{"id":61470,"date":"2026-04-24T22:45:14","date_gmt":"2026-04-24T14:45:14","guid":{"rendered":"https:\/\/blog-admin.thethinkacademy.com\/?p=61470"},"modified":"2026-04-27T22:26:26","modified_gmt":"2026-04-27T14:26:26","slug":"pythagorean-theorem-worksheet-amc-8practice","status":"publish","type":"post","link":"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/04\/24\/pythagorean-theorem-worksheet-amc-8practice\/","title":{"rendered":"Pythagorean Theorem Worksheet: Practice Problems for AMC 8 Students"},"content":{"rendered":"\n<p>The Pythagorean theorem is one of the most important concepts in geometry and one of the most consistently tested topics across every level of the AMC math competition. This Pythagorean theorem worksheet gives <a href=\"https:\/\/www.thinkacademy.ca\/blog\/blog\/2026\/04\/15\/amc-8-math-competition-guide-for-parents\/\" target=\"_blank\" rel=\"noreferrer noopener\">AMC 8<\/a> students a complete set of practice problems \u2014 from straightforward calculations finding missing side lengths, through Pythagorean theorem word problems set in real contexts, to multi-step competition geometry questions of the kind that appear in AMC 8 past papers. Every problem comes with a full worked solution and explanation so students understand not just the answer but the reasoning behind it.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><a href=\"https:\/\/www.thinkacademy.ca\/amc-free-trial?source_id=6172&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\" target=\"_blank\" rel=\" noreferrer noopener\"><img loading=\"lazy\" decoding=\"async\" width=\"744\" height=\"298\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.40-PM.png\" alt=\"Think Academy AMC 8 free evaluation banner \u2014 blue background with gold button reading &quot;Take the free evaluation&quot;\" class=\"wp-image-60940\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.40-PM.png 744w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.40-PM-300x120.png 300w\" sizes=\"auto, (max-width: 744px) 100vw, 744px\" \/><\/a><figcaption class=\"wp-element-caption\">Not sure where your child stands? Take Think Academy&#8217;s free math evaluation to find out.<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>What is the Pythagorean theorem?<\/strong><\/h2>\n\n\n\n<p>The Pythagorean theorem states that in any right-angled triangle, the square of the length of the hypotenuse equals the sum of the squares of the other two sides.<\/p>\n\n\n\n<p>Written as a formula:<\/p>\n\n\n\n<p><strong>a\u00b2 + b\u00b2 = c\u00b2<\/strong><\/p>\n\n\n\n<p>Where a and b are the two shorter sides (called legs) and c is the hypotenuse \u2014 the longest side, always opposite the right angle.<\/p>\n\n\n\n<p>This means:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If you know two sides of a right triangle you can always find the third<\/li>\n\n\n\n<li>If a\u00b2 + b\u00b2 = c\u00b2 is satisfied by three numbers, those numbers form a right triangle<\/li>\n\n\n\n<li>The hypotenuse is always the longest side<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Why the Pythagorean theorem matters for AMC 8<\/strong><\/h3>\n\n\n\n<p>Geometry questions make up a significant portion of every AMC 8 paper. The Pythagorean theorem appears directly \u2014 in problems asking for a missing side length \u2014 and indirectly, embedded in problems about areas, diagonals of rectangles, distances between points, and three-dimensional figures. Students who can apply the theorem quickly and accurately, and who recognise when a problem requires it without being told explicitly, have a meaningful advantage in competition conditions.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Pythagorean triples \u2014 the shortcuts worth memorising<\/strong><\/h3>\n\n\n\n<p>A Pythagorean triple is a set of three whole numbers that satisfy a\u00b2 + b\u00b2 = c\u00b2. Recognising common triples saves significant time in AMC 8 problems because you can identify the missing side instantly without any calculation.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Triple<\/th><th>Multiples to know<\/th><\/tr><\/thead><tbody><tr><td>3, 4, 5<\/td><td>6-8-10, 9-12-15, 12-16-20<\/td><\/tr><tr><td>5, 12, 13<\/td><td>10-24-26<\/td><\/tr><tr><td>8, 15, 17<\/td><td>16-30-34<\/td><\/tr><tr><td>7, 24, 25<\/td><td>14-48-50<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The 3-4-5 triple and its multiples appear more frequently in AMC 8 problems than any other. Students who recognise a 3-4-5 triangle immediately \u2014 without needing to calculate \u2014 solve these problems in seconds rather than minutes.