$

$5\pi+12$

$3\pi+12$

$\frac{\pi}{2}+6$

Correct answer:

$\frac{5}{2}\pi+12$

Explanation:

As you see that a right triangle as sides 3 and 4, you should always remember that, this right triangle is a Pythagorean Triple, in other words, its sides will be in the ratio 3n,4n,5n where is a constant, this will save you a lot of time. Here we can say that the hypotenuse will be 5. Therefore, the circumference will be $5\pi$ . To get the final answer, we should just divide the circumference by 2 and add the perimeter, being 3+4+5 or 12.

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Example Question #21 : Triangles

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

Possible Answers:

Correct answer:

Explanation:

Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ =\sqrt{2^2+2^2}=\sqrt{8}$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $\sqrt{ (2^2-(\sqrt2)^2)}=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)*\sqrt2*\sqrt2=1$

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Example Question #1 : Calculating The Area Of A Right Triangle

If a right triangle has a hypotenuse that is 10, and one side is 6, find the area of the triangle.

Possible Answers:

Correct answer:

Explanation:

Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

$A=\frac{1}{2}L*W=\frac{1}{2}*6*8=24.$

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Example Question #23 : Triangles

The hypotenuse of a $30 ^{\circ } - 60^{\circ } -90^{\circ }$ triangle is equal to the sidelength of a square. Give the ratio of the area of the square to that of the triangle.

Possible Answers:

4:1

8 :3

16 :3

$16\sqrt{3}:3$

$8\sqrt{3}:3$

Correct answer:

$8\sqrt{3}:3$

Explanation:

Let be the sidelength of the square. Then its area is $s ^{2}$ .

If the hypotenuse of a $30 ^{\circ } - 60^{\circ } -90^{\circ }$ triangle is , its shorter leg is half that, or $\frac{1}{2 }s$ ; its longer leg is $\sqrt{3}$ times the shorter leg, or $\frac{1}{2 }s \sqrt{3}$ . The area of the triangle is half the product of the legs, or

$\frac{1}{2} \cdot \frac{1}{2}s \cdot\frac{1}{2 }s \sqrt{3} =\frac{ \sqrt{3}}{8} s ^{2}$

The ratio of the area of the square to that of the triangle is

$s ^{2}: \frac{ \sqrt{3}}{8} s$ or $1: \frac{ \sqrt{3}}{8}$

or $8\sqrt{3}:3$

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Example Question #24 : Triangles

A triangle on the coordinate plane has vertices (0,0), (c+1,0 ), (0,c-1) .

Which of the following expressions is equal to the area of the triangle?

Possible Answers:

$\frac{1}{2} c^{2} -\frac{1}{2}$

$c^{2} -1$

$c^{2}$

$\frac{1}{2} c^{2} +\frac{1}{2}$

$\frac{1}{2} c^{2}$

Correct answer:

$\frac{1}{2} c^{2} -\frac{1}{2}$

Explanation:

This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length c-1 ; the horizontal leg has length c+ 1 .

Now calculate the area:

$A = \frac{1}{2} bh = \frac{1}{2} (c-1)(c+1) = \frac{1}{2} (c^{2}-1) = \frac{1}{2} c^{2}- \frac{1}{2}$

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GMAT Math : Right Triangles

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating The Perimeter Of A Right Triangle

Example Question #2 : Calculating The Perimeter Of A Right Triangle

Example Question #21 : Triangles

Example Question #1 : Calculating The Perimeter Of A Right Triangle

Example Question #5 : Calculating The Perimeter Of A Right Triangle

Example Question #6 : Calculating The Perimeter Of A Right Triangle

Example Question #21 : Triangles

Example Question #1 : Calculating The Area Of A Right Triangle

Example Question #23 : Triangles

Example Question #24 : Triangles

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