Pythagorean Theorem: Practice Problems with Answers

The Pythagorean theorem ( $a^2 + b^2 = c^2$ ) is one of the cornerstones of geometry. It applies to every right triangle, where c is the hypotenuse (the side opposite the right angle) and a and b are the two legs. The 15 practice problems below cover every major application students and exam takers encounter: finding the hypotenuse, finding a missing leg, recognizing Pythagorean triples, using the converse to classify triangles, computing distances on the coordinate plane, and solving real-world word problems in construction, navigation, and geometry. Problems progress from easy to hard within each section.

Contents6 sections

Finding the Hypotenuse

These problems give you both legs and ask for the hypotenuse using $c = \sqrt{a^2 + b^2}$ . This is the most searched direction of the Pythagorean theorem and the essential first skill to master.

Problem 1: Whole-Number Legs

Easy

A right triangle has legs $a = 6$ cm and $b = 8$ cm.

Write the Pythagorean theorem and substitute the known values.
Calculate the exact length of the hypotenuse $c$ .

Hint

Square each leg, add the results, then take the square root. Check whether the result is a perfect square before writing a decimal approximation.

View Solution

Solution to question 1:

a^2 + b^2 = c^2 \implies 6^2 + 8^2 = c^2 \implies 36 + 64 = c^2 \implies c^2 = 100

Solution to question 2:

c = \sqrt{100} = 10 \text{ cm}

The hypotenuse is 10 cm. The set

(6, 8, 10)

is a scalar multiple of the classic

(3, 4, 5)

Pythagorean triple.

Problem 2: Irrational Hypotenuse

Easy

A right triangle has legs $a = 5$ m and $b = 7$ m.

Set up and solve for $c^2$ .
Express the hypotenuse in simplest radical form and as a decimal rounded to the nearest hundredth.

Hint

The sum

25 + 49 = 74

is not a perfect square. Simplify

\sqrt{74}

by checking whether 74 has any perfect-square factors.

View Solution

Solution to question 1:

c^2 = 5^2 + 7^2 = 25 + 49 = 74

Solution to question 2:

c = \sqrt{74} \approx 8.60 \text{ m}

Since

74 = 2 \times 37

and neither factor is a perfect square,

\sqrt{74}

is already in simplest radical form.

Problem 3: Algebraic Sides

Hard

A right triangle has legs $a = x$ and $b = x + 3$ , and its hypotenuse is $c = x + 5$ , where $x > 0$ .

Apply the Pythagorean theorem to set up an equation in $x$ .
Solve for $x$ and state all three side lengths.

Hint

Expand

(x+3)^2

and

(x+5)^2

, collect all terms on one side, and solve the resulting quadratic equation. Reject any negative solution.

View Solution

Solution to question 1:

x^2 + (x+3)^2 = (x+5)^2

x^2 + x^2 + 6x + 9 = x^2 + 10x + 25

2x^2 + 6x + 9 = x^2 + 10x + 25 \implies x^2 - 4x - 16 = 0

Solution to question 2:

x = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5}

Since

x > 0

, take

x = 2 + 2\sqrt{5} \approx 6.47

. The sides are approximately

6.47,\ 9.47,\ 11.47

Finding a Missing Leg

When the hypotenuse and one leg are known, rearrange the formula to $a = \sqrt{c^2 - b^2}$ . This direction catches many students off-guard: the key is to subtract, not add, before taking the square root.

Problem 4: Integer Missing Leg

Easy

A right triangle has a hypotenuse of $c = 13$ in and one leg $b = 5$ in.

Rearrange the Pythagorean theorem to isolate $a^2$ .
Find the exact length of the missing leg $a$ .

Hint

Compute

13^2 - 5^2 = 169 - 25

and check whether the result is a perfect square.

View Solution

Solution to question 1:

a^2 = c^2 - b^2 = 13^2 - 5^2 = 169 - 25 = 144

Solution to question 2:

a = \sqrt{144} = 12 \text{ in}

The triple

(5, 12, 13)

is a fundamental Pythagorean triple.

Problem 5: Radical-Form Sides

Hard

A right triangle has hypotenuse $c = \sqrt{50}$ and one leg $b = \sqrt{18}$ .

Compute $a^2 = c^2 - b^2$ using properties of radicals.
Simplify $a$ completely and identify any special triangle relationship.

Hint

Recall that

(\sqrt{n})^2 = n

. Subtract directly:

50 - 18 = 32

. Then simplify

\sqrt{32}

by factoring out the largest perfect-square factor.

View Solution

Solution to question 1:

a^2 = (\sqrt{50})^2 - (\sqrt{18})^2 = 50 - 18 = 32

Solution to question 2:

a = \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}

The three sides are

3\sqrt{2},\ 4\sqrt{2},\ 5\sqrt{2}

. Dividing by

\sqrt{2}

yields the ratio

3:4:5

, confirming this is a scaled

(3, 4, 5)

right triangle.