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pythagorean theorem worksheet practice problems \u2014 Level 1<\/strong><\/h2>\n\n\n\n<p>These problems test direct application of the theorem to find a missing side. They are equivalent in difficulty to questions 1 to 10 on the AMC 8.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 1<\/strong><\/p>\n\n\n\n<p>A right triangle has legs of length 6 and 8. What is the length of the hypotenuse?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>6\u00b2 + 8\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>36 + 64 = c\u00b2 <\/p>\n\n\n\n<p>100 = c\u00b2 <\/p>\n\n\n\n<p>c = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>10<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Recognise this as a 3-4-5 triple scaled by 2. Once you have seen enough Pythagorean triples, 6-8 immediately suggests the hypotenuse is 10 without any calculation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 2<\/strong><\/p>\n\n\n\n<p>A right triangle has a hypotenuse of 13 and one leg of length 5. What is the length of the other leg?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>5\u00b2 + b\u00b2 = 13\u00b2 <\/p>\n\n\n\n<p>25 + b\u00b2 = 169 <\/p>\n\n\n\n<p>b\u00b2 = 144 <\/p>\n\n\n\n<p>b = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>12<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is the 5-12-13 Pythagorean triple. Recognising it immediately is faster than using the formula. Memorising the common triples in the table above pays dividends throughout competition preparation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 3<\/strong><\/p>\n\n\n\n<p>A right triangle has legs of length 9 and 12. What is the length of the hypotenuse?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>9\u00b2 + 12\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>81 + 144 = c\u00b2 <\/p>\n\n\n\n<p>225 = c\u00b2 <\/p>\n\n\n\n<p>c = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>15<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is a 3-4-5 triple scaled by 3. 9 = 3 x 3, 12 = 4 x 3, so the hypotenuse is 5 x 3 = 15.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 4<\/strong><\/p>\n\n\n\n<p>A right triangle has a hypotenuse of 17 and one leg of length 8. What is the length of the other leg?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>8\u00b2 + b\u00b2 = 17\u00b2 <\/p>\n\n\n\n<p>64 + b\u00b2 = 289 <\/p>\n\n\n\n<p>b\u00b2 = 225 <\/p>\n\n\n\n<p>b = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>15<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is the 8-15-17 Pythagorean triple. If you recognise it you can write the answer immediately. If not, the calculation is straightforward \u2014 the key is making sure you subtract correctly: 289 minus 64 equals 225, and the square root of 225 is 15.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 5<\/strong><\/p>\n\n\n\n<p>A right triangle has legs of length 7 and 24. What is the length of the hypotenuse?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> 7\u00b2 + 24\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>49 + 576 = c\u00b2 <\/p>\n\n\n\n<p>625 = c\u00b2 <\/p>\n\n\n\n<p>c = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>25<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is the 7-24-25 Pythagorean triple \u2014 less well known than 3-4-5 but worth memorising since it appears in AMC 8 past papers. The square root of 625 is 25 \u2014 knowing that 25\u00b2 = 625 is a useful number fact.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pythagorean theorem worksheet practice problems \u2014 Level 2<\/strong><\/h2>\n\n\n\n<p>These problems require applying the theorem in a geometric context \u2014 finding diagonals, heights, and distances. They are equivalent in difficulty to questions 10 to 18 on the AMC 8.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 6 \u2014 Diagonal of a rectangle<\/strong><\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"587\" height=\"339\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.28.21-PM.png\" alt=\"rectangle math problem pythagorean theorem worksheet \" class=\"wp-image-61474\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.28.21-PM.png 587w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.28.21-PM-300x173.png 300w\" sizes=\"auto, (max-width: 587px) 100vw, 587px\" \/><\/figure>\n<\/div>\n\n\n<p>A rectangle has length 16 and width 12. What is the length of its diagonal?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> The diagonal of a rectangle divides it into two right triangles. The diagonal is the hypotenuse and the sides are the legs.<\/p>\n\n\n\n<p>a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>12\u00b2 + 16\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>144 + 256 = c\u00b2 <\/p>\n\n\n\n<p>400 = c\u00b2 <\/p>\n\n\n\n<p>c = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>20<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Recognise this as a 3-4-5 triple scaled by 4. 