Here are 30 Pythagorean theorem practice problems with worked solutions, in three levels. Level 1 covers finding a hypotenuse or a leg, Level 2 adds coordinates, special shapes and the converse, and Level 3 brings in algebra and multi-step reasoning. Work each problem on paper before you open its solution.
Most answers are exact (whole numbers or simplest radical form) with a decimal where it helps. Round decimals to the nearest hundredth unless a problem says otherwise. If you get stuck on the basics, the worked examples walk through every problem type slowly.
Before you start: three moves
Almost every Pythagorean theorem question comes down to one of three moves:
- Find the hypotenuse: c = √a2 + b2. Add the squares of the legs. (More on the hypotenuse.)
- Find a leg: a = √c2 − b2. Subtract the square of the known leg from the square of the hypotenuse. (More on missing legs.)
- Test a triangle: compare a2 + b2 with c2, where c is the longest side. Equal means right; less means obtuse; greater means acute. This is the converse of the theorem.
Level 1: basic problems
One right triangle, two known sides, one unknown. Problems 5 to 7 have radical answers; simplify them using the largest perfect-square factor (see square roots and radicals).
Problem 1 Level 1
The legs of a right triangle are 16 and 30. Find the hypotenuse.
Show solution
c2= 162 + 302 = 256 + 900 = 1156c= √1156 = 34Answer: 34. This is the 8-15-17 triple doubled.
Problem 2 Level 1
Find the hypotenuse of a right triangle with legs 13 and 84.
Show solution
c2= 132 + 842 = 169 + 7056 = 7225c= √7225 = 85Answer: 85.
Problem 3 Level 1
A right triangle has hypotenuse 26 and one leg 10. Find the other leg.
Show solution
b2= 262 − 102 = 676 − 100 = 576b= √576 = 24Answer: 24. The hypotenuse is given, so subtract.
Problem 4 Level 1
The hypotenuse of a right triangle is 50 and one leg is 14. Find the other leg.
Show solution
a2= 502 − 142 = 2500 − 196 = 2304a= √2304 = 48Answer: 48. (14-48-50 is 7-24-25 doubled.)
Problem 5 Level 1
Find the hypotenuse of a right triangle with legs 5 and 7. Give the exact answer and a decimal.
Show solution
c2= 25 + 49 = 74c= √74 ≈ 8.60Answer: √74 ≈ 8.60. 74 = 2 × 37 has no square factor, so √74 is already in simplest form.
Problem 6 Level 1
Both legs of a right triangle are 3. Find the hypotenuse in simplest radical form.
Show solution
c2= 9 + 9 = 18c= √18 = √9 · 2 = 3√2 ≈ 4.24Answer: 3√2. Equal legs always give leg ×√2.
Problem 7 Level 1
A right triangle has hypotenuse 6 and one leg 2. Find the other leg in simplest radical form.
Show solution
b2= 36 − 4 = 32b= √32 = √16 · 2 = 4√2 ≈ 5.66Answer: 4√2. A common slip is to stop at 2√8, which is not fully simplified.
Problem 8 Level 1
The legs of a right triangle are 4.4 and 11.7. Find the hypotenuse.
Show solution
c2= 4.42 + 11.72= 19.36 + 136.89 = 156.25c= 12.5Answer: 12.5, exactly. (It's the 44-117-125 triple divided by 10.)
Problem 9 Level 1
Is a triangle with sides 18, 80 and 82 a right triangle?
Show solution
182 + 802= 324 + 6400 = 6724822= 6724Answer: yes. The two shorter sides' squares add up to the longest side's square, so the angle opposite the 82 side is 90°.
Problem 10 Level 1
Is a triangle with sides 9, 11 and 14 a right triangle? If not, is it acute or obtuse?
Show solution
92 + 112= 81 + 121 = 202142= 196Answer: no, it is acute. 202 ≠ 196, so it isn't right; since 202 > 196, the largest angle is less than 90°.
Level 2: intermediate problems
Now the right triangle may be hiding inside another shape, or on a coordinate grid. Draw it in first, then use one of the three moves.
Problem 11 Level 2
Find the distance between the points (−6, −1) and (9, 7).
