Pythagorean theorem

1. The legs are a = 6 m and b = 8 m; c is the hypotenuse.
2. a^2 + b^2 = c^2 (on the formula sheet).
3. 6^2 + 8^2 = 36 + 64 = 100, so c^2 = 100.
4. c = \sqrt{100} = 10.

The hypotenuse is 10 m.

Check: 10 is longer than each leg but shorter than 6 + 8 = 14 ✓. Also, 6-8-10 is the 3-4-5 triple doubled ✓.

Tip: Square → add → square root. Stopping before the square root (100) or skipping the squares (14) are the two classic slips.

1. The right angle is at vertex Q.
2. The two sides that meet at Q are \overline{PQ} and \overline{QR}. These are the legs.
3. The hypotenuse is the side opposite the right angle — the only side that does not touch Q. That side connects the other two vertices: \overline{PR}.

The hypotenuse is \overline{PR}.

Check: a quick test — the hypotenuse's name never contains the letter of the right-angle vertex. PR has no Q ✓.

Tip: On the GED, triangles are often named by letters with no picture. Find the right-angle letter; the hypotenuse is named by the other two letters. In a^2 + b^2 = c^2, that side is c.

1. The legs are a = 9 and b = 12; c is across from the right angle.
2. Pythagorean theorem: c^2 = a^2 + b^2 = 9^2 + 12^2 = 81 + 144 = 225.
3. Square root: c = \sqrt{225} = 15.

c = 15 cm.

Check: 9 = 3 \times 3 and 12 = 4 \times 3, so this is the 3-4-5 triangle times 3, and 5 \times 3 = 15 ✓. Also 15 is longer than each leg but shorter than 9 + 12 = 21 ✓.

Tip: Memorize the common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25, and their multiples (6-8-10, 9-12-15). They save time and are a fast check.

1. The right angle is at K, so the side opposite it, \overline{JL} = 17, is the hypotenuse c. Side \overline{JK} = 8 is a leg a, and \overline{KL} is the other leg b.
2. a^2 + b^2 = c^2: 8^2 + b^2 = 17^2, so 64 + b^2 = 289.
3. Subtract: b^2 = 289 - 64 = 225.
4. Square root: b = \sqrt{225} = 15.

The other leg is 15 m.

Check: 8^2 + 15^2 = 64 + 225 = 289 = 17^2 ✓. 8-15-17 is a Pythagorean triple ✓.

Tip: Finding a leg is "square, subtract, root"; finding the hypotenuse is "square, add, root". A leg must always come out shorter than the hypotenuse — if your answer is longer than 17, you added.

1. The wall and ground form the right angle, so the ladder is the hypotenuse: c = 20. The ground distance is a leg: b = 6.
2. Missing leg: h^2 = c^2 - b^2 = 20^2 - 6^2 = 400 - 36 = 364.
3. h = \sqrt{364} \approx 19.079.
4. To the nearest tenth: 19.1.

The ladder reaches about 19.1 ft up the wall.

Check: 19.1^2 + 6^2 = 364.81 + 36 = 400.81 \approx 400 = 20^2 ✓. The height is a little less than the ladder's length, which makes sense for a steep ladder ✓.

Tip: A leg is always shorter than the hypotenuse. If you get 20.9 ft for a 20 ft ladder, you added squares when you should have subtracted.

1. Make a right triangle: go across from A, then up to B.
2. Horizontal leg: 3 - (-3) = 3 + 3 = 6 units.
3. Vertical leg: 6 - (-2) = 6 + 2 = 8 units.
4. Distance: d^2 = 6^2 + 8^2 = 36 + 64 = 100, so d = \sqrt{100} = 10.

The distance is 10 units.

Check: count grid squares: from x = -3 to x = 3 is 6 squares, from y = -2 to y = 6 is 8 squares ✓. 6-8-10 is a triple ✓.

Tip: On a grid, the legs are the change in x and the change in y. Subtracting a negative coordinate means adding — or just count the squares.