12 = 3 x 4, 16 = 4 x 4, diagonal = 5 x 4 = 20. Rectangle diagonal problems are a standard AMC 8 geometry type \u2014 expect to see them regularly in past papers.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 7 \u2014 Height of an isosceles triangle<\/strong><\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"553\" height=\"440\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.30.52-PM.png\" alt=\"isosceles triangle math question pythagorean theorem worksheet \" class=\"wp-image-61475\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.30.52-PM.png 553w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.30.52-PM-300x239.png 300w\" sizes=\"auto, (max-width: 553px) 100vw, 553px\" \/><\/figure>\n<\/div>\n\n\n<p>An isosceles triangle has two equal sides of length 10 and a base of length 12. What is its height?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Drop a perpendicular from the apex to the base. This creates two right triangles. The height is the vertical leg, half the base (6) is the horizontal leg, and the equal side (10) is the hypotenuse.<\/p>\n\n\n\n<p>a\u00b2 + b\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>6\u00b2 + h\u00b2 = 10\u00b2 <\/p>\n\n\n\n<p>36 + h\u00b2 = 100 <\/p>\n\n\n\n<p>h\u00b2 = 64 <\/p>\n\n\n\n<p>h = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>8<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> When a problem involves an isosceles triangle and asks for the height, always drop the perpendicular to the base first. This splits the triangle into two right triangles and the Pythagorean theorem becomes applicable. This technique appears frequently in AMC 8 geometry.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 8 \u2014 Distance between two points<\/strong><\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"534\" height=\"610\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.34.37-PM.png\" alt=\"grid math problem pythagorean theorem worksheet \" class=\"wp-image-61477\" style=\"width:450px;height:auto\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.34.37-PM.png 534w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-24-at-3.34.37-PM-263x300.png 263w\" sizes=\"auto, (max-width: 534px) 100vw, 534px\" \/><\/figure>\n<\/div>\n\n\n<p>Two points are plotted on a coordinate grid. Point A is at (1, 2) and point B is at (7, 10). What is the distance between them?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> The horizontal distance between the points is 7 minus 1 equals 6. The vertical distance is 10 minus 2 equals 8. These form the legs of a right triangle and the distance between the points is the hypotenuse.<\/p>\n\n\n\n<p>6\u00b2 + 8\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>36 + 64 = c\u00b2 <\/p>\n\n\n\n<p>100 = c\u00b2 <\/p>\n\n\n\n<p>c = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>10<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Distance between two points on a coordinate grid is always a Pythagorean theorem problem. The horizontal and vertical differences are the legs. Recognise this immediately rather than trying to use the distance formula \u2014 it is the same thing and understanding why it works is more useful than memorising a formula.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 9 \u2014 Ladder against a wall<\/strong><\/p>\n\n\n\n<p>A ladder 15 metres long leans against a vertical wall. The base of the ladder is 9 metres from the wall. How high up the wall does the ladder reach?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> The ladder is the hypotenuse, the distance from the wall is one leg, and the height up the wall is the other leg.<\/p>\n\n\n\n<p>9\u00b2 + h\u00b2 = 15\u00b2 <\/p>\n\n\n\n<p>81 + h\u00b2 = 225 <\/p>\n\n\n\n<p>h\u00b2 = 144 <\/p>\n\n\n\n<p>h = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>12 metres<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is the 3-4-5 triple scaled by 3. 9 = 3 x 3, 15 = 5 x 3, height = 4 x 3 = 12. Ladder problems are a classic Pythagorean theorem word problem type and appear across many AMC 8 past papers in slightly different forms.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 10 \u2014 Diagonal of a square<\/strong><\/p>\n\n\n\n<p>A square has side length 5. What is the length of its diagonal? Give your answer in simplest radical form.<\/p>\n\n\n\n<p><strong>Solution:<\/strong> The diagonal divides the square into two right isosceles triangles. Both legs have length 5.