Pythagorean Triples

A Pythagorean triple is a set of three positive integers satisfying $a^2 + b^2 = c^2$ . Recognizing triples such as $(3,4,5)$ , $(5,12,13)$ , $(8,15,17)$ , and $(7,24,25)$ (and all their multiples) is a major time-saver in timed exams and a key SEO cluster for this topic.

Problem 6: Identifying Pythagorean Triples

Easy

Determine whether each set of side lengths is a Pythagorean triple.

$(9,\ 40,\ 41)$
$(10,\ 24,\ 26)$
$(6,\ 9,\ 12)$

Hint

Assign the largest number to

c

, compute

a^2 + b^2

, and compare to

c^2

. Also check whether a set is a whole-number multiple of a known triple.

View Solution

Solution to question 1:

9^2 + 40^2 = 81 + 1600 = 1681 = 41^2 \quad \checkmark

Yes:

(9, 40, 41)

is a Pythagorean triple. Solution to question 2:

10^2 + 24^2 = 100 + 576 = 676 = 26^2 \quad \checkmark

Yes: it equals

2 \times (5, 12, 13)

. Solution to question 3:

6^2 + 9^2 = 36 + 81 = 117 \neq 144 = 12^2 \quad \times

No:

(6, 9, 12)

is not a Pythagorean triple.

Problem 7: Unknown Multiplier in a Scaled Triple

Medium

A right triangle’s sides are in the ratio $7 : 24 : 25$ . The hypotenuse measures $100$ units.

Find the scale factor $k$ and state all three side lengths.
Calculate the area of the triangle.

Hint

Set

25k = 100

to find

k

, then multiply each ratio value by

k

. The area of a right triangle is

\tfrac{1}{2} \times \text{leg}_1 \times \text{leg}_2

View Solution

Solution to question 1:

k = \frac{100}{25} = 4 \implies \text{sides: } 28,\ 96,\ 100

Solution to question 2:

\text{Area} = \frac{1}{2} \times 28 \times 96 = 1344 \text{ sq units}

Converse of the Pythagorean Theorem

The converse states: if $a^2 + b^2 = c^2$ , the triangle is a right triangle. Comparing $a^2 + b^2$ to $c^2$ also classifies triangles as acute (greater) or obtuse (less). This concept is critical for proofs, construction checks, and classification problems.

Problem 8: Classify as Acute, Right, or Obtuse

Medium

Classify each triangle as acute, right, or obtuse.

Sides: $5,\ 6,\ 9$
Sides: $9,\ 40,\ 41$
Sides: $6,\ 8,\ 9$

Hint

Always assign the largest side to

c

. If

a^2+b^2 = c^2

→ right;

> c^2

→ acute;

< c^2

→ obtuse.

View Solution

Solution to question 1:

5^2 + 6^2 = 61 < 81 = 9^2 \implies \textbf{Obtuse}

Solution to question 2:

9^2 + 40^2 = 1681 = 41^2 \implies \textbf{Right}

Solution to question 3:

6^2 + 8^2 = 100 > 81 = 9^2 \implies \textbf{Acute}

Problem 9: Construction (Verifying a Right Angle)

Hard

A contractor lays a rectangular foundation with two adjacent walls of $30$ ft and $40$ ft. Measuring the diagonal yields $51$ ft.

Use the converse of the Pythagorean theorem to determine whether the corner is a true right angle.
If not, should the contractor move the corner inward or outward? Explain using the inequality $a^2 + b^2$ vs $c^2$ .

Hint

Compute

30^2 + 40^2

and compare to

51^2

. The correct diagonal for a right angle is

\sqrt{30^2 + 40^2}

View Solution

Solution to question 1:

30^2 + 40^2 = 900 + 1600 = 2500 \quad \text{but} \quad 51^2 = 2601

Since

2500 \neq 2601

, the corner is not a right angle. The correct diagonal would be

\sqrt{2500} = 50

ft. Solution to question 2: Since

a^2 + b^2 = 2500 < 2601 = c^2

, the triangle is obtuse: the diagonal is too long. The contractor must move the corner inward until the diagonal reads exactly 50 ft.

Distance on the Coordinate Plane

The Pythagorean theorem is the geometric foundation of the distance formula. Given two points $A(x_1, y_1)$ and $B(x_2, y_2)$ , the horizontal and vertical separations form the legs of a right triangle, and the distance $AB$ is the hypotenuse: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ .

Problem 10: Distance Between Two Points

Easy

Find the exact distance between $A(1, 2)$ and $B(5, 5)$ .