1. Along two sides: 300 + 400 = 700 yd.
2. The diagonal is the hypotenuse of a right triangle with legs 300 and 400: d^2 = 300^2 + 400^2 = 90{,}000 + 160{,}000 = 250{,}000, so d = \sqrt{250{,}000} = 500 yd.
3. Savings: 700 - 500 = 200 yd.

Kim's walk is 200 yd shorter.

Check: 300-400-500 is the 3-4-5 triple times 100 ✓, and the diagonal (500) is less than 700 ✓.

Tip: Read the last sentence carefully: "how much shorter" asks for a difference. Finding the diagonal is only the middle step.

1. The ground and the kite's height meet at a right angle, so they are the legs. The 100-ft string is the hypotenuse.
2. 60^2 + h^2 = 100^2, so 3{,}600 + h^2 = 10{,}000.
3. h^2 = 10{,}000 - 3{,}600 = 6{,}400.
4. h = \sqrt{6{,}400} = 80.

The kite is 80 ft high.

Check: 60-80-100 is the 3-4-5 triple times 20 ✓. The height (80) is less than the string (100) ✓.

Tip: Ladders, strings, wires and ramps are slanted, so they are almost always the hypotenuse. If the slanted length is given, you are finding a leg: subtract the squares.

1. The width and height of the gate meet at a right angle, so they are the legs: 6 ft and 4 ft. The brace is the hypotenuse.
2. b^2 = 6^2 + 4^2 = 36 + 16 = 52.
3. b = \sqrt{52} \approx 7.211.
4. Nearest tenth: 7.2.

The brace is about 7.2 ft long.

Check: 7^2 = 49 and 8^2 = 64, so \sqrt{52} is between 7 and 8, closer to 7 ✓. The brace is longer than the 6-ft width but shorter than 6 + 4 = 10 ✓.

Tip: The diagonal of any rectangle is \sqrt{\text{length}^2 + \text{width}^2}. Estimate between two perfect squares to check your calculator answer.

1. Make a right triangle: go across from P and then up to Q.
2. Horizontal leg: 3 - (-2) = 5 units.
3. Vertical leg: 9 - (-3) = 12 units.
4. Distance: d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

The distance is 13 units.

Check: 5-12-13 is a Pythagorean triple ✓. The distance is longer than the longer leg (12) but shorter than walking the legs, 5 + 12 = 17 ✓.

Tip: Subtract coordinates to get the legs: x_2 - x_1 and y_2 - y_1. With a negative, subtracting means adding: 9 - (-3) = 12. The GED formula sheet has no separate distance formula, so build this right triangle and use a^2 + b^2 = c^2 (which is on the sheet).

Test each set with a^2 + b^2 \overset{?}{=} c^2, where c is the longest side.

1. 7, 24, 25: 49 + 576 = 625 = 25^2 ✓ right triangle.
2. 10, 24, 26: 100 + 576 = 676 = 26^2 ✓ right triangle (5-12-13 times 2).
3. 12, 16, 20: 144 + 256 = 400 = 20^2 ✓ right triangle (3-4-5 times 4).
4. 6, 9, 11: 36 + 81 = 117, but 11^2 = 121. 117 \ne 121 ✗.

The set 6, 9, 11 is not a right triangle.

Check: since 117 < 121, the longest side is a little too long for a right angle, so the angle opposite it is slightly more than 90° ✓.

Tip: The converse of the Pythagorean theorem: if a^2 + b^2 = c^2 exactly, the triangle is right. Always use the longest side as c, and watch for "not" in the question.

1. The pole and the ground meet at a right angle, so h and 18 are the legs.
2. The 30-ft wire is across from the right angle, so it is the hypotenuse.
3. Pythagorean theorem: (\text{leg})^2 + (\text{leg})^2 = (\text{hypotenuse})^2, so h^2 + 18^2 = 30^2.
4. (Solving: h^2 = 900 - 324 = 576, so h = 24 ft.)

The equation is h^2 + 18^2 = 30^2.