<\/p>\n\n\n\n<p>5\u00b2 + 5\u00b2 = c\u00b2 <\/p>\n\n\n\n<p>25 + 25 = c\u00b2 <\/p>\n\n\n\n<p>50 = c\u00b2 <\/p>\n\n\n\n<p>c = \u221a50 = \u221a(25 x 2) = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>5\u221a2<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> The diagonal of any square with side length s is s\u221a2. This result is worth memorising directly for AMC 8 purposes. Deriving it once from the Pythagorean theorem gives you the formula, and from then on you can apply it immediately.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pythagorean theorem worksheet word problems \u2014 Level 3<\/strong><\/h2>\n\n\n\n<p>These Pythagorean theorem word problems require identifying that the theorem is needed without being told explicitly, then setting up and solving the problem correctly. They are equivalent in difficulty to questions 15 to 22 on the AMC 8.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 11 \u2014 Area of a triangle<\/strong><\/p>\n\n\n\n<p>A triangle has sides of length 10, 10, and 12. What is its area?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> This is an isosceles triangle. Drop a perpendicular from the apex to the base to find the height.<\/p>\n\n\n\n<p>Half the base = 12 divided by 2 = 6 One equal side = 10 = hypotenuse<\/p>\n\n\n\n<p>6\u00b2 + h\u00b2 = 10\u00b2 <\/p>\n\n\n\n<p>36 + h\u00b2 = 100 <\/p>\n\n\n\n<p>h\u00b2 = 64 <\/p>\n\n\n\n<p>h = 8<\/p>\n\n\n\n<p>Area = half base times height = (1\/2) x 12 x 8 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>48<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Finding the area of an isosceles triangle almost always requires finding the height first using the Pythagorean theorem. The process is always the same: drop the perpendicular, identify the right triangle formed, apply the theorem.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 12 \u2014 Perimeter after finding a missing side<\/strong><\/p>\n\n\n\n<p>A right triangle has one leg of length 5 and a hypotenuse of length 13. What is the perimeter of the triangle?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Find the missing leg first.<\/p>\n\n\n\n<p>5\u00b2 + b\u00b2 = 13\u00b2 <\/p>\n\n\n\n<p>25 + b\u00b2 = 169 <\/p>\n\n\n\n<p>b\u00b2 = 144 <\/p>\n\n\n\n<p>b = 12<\/p>\n\n\n\n<p>Perimeter = 5 + 12 + 13 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>30<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Two-step problems like this are common on the AMC 8. The first step finds the missing side, the second step uses it to answer the actual question. Students who only find the missing side and stop have done the harder part but missed the final step \u2014 always re-read the question before writing down the answer.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 13 \u2014 Three-dimensional distance<\/strong><\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"683\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-1024x683.png\" alt=\"pythagorean theorem worksheet problem question box\" class=\"wp-image-61479\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-1024x683.png 1024w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-300x200.png 300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-768x512.png 768w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-1300x867.png 1300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1-800x533.png 800w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/ChatGPT-Image-Apr-24-2026-03_35_35-PM-1.png 1536w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n<\/div>\n\n\n<p>A rectangular box has length 12, width 4, and height 3. What is the length of the longest diagonal of the box \u2014 the one connecting opposite corners?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Use the Pythagorean theorem twice.<\/p>\n\n\n\n<p>Step one: find the diagonal of the base rectangle. Base diagonal\u00b2 = 12\u00b2 + 4\u00b2 = 144 + 16 = 160 Base diagonal = \u221a160 = 4\u221a10<\/p>\n\n\n\n<p>Step two: use this diagonal as one leg of a new right triangle with the height as the other leg. Longest diagonal\u00b2 = (4\u221a10)\u00b2 + 3\u00b2 = 160 + 9 = 169 Longest diagonal = \u221a169 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>13<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Three-dimensional diagonal problems require applying the Pythagorean theorem twice \u2014 once to find the diagonal of the base, then again to find the space diagonal. The formula for the space diagonal of a rectangular box is \u221a(l\u00b2 + w\u00b2 + h\u00b2). Recognising this pattern across different problems is a valuable AMC skill.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 14 \u2014 Finding area using an embedded right triangle<\/strong><\/p>\n\n\n\n<p>A circle has a chord of length 16. The chord is 6 units from the centre of the circle. What is the radius of the circle?