Show solution
The horizontal leg is 9 − (−6) = 15 and the vertical leg is 7 − (−1) = 8.
d = √152 + 82 = √225 + 64 = √289 = 17Answer: 17. This is the distance formula in action.
Problem 12 Level 2
Find the distance between (−1, 4) and (4, −2), exactly and to the nearest hundredth.
Show solution
d= √(4 − (−1))2 + (−2 − 4)2= √52 + (−6)2 = √25 + 36= √61 ≈ 7.81Answer: √61 ≈ 7.81. Squaring removes the negative sign: (−6)2 = 36.
Problem 13 Level 2
The diagonal of a square is 14. Find the side length and the area.
Show solution
s2 + s2= 1422s2= 196s2= 98s= √98 = 7√2 ≈ 9.90Answer: side 7√2 ≈ 9.90; area 98. Notice that the area is s2 = 98 directly, so you don't need the square root for that part. In general, a square's area is half its diagonal squared.
Problem 14 Level 2
An isosceles triangle has equal sides of 29 and a base of 40. Find its height and its area.
Show solution
The height cuts the base in half, leaving a right triangle with hypotenuse 29 and leg 20.
h= √292 − 202 = √841 − 400 = √441 = 21Area= 1 over 2 · 40 · 21 = 420Answer: height 21, area 420.
Problem 15 Level 2
Find the height and the area of an equilateral triangle with side 10.
Show solution
h= √102 − 52 = √75 = 5√3 ≈ 8.66Area= 1 over 2 · 10 · 5√3 = 25√3 ≈ 43.30Answer: height 5√3 ≈ 8.66, area 25√3 ≈ 43.30. Each half is a 30-60-90 triangle.
Problem 16 Level 2
A rectangle is 48 wide and has a diagonal of 73. Find its length, its area and its perimeter.
Show solution
ℓ= √732 − 482 = √5329 − 2304= √3025 = 55Area = 48 × 55 = 2640; perimeter = 2(48 + 55) = 206.
Answer: length 55, area 2640, perimeter 206.
Problem 17 Level 2
Classify the triangle with sides 7, 8 and 11 as acute, right or obtuse.
Show solution
72 + 82= 49 + 64 = 113112= 121Answer: obtuse. 113 < 121: the longest side is too long for a right angle, so the angle opposite it is greater than 90°. See acute, right or obtuse.
Problem 18 Level 2
A right triangle has legs x and 20, and its hypotenuse is x + 8. Find x.
Show solution
x2 + 202= (x + 8)2x2 + 400= x2 + 16x + 64336= 16xx= 21Answer: x = 21, so the sides are 20, 21 and 29. Check: 400 + 441 = 841 = 292. The x2 terms cancel, which leaves a simple linear equation.
Problem 19 Level 2
A right triangle has legs 2 and 3. Its hypotenuse is used as one leg of a second right triangle whose other leg is 6. Find the hypotenuse x of the second triangle.
Show solution
First triangle: hypotenuse = √4 + 9 = √13. Keep it exact.
x2= (√13)2 + 62 = 13 + 36 = 49x= 7Answer: 7. If you had rounded √13 to 3.6 first, you'd get √48.96 ≈ 6.997 instead of exactly 7.
Problem 20 Level 2
A right triangle has hypotenuse 2√13 and one leg 4. Find the other leg.
Show solution
b2= (2√13)2 − 42= 4 · 13 − 16 = 52 − 16 = 36b= 6Answer: 6. Square both parts: (2√13)2 = 22 · 13 = 52, not 26.
Level 3: advanced problems
These need algebra, two triangles, or a decision about which side is the hypotenuse. They are the kind of questions that appear at the end of a test. For even harder ones, see the challenge problems.
Problem 21 Level 3
The sides of a right triangle are x, x + 7 and x + 8. Find x and the three sides.
Show solution
The longest side, x + 8, is the hypotenuse.
x2 + (x + 7)2= (x + 8)2x2 + x2 + 14x + 49= x2 + 16x + 64x2 − 2x − 15= 0(x − 5)(x + 3)= 0So x = 5 or x = −3. A side can't be negative, so x = 5.
Answer: x = 5; the sides are 5, 12 and 13. Check: 25 + 144 = 169.