Check: 24^2 + 18^2 = 576 + 324 = 900 = 30^2 ✓. 18-24-30 is 3-4-5 times 6 ✓.

Tip: Before choosing an equation, label the hypotenuse. It must be the number that stands alone on one side of a^2 + b^2 = c^2.

1. In an isosceles triangle, the height meets the base at its midpoint, so it splits the 24 ft base into two 12 ft halves.
2. Each half is a right triangle with legs 12 ft and 5 ft; the rafter r is the hypotenuse.
3. r^2 = 12^2 + 5^2 = 144 + 25 = 169.
4. r = \sqrt{169} = 13.

Each rafter is 13 ft long.

Check: 5-12-13 is a Pythagorean triple ✓. Two rafters total 26 ft, a bit more than the 24 ft span, which fits a low roof ✓.

Tip: When a height is drawn in an isosceles triangle, use half the base as a leg. Using the whole base is the most common error on roof and tent problems.

1. Along the edges: 120 + 50 = 170 m.
2. Diagonal (hypotenuse): d = \sqrt{120^2 + 50^2} = \sqrt{14{,}400 + 2{,}500} = \sqrt{16{,}900} = 130 m.
3. Difference: 170 - 130 = 40 m.

The diagonal walk is 40 m shorter.

Check: 50-120-130 is the 5-12-13 triple times 10 ✓. The saving (40 m) is less than the shorter side (50 m) — reasonable ✓.

Tip: "How much shorter / farther / longer" questions always end with a subtraction. Read the last line again before you pick: 130 is the diagonal, not the saving.

1. The height h goes from the top straight down to the middle of the base, so it splits the base into two halves of 12 \div 2 = 6 in.
2. Each half is a right triangle with hypotenuse 10 and one leg 6: h^2 + 6^2 = 10^2 \Rightarrow h^2 = 100 - 36 = 64 \Rightarrow h = 8 in.
3. Area of the pennant: \tfrac{1}{2} \times 12 \times 8 = 48\text{ in}^2.

Each pennant has an area of 48 in².

Check: 6-8-10 is the 3-4-5 triple times 2 ✓. The pennant fills exactly half of the 12 \times 8 = 96 rectangle around it ✓.

Tip: In an isosceles triangle, the height cuts the base in half. Use the theorem to get the height, then go back and finish the area with the full base.

1. Before: base 7 ft, ladder 25 ft (hypotenuse). Height: \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 ft.
2. After: the base is 7 + 8 = 15 ft from the wall. Height: \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 ft.
3. Slide: 24 - 20 = 4 ft.

The top slides down 4 ft.

Check: 7-24-25 and 15-20-25 are both right triangles: 49 + 576 = 625 ✓ and 225 + 400 = 625 ✓.

Tip: In "before and after" problems, solve two right triangles and subtract. Don't assume the top moves the same distance as the bottom — it doesn't.

1. Total north: 2 + 7 = 9 km.
2. Total west: 3 + 9 = 12 km.
3. North and west are at right angles, so these are the legs of one right triangle.
4. d = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 km.

The hiker is 15 km from camp.

Check: 9-12-15 is 3-4-5 times 3 ✓. The straight-line distance (15) is less than the total walked, 2 + 3 + 7 + 9 = 21 km ✓.

Tip: For zig-zag trips, combine all moves in the same direction first (subtract moves in opposite directions). Then there is only one right triangle to solve.

1. Wire A: legs 16 (pole) and 12 (ground). A = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 ft.
2. Wire B: legs 16 and 30. B = \sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1{,}156} = 34 ft.
3. Total wire: 20 + 34 = 54 ft.

54 ft of wire is needed.

Check: 12-16-20 is 3-4-5 times 4 ✓, and 16-30-34 is 8-15-17 times 2 ✓. Each wire is longer than its ground distance, as a hypotenuse must be ✓.

Tip: When two right triangles share a side, solve them one at a time, then combine. Label each triangle's legs and hypotenuse separately so you do not mix them up.