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Draw a line from the centre perpendicular to the chord. This line bisects the chord, creating a right triangle with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Half the chord length = 8 as one leg<\/li>\n\n\n\n<li>Distance from centre to chord = 6 as the other leg<\/li>\n\n\n\n<li>Radius = hypotenuse<\/li>\n<\/ul>\n\n\n\n<p>6\u00b2 + 8\u00b2 = r\u00b2 <\/p>\n\n\n\n<p>36 + 64 = r\u00b2 <\/p>\n\n\n\n<p>100 = r\u00b2 <\/p>\n\n\n\n<p>r = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>10<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> The perpendicular from the centre of a circle to a chord always bisects the chord. Knowing this geometric fact turns what appears to be a circles problem into a straightforward Pythagorean theorem calculation. Many AMC 8 problems combine one geometric theorem with the Pythagorean theorem in exactly this way.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 15 \u2014 Multi-step word problem<\/strong><\/p>\n\n\n\n<p>A rectangular field measures 40 metres by 30 metres. A path runs diagonally across the field from one corner to the opposite corner. How much shorter is the diagonal path than walking around two sides of the field?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Step one: find the diagonal. 40\u00b2 + 30\u00b2 = c\u00b2 1600 + 900 = c\u00b2 2500 = c\u00b2 c = 50 metres<\/p>\n\n\n\n<p>Step two: find the distance walking around two sides. 40 + 30 = 70 metres<\/p>\n\n\n\n<p>Step three: find the difference. 70 minus 50 = ? metres<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>The diagonal path is 20 metres shorter<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is a 3-4-5 triple scaled by 10. 30 = 3 x 10, 40 = 4 x 10, diagonal = 5 x 10 = 50. Real-world context word problems like this are common on the AMC 8 \u2014 the mathematics is the same as the direct problems above but students must first extract the relevant numbers and structure from the context.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><a href=\"https:\/\/www.thinkacademy.ca\/amc-free-trial?source_id=6172&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\" target=\"_blank\" rel=\" noreferrer noopener\"><img loading=\"lazy\" decoding=\"async\" width=\"728\" height=\"293\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.45-PM.png\" alt=\"Think Academy AMC 8 course banner \u2014 green background with teal button reading &quot;Find the right course&quot;\" class=\"wp-image-60941\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.45-PM.png 728w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/Screenshot-2026-04-14-at-12.30.45-PM-300x121.png 300w\" sizes=\"auto, (max-width: 728px) 100vw, 728px\" \/><\/a><figcaption class=\"wp-element-caption\">Think Academy students have earned 1,700+ AMC 8 medals since 2021. Find the right course level for your child.<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pythagorean theorem worksheet questions \u2014 AMC 8 past contest style<\/strong><\/h2>\n\n\n\n<p>These problems are written in the style of actual AMC 8 past contest questions. They are harder than the problems above and require recognising which geometric technique to use without being guided.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 16 \u2014 AMC 8 style<\/strong><\/p>\n\n\n\n<p>The base of a right triangle is 24 and the hypotenuse is 26. What is the area of the triangle?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Find the height (the other leg) using the Pythagorean theorem.<\/p>\n\n\n\n<p>a\u00b2 + 24\u00b2 = 26\u00b2 <\/p>\n\n\n\n<p>a\u00b2 + 576 = 676 <\/p>\n\n\n\n<p>a\u00b2 = 100 <\/p>\n\n\n\n<p>a = 10<\/p>\n\n\n\n<p>Area = (1\/2) x base x height = (1\/2) x 24 x 10 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>120<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This is the 5-12-13 triple scaled by 2. 24 = 12 x 2, 26 = 13 x 2, missing leg = 5 x 2 = 10. Once the missing side is found, area is straightforward.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 17 \u2014 AMC 8 style<\/strong><\/p>\n\n\n\n<p>Square ABCD has side length 6. Point E is the midpoint of side CD. What is the length of AE?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Place the square on a coordinate grid. Let A be at (0, 6), B at (6, 6), C at (6, 0), D at (0, 0). Point E is the midpoint of CD so E is at (3, 0).<\/p>\n\n\n\n<p>AE\u00b2 = (6-3)\u00b2 + (6-0)\u00b2 <\/p>\n\n\n\n<p>AE\u00b2 = 9 + 36 <\/p>\n\n\n\n<p>AE\u00b2 = 45 <\/p>\n\n\n\n<p>AE = \u221a45 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>3\u221a5<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> Placing geometric figures on a coordinate grid and using the Pythagorean theorem to find distances is a powerful and flexible technique. When a problem involves a distance inside a square, rectangle, or other figure and no obvious right triangle is drawn, create one by adding a horizontal and vertical line.