GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #12 : Right Triangles

What is the length of a right triangle hypotenuse if the the other two sides are and units long?

Possible Answers:

Correct answer:

Explanation:

To find a length of a side in a right triangle, we can use the Pythagorean theorem:

a^2 + b^2 = c^2

5^2 + 12^2 = c^2

25 + 144 = c^2

169 = c^2

13 = c

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Example Question #501 : Problem Solving Questions

What is the hypotenuse of a right triangle with sides 14 and 48?

Possible Answers:

$\dpi{100} \small 52$

$\dpi{100} \small 55$

$\dpi{100} \small 48$

$\dpi{100} \small 25$

$\dpi{100} \small 50$

Correct answer:

$\dpi{100} \small 50$

Explanation:

For starters, we know that the hypotenuse is the longest side of a triangle, so we can immediately cross off 25 as an answer choice because it is smaller than one of the legs, 48. To find the hypotenuse, we can use the Pythagorean Theorem.

$hypotenuse^{^{2}} = 14^{2} + 48^{2} = 2500$

$hypotenuse = 50$

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Example Question #14 : Right Triangles

A right triangle has a height of 8 and a base of 15. What is the length of its hypotenuse?

Possible Answers:

Correct answer:

Explanation:

We are given the length of the height and the base, so we can use the Pythagorean theorem to calculate the length of the hypotenuse:

a^2+b^2=c^2

8^2+15^2=c^2

c^2=64+225

$c^2=289\rightarrow c=\sqrt{289}=17$

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Example Question #12 : Right Triangles

What is the hypotenuse of a right triangle with legs of length and ?

Possible Answers:

$\sqrt{21}$

$\sqrt{74}$

$\sqrt{30}$

$\sqrt{15}$

$\sqrt{75}$

Correct answer:

$\sqrt{74}$

Explanation:

In order to find the value of the hypotenuse , we need to plug the values of the legs and into the Pythagorean Theorem: $a^{2}+b^{2}=c^{2}$

$5^{2}+7^{2}=c^{2}$

$25+49=c^{2}$

$74=c^{2}$

$\sqrt{74}=c$

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Example Question #16 : Triangles

What is the hypotenuse of a right triangle with an area of and a leg length of ?

Possible Answers:

Correct answer:

Explanation:

In order to find the value of the hypotenuse given area and leg length , we first need to calculate the length of the other leg length . Given the area equation $A=\frac{1}{2}ab$ :

2A=ab

$\frac{2A}{b}=a$

Plugging in and :

$\frac{2(30)}{12}=a$

$\frac{60}{12}=a$

5=a

Now we need to plug the values of the legs and into the Pythagorean Theorem: $a^{2}+b^{2}=c^{2}$

$5^{2}+12^{2}=c^{2}$

$25+144=c^{2}$

$169=c^{2}$

13=c

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Example Question #4 : Calculating The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

What is the hypotenuse of a right triangle with legs of lengths 7cm and 9cm ?

Possible Answers:

$\sqrt{65}cm$

15cm

$\sqrt{130}cm$

$\sqrt{70}cm$

12cm

Correct answer:

$\sqrt{130}cm$

Explanation:

In order to find the value of the hypotenuse , we need to plug the values of the legs and into the Pythagorean Theorem: $a^{2}+b^{2}=c^{2}$

$(7cm)^{2}+(9cm)^{2}=c^{2}$

$49cm+81cm=c^{2}$

$130cm=c^{2}$

$\sqrt{130}cm=c$

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Example Question #3 : Calculating The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Each leg of a $\textup{45-45-90}$ triangle has length $2 ^{K}$ . Give the length of its hypotenuse.

Possible Answers:

$2 ^{K +1}$

$2 ^{ 2K }$

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

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

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

Correct answer:

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

Explanation:

The hypotenuse of a $\textup{45-45-90}$ triangle has length $\sqrt{2} = 2^{\frac{1}{2}}$ times that of a leg, which here is

$2 ^{K} \cdot 2^{\frac{1}{2}} = 2 ^{K+ \frac{1}{2}}$ .

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Example Question #262 : Geometry

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 #21 : Right Triangles

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:

$1,400\textrm{ in}$

$1,080\textrm{ in}$

$960\textrm{ in}$

$1,040\textrm{ in}$

$1,200\textrm{ in}$

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 #3 : Calculating The Perimeter Of A Right Triangle

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:

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

$6 \textrm{ yd}$

$7 \textrm{ yd}$

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

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

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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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #12 : Right Triangles

Example Question #501 : Problem Solving Questions

Example Question #14 : Right Triangles

Example Question #12 : Right Triangles

Example Question #16 : Triangles

Example Question #4 : Calculating The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #3 : Calculating The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #262 : Geometry

Example Question #21 : Right Triangles

Example Question #3 : Calculating The Perimeter Of A Right Triangle

All GMAT Math Resources

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