Problem 22 Level 3
The legs of a right triangle are x and x + 17, and the hypotenuse is x + 25. Find the sides.
Show solution
x2 + (x + 17)2= (x + 25)22x2 + 34x + 289= x2 + 50x + 625x2 − 16x − 336= 0(x − 28)(x + 12)= 0Reject x = −12, so x = 28.
Answer: 28, 45 and 53. Check: 784 + 2025 = 2809 = 532.
Problem 23 Level 3
Two sides of a right triangle are 5 and 9. Find every possible length of the third side.
Show solution
The problem doesn't say which side is the hypotenuse, so there are two cases.
If x is the hypotenuse: x = √25 + 81 = √106 ≈ 10.30.
If 9 is the hypotenuse: x = √81 − 25 = √56 = 2√14 ≈ 7.48.
Answer: √106 or 2√14. The 5 can't be the hypotenuse because it is shorter than 9.
Problem 24 Level 3
Show that the triangle with vertices A(−2, 1), B(4, 3) and C(2, 9) is a right triangle. Where is the right angle, and what is the area?
Show solution
Find the squared lengths (no square roots needed):
AB2= 62 + 22 = 40BC2= (−2)2 + 62 = 40AC2= 42 + 82 = 80AB2 + BC2 = 80 = AC2, so by the converse the right angle is at B, opposite the longest side AC. The legs are both √40, so the area is 1 over 2 · √40 · √40 = 20.
Answer: right angle at B; area 20. It's also isosceles, since AB = BC.
Problem 25 Level 3
A right triangle has legs 24 and 45. Find the length of the altitude from the right angle to the hypotenuse.
Show solution
The hypotenuse is √576 + 2025 = √2601 = 51. Now write the area two ways: half the product of the legs, and half the hypotenuse times the altitude.
1 over 2 · 24 · 45= 1 over 2 · 51 · hh= 1080 over 51 = 360 over 17 ≈ 21.18Answer: 360 over 17 ≈ 21.18. In any right triangle, altitude = ab/c. The geometric mean theorem gives more relationships like this.
Problem 26 Level 3
A cube has edges of length 4. Find the length of a face diagonal and of the space diagonal (corner to opposite corner through the inside).
Show solution
Face diagonal: √42 + 42 = √32 = 4√2 ≈ 5.66.
d= √(4√2)2 + 42 = √32 + 16= √48 = 4√3 ≈ 6.93Answer: 4√2 and 4√3. For any cube with edge s, the space diagonal is s√3. See the theorem in 3D.
Problem 27 Level 3
An isosceles trapezoid has parallel sides 11 and 41, and its two slanted sides are 17. Find its height and its area.
Show solution
Drop a height from each end of the short side. The long side splits into 15 + 11 + 15, since (41 − 11) ÷ 2 = 15.
h= √172 − 152 = √289 − 225 = √64 = 8Area= 1 over 2(11 + 41) · 8 = 208Answer: height 8, area 208.
Problem 28 Level 3
The diagonals of a rhombus are 24 and 70. Find the side length, the perimeter and the area.
Show solution
A rhombus's diagonals bisect each other at right angles, making four right triangles with legs 12 and 35.
s = √122 + 352 = √144 + 1225 = √1369 = 37Perimeter = 4 × 37 = 148; area = 1 over 2 · 24 · 70 = 840.
Answer: side 37, perimeter 148, area 840.
Problem 29 Level 3
A right triangle has a perimeter of 132 and a hypotenuse of 61. Find the two legs.
Show solution
The legs add to 132 − 61 = 71, and a2 + b2 = 612 = 3721. Square the sum:
(a + b)2= a2 + 2ab + b2712= 3721 + 2ab5041 − 3721= 2abab= 660So a and b are the roots of t2 − 71t + 660 = 0, that is (t − 11)(t − 60) = 0.
Answer: 11 and 60. Check: 11 + 60 + 61 = 132 and 121 + 3600 = 3721.
Problem 30 Level 3
Find the point P on the x-axis that is the same distance from A(1, 4) and B(7, 2). How far is P from each point?