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 18 \u2014 AMC 8 style<\/strong><\/p>\n\n\n\n<p>A right triangle has legs in the ratio 3:4. The hypotenuse has length 20. What is the area of the triangle?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Let the legs be 3k and 4k for some value k.<\/p>\n\n\n\n<p>(3k)\u00b2 + (4k)\u00b2 = 20\u00b2 <\/p>\n\n\n\n<p>9k\u00b2 + 16k\u00b2 = 400 <\/p>\n\n\n\n<p>25k\u00b2 = 400 <\/p>\n\n\n\n<p>k\u00b2 = 16 <\/p>\n\n\n\n<p>k = 4<\/p>\n\n\n\n<p>Legs are 3 x 4 = 12 and 4 x 4 = 16.<\/p>\n\n\n\n<p>Area = (1\/2) x 12 x 16 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>96<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> When sides are given in ratio form, introduce a variable k and set up the equation using the Pythagorean theorem. This technique works for any ratio and any given side. Solving for k first, then finding the actual side lengths, keeps the working organised.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 19 \u2014 AMC 8 style<\/strong><\/p>\n\n\n\n<p>An equilateral triangle has side length 8. What is its area?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Drop a perpendicular from the apex to the base. This creates two right triangles with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hypotenuse = 8 (the equal side)<\/li>\n\n\n\n<li>One leg = 4 (half the base)<\/li>\n\n\n\n<li>Other leg = height h<\/li>\n<\/ul>\n\n\n\n<p>4\u00b2 + h\u00b2 = 8\u00b2 16 + h\u00b2 = 64 h\u00b2 = 48 h = \u221a48 = 4\u221a3<\/p>\n\n\n\n<p>Area = (1\/2) x base x height = (1\/2) x 8 x 4\u221a3 = 16\u221a3<\/p>\n\n\n\n<p><strong>Answer: 16\u221a3<\/strong><\/p>\n\n\n\n<p><strong>Key insight:<\/strong> The area of any equilateral triangle with side length s is (s\u00b2 \u221a3)\/4. This formula is worth memorising directly for AMC 8 \u2014 it comes up regularly. Understanding that it derives from the Pythagorean theorem applied to the height is more valuable than memorising it blindly.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem 20 \u2014 AMC 8 style (hardest)<\/strong><\/p>\n\n\n\n<p>Inside a rectangle with length 15 and width 9, a point P is located such that it is 12 units from one corner and 9 units from the adjacent corner. How far is P from the corner diagonally opposite to the first corner?<\/p>\n\n\n\n<p><strong>Solution:<\/strong> Place the rectangle with corner A at the origin (0,0), B at (15,0), C at (15,9), D at (0,9). Let P be at coordinates (x, y).<\/p>\n\n\n\n<p>From corner A: x\u00b2 + y\u00b2 = 144 (distance 12 from A) <\/p>\n\n\n\n<p>From corner B: (x-15)\u00b2 + y\u00b2 = 81 (distance 9 from B)<\/p>\n\n\n\n<p>Expand the second equation: x\u00b2 &#8211; 30x + 225 + y\u00b2 = 81<\/p>\n\n\n\n<p>Substitute x\u00b2 + y\u00b2 = 144: <\/p>\n\n\n\n<p>144 &#8211; 30x + 225 = 81 <\/p>\n\n\n\n<p>369 &#8211; 30x = 81 <\/p>\n\n\n\n<p>30x = 288 x = 9.6<\/p>\n\n\n\n<p>Find y: <\/p>\n\n\n\n<p>9.6\u00b2 + y\u00b2 = 144 <\/p>\n\n\n\n<p>92.16 + y\u00b2 = 144 <\/p>\n\n\n\n<p>y\u00b2 = 51.84 y = 7.2<\/p>\n\n\n\n<p>Distance from P to corner C at (15, 9): <\/p>\n\n\n\n<p>PC\u00b2 = (15 &#8211; 9.6)\u00b2 + (9 &#8211; 7.2)\u00b2 <\/p>\n\n\n\n<p>PC\u00b2 = 5.4\u00b2 + 1.8\u00b2 <\/p>\n\n\n\n<p>PC\u00b2 = 29.16 + 3.24 <\/p>\n\n\n\n<p>PC\u00b2 = 32.4 <\/p>\n\n\n\n<p>PC = \u221a32.4 = ?<\/p>\n\n\n\n<details class=\"wp-block-details is-layout-flow wp-block-details-is-layout-flow\"><summary><strong>Answer: <\/strong><\/summary>\n<p><strong>1.8\u221a10<\/strong><\/p>\n<\/details>\n\n\n\n<p><strong>Key insight:<\/strong> This problem requires setting up a coordinate system, writing simultaneous equations, and applying the Pythagorean theorem three times. It is deliberately harder than most AMC 8 questions \u2014 it is closer to AMC 10 level. If your student can follow and reproduce this solution they are very well prepared for competition geometry.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Pythagorean theorem worksheet reference sheet<\/strong><\/h2>\n\n\n\n<p>Use this as a quick reference when working through the practice problems above.