State the horizontal and vertical distances (the two legs).
Apply the Pythagorean theorem to find $AB$ .

Hint

The horizontal leg is

|5 - 1|

and the vertical leg is

|5 - 2|

. Look for a familiar Pythagorean triple in the result.

View Solution

Solution to question 1:

\Delta x = |5-1| = 4, \quad \Delta y = |5-2| = 3

Solution to question 2:

AB = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5 \text{ units}

Problem 11: Points in Different Quadrants

Medium

Find the distance between $P(-3, 4)$ and $Q(5, -2)$ .

Calculate $\Delta x$ and $\Delta y$ , paying careful attention to signs.
State the exact distance and a decimal approximation to two places.

Hint

Use absolute values:

|5-(-3)|

and

|-2-4|

. Negative coordinates do not change the procedure, only the arithmetic.

View Solution

Solution to question 1:

\Delta x = |5-(-3)| = 8, \quad \Delta y = |-2-4| = 6

Solution to question 2:

PQ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units}

An exact whole number, since

(6, 8, 10)

is a

(3,4,5)

multiple.

Problem 12: Perimeter of a Triangle on a Grid

Hard

Triangle $ABC$ has vertices $A(0,0)$ , $B(6,0)$ , and $C(2,5)$ .

Calculate the length of each side using the Pythagorean theorem.
Find the exact perimeter of the triangle.

Hint

For each side find

\Delta x

and

\Delta y

, then compute the hypotenuse. Side

AB

lies along the x-axis, so its length is simply

|6-0|

View Solution

Solution to question 1:

AB: \quad \sqrt{6^2+0^2} = 6

BC: \quad \sqrt{(6-2)^2+(0-5)^2} = \sqrt{16+25} = \sqrt{41}

CA: \quad \sqrt{(2-0)^2+(5-0)^2} = \sqrt{4+25} = \sqrt{29}

Solution to question 2:

P = 6 + \sqrt{41} + \sqrt{29} \approx 6 + 6.40 + 5.39 \approx 17.79 \text{ units}

Real-World Word Problems

Every standardized test includes real-life Pythagorean theorem word problems. The skill is identifying the hidden right triangle within a scenario, a ladder against a wall, the diagonal of a room, the height of a triangle, or the space diagonal of a 3D box,before applying $a^2 + b^2 = c^2$ .

Problem 13: Ladder Against a Wall

Easy

A ladder leans against a vertical wall. Its base is $5$ ft from the wall, and the ladder reaches $12$ ft up the wall.

Identify and label the right triangle formed by the ladder, wall, and ground.
Find the exact length of the ladder.

Hint

The wall is vertical, the ground is horizontal; they meet at a right angle. The ladder is the hypotenuse. Look for a common Pythagorean triple.

View Solution

Solution to question 1: Legs: ground distance

a = 5

ft, wall height

b = 12

ft. The right angle is at the base of the wall. The ladder

c

is the hypotenuse. Solution to question 2:

c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ ft}

The ladder is 13 ft long. This is the

(5, 12, 13)

Pythagorean triple.

Problem 14: Height of an Isosceles Triangle

Medium

An isosceles triangle has two equal sides of $13$ cm and a base of $10$ cm.

Explain why a perpendicular from the apex to the base creates two congruent right triangles.
Find the height and the area of the triangle.

Hint

The perpendicular bisects the base, giving each right triangle a base leg of

5

cm and hypotenuse of

13

cm. Solve for the height leg.

View Solution

Solution to question 1: The perpendicular from the apex is also the axis of symmetry of the isosceles triangle. It bisects the base into two equal parts of 5 cm, forming two congruent right triangles. Solution to question 2:

h^2 + 5^2 = 13^2 \implies h^2 = 169 - 25 = 144 \implies h = 12 \text{ cm}

\text{Area} = \frac{1}{2} \times 10 \times 12 = 60 \text{ cm}^2

Problem 15: Space Diagonal of a Rectangular Box

Hard

A rectangular box has length $l = 12$ cm, width $w = 4$ cm, and height $h = 3$ cm.

Find the diagonal of the rectangular base using the Pythagorean theorem.
Use that result as a leg of a second right triangle to find the full space diagonal of the box.

Hint

Base diagonal:

d_1 = \sqrt{l^2 + w^2}

. Space diagonal:

d = \sqrt{d_1^2 + h^2}

. You can combine these into one step as

d = \sqrt{l^2 + w^2 + h^2}

View Solution

Solution to question 1:

d_1 = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \text{ cm}

Solution to question 2:

d = \sqrt{d_1^2 + h^2} = \sqrt{160 + 9} = \sqrt{169} = 13 \text{ cm}

The space diagonal of the box is exactly 13 cm.