Show solution
Let P = (x, 0) and set PA2 = PB2:
(x − 1)2 + 42= (x − 7)2 + 22x2 − 2x + 17= x2 − 14x + 5312x= 36x= 3PA = √22 + 42 = √20 = 2√5 ≈ 4.47, and PB = √42 + 22 = √20 as well.
Answer: P(3, 0), at distance 2√5 ≈ 4.47 from both.
Answer key
Use this to mark a whole set quickly. Open the solution for any problem you missed: the method matters more than the number.
| # | Answer | # | Answer | # | Answer |
|---|---|---|---|---|---|
| 1 | 34 | 11 | 17 | 21 | 5, 12, 13 |
| 2 | 85 | 12 | √61 ≈ 7.81 | 22 | 28, 45, 53 |
| 3 | 24 | 13 | 7√2; area 98 | 23 | √106 or 2√14 |
| 4 | 48 | 14 | h = 21; area 420 | 24 | right angle at B; area 20 |
| 5 | √74 ≈ 8.60 | 15 | 5√3; 25√3 | 25 | 360 over 17 ≈ 21.18 |
| 6 | 3√2 | 16 | 55; 2640; 206 | 26 | 4√2; 4√3 |
| 7 | 4√2 | 17 | obtuse | 27 | h = 8; area 208 |
| 8 | 12.5 | 18 | x = 21 | 28 | 37; 148; 840 |
| 9 | yes | 19 | 7 | 29 | 11 and 60 |
| 10 | no, acute | 20 | 6 | 30 | P(3, 0); 2√5 |
What to do with your score
- Missed Level 1 problems? Re-read how to do the Pythagorean theorem, then try the hypotenuse and missing side worksheets.
- Missed Level 2? Most slips here come from forgetting to halve a base, sign errors on the grid, or comparing the wrong sides in the converse. Check the common mistakes.
- Got Level 3 right? Try the quiz, the SAT and ACT questions or the challenge problems.
Redo any problem you missed a few days later without looking at the solution. Getting it right the second time, from memory, is what makes it stick.
Frequently asked questions
What kinds of Pythagorean theorem questions come up on tests?
Finding a missing side (the hypotenuse or a leg), deciding whether three lengths form a right triangle, finding distances on a coordinate grid, and word problems. Harder tests add special triangles, 3D diagonals and algebra with variable sides, like problems 21 and 22.
Should my answers be exact or decimals?
Give the exact form, such as 4√2, unless the question asks for a decimal or a real-world measurement. If you give a decimal, say how you rounded it, for example to the nearest hundredth.
How do I know which side is the hypotenuse?
It is the side opposite the right angle and always the longest side. If you are only given lengths and no diagram, the longest length must be the hypotenuse, unless the problem leaves it open, as in problem 23.
How many problems should I do to get good at this?
Quality beats quantity. Ten well-chosen problems with every mistake understood are worth more than fifty done carelessly. A good routine is one level per session, then a short mixed set a few days later.
Where can I get more practice with answers?
The practice hub lists every resource on the site by level, including printable worksheets with answer keys, a quiz that gives new numbers each time, and word problems.
Keep exploring
- PracticePythagorean Theorem ExamplesFifteen Pythagorean theorem examples solved step by step, from finding the hypotenuse to 3D diagonals, coordinates and algebra, each with a diagram and answer.
- PracticePythagorean Theorem Challenge Problems12 hard Pythagorean theorem challenge problems: an ant on a box, folded paper, sliding and crossed ladders, Dudeney's spider, with hints and full solutions.
- PracticePythagorean Theorem QuizTake a free Pythagorean theorem quiz: 10 questions on the hypotenuse, missing sides, triples, the converse and distance, with instant worked solutions.
- PracticePythagorean Theorem WorksheetsFree Pythagorean theorem worksheets with answers: a printable worksheet generator plus ready-made sheets on the hypotenuse, missing sides and word problems.
- PracticePythagorean Theorem Word Problems15 Pythagorean theorem word problems from real life (ladders, TVs, ramps, ships, kites, roofs) with a 4-step strategy, a sketch for each and full solutions.
- PracticeThe Pythagorean Theorem on the SAT and ACTHow the Pythagorean theorem appears on the SAT and ACT, what each test gives you, the triple and special-triangle shortcuts, and 10 original practice questions.