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Concept<\/th><th>Formula or fact<\/th><\/tr><\/thead><tbody><tr><td>Pythagorean theorem<\/td><td>a\u00b2 + b\u00b2 = c\u00b2<\/td><\/tr><tr><td>Finding the hypotenuse<\/td><td>c = \u221a(a\u00b2 + b\u00b2)<\/td><\/tr><tr><td>Finding a leg<\/td><td>a = \u221a(c\u00b2 &#8211; b\u00b2)<\/td><\/tr><tr><td>3-4-5 triple<\/td><td>And all multiples: 6-8-10, 9-12-15, 12-16-20<\/td><\/tr><tr><td>5-12-13 triple<\/td><td>And multiples: 10-24-26<\/td><\/tr><tr><td>8-15-17 triple<\/td><td>And multiples: 16-30-34<\/td><\/tr><tr><td>7-24-25 triple<\/td><td>And multiples: 14-48-50<\/td><\/tr><tr><td>Diagonal of a square<\/td><td>d = s\u221a2<\/td><\/tr><tr><td>Height of equilateral triangle<\/td><td>h = (s\u221a3)\/2<\/td><\/tr><tr><td>Area of equilateral triangle<\/td><td>A = (s\u00b2\u221a3)\/4<\/td><\/tr><tr><td>Space diagonal of a box<\/td><td>d = \u221a(l\u00b2 + w\u00b2 + h\u00b2)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>How to use the worksheet effectively<\/strong><\/h2>\n\n\n\n<p>Working through practice problems without a plan produces slower improvement than deliberate practice. Here is how to get the most from this Pythagorean theorem worksheet.<\/p>\n\n\n\n<p><strong>Attempt every problem before reading the solution.<\/strong> Even if you are not sure of the approach, writing down what you do know \u2014 the theorem, the known values, what you are looking for \u2014 is more useful than reading the solution immediately. The struggle of attempting a problem first is what builds problem-solving intuition.<\/p>\n\n\n\n<p><strong>Identify which Pythagorean triple is involved before calculating.<\/strong> For every problem in this worksheet, check whether the numbers are a multiple of 3-4-5, 5-12-13, 8-15-17, or 7-24-25 before starting any calculation. Students who build the habit of checking for Pythagorean triples first develop significantly faster problem-solving speed over time.<\/p>\n\n\n\n<p><strong>Draw a diagram for every word problem.<\/strong> Pythagorean theorem word problems become significantly easier once you have drawn the right triangle that the problem describes. Label the known sides and the unknown side before setting up the equation.<\/p>\n\n\n\n<p><strong>Review every problem you got wrong until you can reproduce the solution without looking.<\/strong> Getting a problem wrong is only useful if you understand why and can solve it correctly next time. For each incorrect answer read the solution, put it away, and try to reproduce the working from scratch. If you cannot, read it again and try once more.<\/p>\n\n\n\n<p>This Pythagorean theorem worksheet is designed to be worked through in stages rather than all at once. Start with the Level 1 problems in one session, move to Level 2 in the next, and save the AMC 8 style questions for when the earlier problems feel comfortable. Students who return to this Pythagorean theorem worksheet regularly \u2014 rather than completing it once and moving on \u2014 build the kind of fluency that competition conditions demand. You can also use this Pythagorean theorem worksheet alongside AMC 8 past papers, treating the two as complementary resources: the past papers show you the exact question style, and this Pythagorean theorem worksheet gives you the focused practice on the specific skill those questions test.<\/p>\n\n\n\n<p>Bookmark this worksheet and come back to it as your child&#8217;s AMC 8 preparation progresses. A problem that feels difficult at the start of preparation often feels straightforward three months later \u2014 working through this Pythagorean theorem worksheet at different stages is a useful way to measure how much your child&#8217;s geometry skills have developed. If your child is working with a Think Academy teacher, share this Pythagorean theorem worksheet with them so they can identify which problem types still need work and adjust the course focus accordingly.<\/p>\n\n\n\n<p><a href=\"https:\/\/www.thinkacademy.ca\/blog\/blog\/2026\/04\/15\/amc-8-math-competition-guide-for-parents\/\">AMC 8 Math Competition: The Complete Guide for Canadian Students<\/a><\/p>\n\n\n\n<p><a href=\"https:\/\/www.thinkacademy.ca\/blog\/blog\/2026\/04\/24\/factors-of-twenty-four-and-forty-five-amc-guide\/\" target=\"_blank\" rel=\"noreferrer noopener\">Factors of 24 and 45: How to Find All Factors AMC 8 Guide<\/a><\/p>\n\n\n\n<p><a href=\"https:\/\/www.thinkacademy.ca\/blog\/blog\/2026\/04\/23\/what-is-a-broken-line-graph-examples-amc\/\" target=\"_blank\" rel=\"noreferrer noopener\">What is a Broken Line Graph? Examples, How to Read One and AMC Practice<\/a><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Frequently Asked Questions<\/strong><\/h2>\n\n\n\n<p><strong>What is the Pythagorean theorem?<\/strong> The Pythagorean theorem states that in any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: a\u00b2 + b\u00b2 = c\u00b2, where c is the hypotenuse. It is used to find any missing side of a right triangle when the other two sides are known.<\/p>\n\n\n\n<p><strong>What are the most common Pythagorean theorem problems?<\/strong> The most common types are: finding a missing side given the other two, finding the diagonal of a rectangle, finding the height of an isosceles or equilateral triangle, calculating distance between two points on a grid, and word problems involving ladders, ramps, or paths across rectangular areas.<\/p>\n\n\n\n<p><strong>What Pythagorean triples should I memorise for AMC 8?<\/strong> The four most important are 3-4-5, 5-12-13, 8-15-17, and 7-24-25, along with their multiples. The 3-4-5 triple and its multiples appear most frequently. Recognising these instantly without calculation saves significant time in competition conditions.<\/p>\n\n\n\n<p><strong>How does the Pythagorean theorem appear in AMC 8 problems?<\/strong> It appears directly in questions asking for a missing side, and indirectly in problems about diagonals of rectangles and squares, heights of triangles, distances between points on a grid, circle chord problems, and three-dimensional box diagonals. Recognising when to apply the theorem without being told is the key skill.<\/p>\n\n\n\n<p><strong>What is a Pythagorean triple?<\/strong> A Pythagorean triple is a set of three whole numbers that satisfy a\u00b2 + b\u00b2 = c\u00b2. The simplest is 3-4-5 because 9 + 16 = 25. Any multiple of a Pythagorean triple is also a Pythagorean triple \u2014 so 6-8-10, 9-12-15, and 12-16-20 are all valid triples derived from 3-4-5.<\/p>\n\n\n\n<p><strong>How do I solve Pythagorean theorem word problems?<\/strong> Draw the right triangle described in the problem first. Label the known sides and the unknown side. Identify which value is the hypotenuse (always the longest side, opposite the right angle). Substitute into a\u00b2 + b\u00b2 = c\u00b2 and solve. Check whether the numbers form a recognisable Pythagorean triple before calculating \u2014 this often makes the arithmetic unnecessary.<\/p>\n\n\n\n<p><strong>What is the Pythagorean theorem used for in real life?<\/strong> The Pythagorean theorem is used in construction (finding diagonal measurements), navigation (calculating straight-line distances), architecture (designing roof angles), and computer graphics (calculating distances between points). In mathematics it extends into trigonometry, coordinate geometry, and vector calculations at higher levels.<\/p>\n\n\n\n<p><strong>How hard are AMC 8 Pythagorean theorem questions?<\/strong> AMC 8 Pythagorean theorem questions range from straightforward missing side calculations in the first ten questions to multi-step geometry problems in questions 18 to 25. The harder problems embed the theorem inside a larger geometric context \u2014 finding the height of a triangle before calculating area, or applying the theorem twice to find a three-dimensional diagonal. Regular practice with problems at increasing difficulty levels is the most effective preparation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/www.thinkacademy.ca\/amc-free-trial?source_id=6172&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\" target=\"_blank\" rel=\" noreferrer noopener\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"341\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/3-1-1024x341.png\" alt=\"amc 8 competition american math competition\" class=\"wp-image-61108\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/3-1-1024x341.png 1024w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/3-1-300x100.png 300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/3-1-768x256.png 768w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/04\/3-1.png 1200w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Pythagorean theorem is one of the most important concepts in geometry and one of the &hellip; <a title=\"Pythagorean Theorem Worksheet: Practice Problems for AMC 8 Students\" class=\"hm-read-more\" href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/04\/24\/pythagorean-theorem-worksheet-amc-8practice\/\"><span class=\"screen-reader-text\">Pythagorean Theorem Worksheet: Practice Problems for AMC 8 Students<\/span>Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":61481,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17153,1752,17160],"tags":[11434,12796,2417,17164,254,17165],"class_list":["post-61470","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-amc-category","category-competitions","category-math-skills","tag-amc-8","tag-math-practice","tag-math-skills","tag-pythagorean-theorem-worksheet","tag-thinkacademy","tag-worksheet"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Pythagorean Theorem Worksheet: Practice Problems for AMC<\/title>\n<meta name=\"description\" content=\"A complete Pythagorean theorem worksheet for AMC 8 students. 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