math-2017:8.G.7
math-2017:8.G.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
Learning goal: Apply the Pythagorean Theorem to determine unknown side lengths in right triangles that appear in real-world and mathematical problems, in two dimensions (ladders, ramps, diagonals of rectangles) and in three dimensions (diagonals of boxes, heights of cones), giving exact answers and sensible approximations when the answer is irrational.
What you may use, and when. In a right triangle with legs a and b and hypotenuse c, a² + b² = c². Before applying the theorem, confirm two things: the triangle really has a right angle, and you know which side is the hypotenuse, the side opposite the right angle. Two different unknowns appear in problems.
Finding the hypotenuse: square both legs, add, then take the square root. Legs 9 and 12 give c² = 81 + 144 = 225, so c = 15.
Finding a leg: square the hypotenuse, subtract the square of the known leg, then take the square root. Hypotenuse 15 and leg 9 give b² = 225 − 81 = 144, so b = 12. Subtraction here is not optional. Adding 15² + 9² = 306 and taking the root gives about 17.5, which is longer than the hypotenuse, and no leg can be longer than the hypotenuse.
Units. All three lengths must be in the same unit before squaring, and the answer carries that unit. If a ladder is measured in feet and its distance from the wall in inches, convert first. Squared quantities are in square units during the calculation, and taking the square root brings the answer back to plain length units.
Irrational answers. When c² or b² is a whole number that is not a perfect square, the exact answer is a square root that cannot be written as a fraction, an irrational number. (A decimal square such as 2.25 can still have a rational root: 2.25 = 9/4, so its root is 3/2 = 1.5.) Give it in two forms: the exact form, such as √56, simplified when possible (√56 = √(4 × 14) = 2√14), and a decimal approximation to a stated precision. To approximate √56 without a calculator, locate it between perfect squares: 49 < 56 < 64, so √56 is between 7 and 8. To see which end it is nearer, test the midpoint: 7.5² = 56.25, slightly too large, so √56 is below 7.5 and nearer 7 (comparing 56 with 49 and 64 would not settle this, because squaring is not evenly spaced). Try 7.48: 7.48² = 55.9504, slightly too small, while 7.49² = 56.1001 is too large, so √56 is between 7.48 and 7.49. To round to the nearest hundredth, test the midpoint: 7.485² = 56.025225 is too large, so √56 is below 7.485 and √56 ≈ 7.48 to the nearest hundredth; a calculator confirms 7.483. In a real-world problem, round to a precision the situation supports; a ladder length to the nearest tenth of a foot is sensible, to the nearest millionth is not.
A two-dimensional model. A guy wire runs from the top of a pole to an anchor in level ground 12 m from the base of the pole. The wire is 20 m long. How tall is the pole? The pole is vertical and the ground is horizontal, so the pole and the ground meet at a right angle; the wire is the hypotenuse because it is opposite that right angle. The pole height h is a leg: h² = 20² − 12² = 400 − 144 = 256, so h = 16 m. Check: 16² + 12² = 256 + 144 = 400 = 20². The step students skip is writing down which side is the hypotenuse before choosing add or subtract.
Rectangles. Every rectangle contains two right triangles cut by a diagonal, so the diagonal is a hypotenuse. A television screen 48 inches wide and 27 inches tall has diagonal d with d² = 48² + 27² = 2304 + 729 = 3033. Since 55² = 3025 and 56² = 3136, d is just over 55 inches; a calculator gives 55.07. Screens are commonly described by their diagonal length, which is why a screen can be called 55-inch while neither side is 55 inches.
Three dimensions. A rectangular box with length l, width w and height h has a space diagonal, the longest straight segment inside it, running from one bottom corner to the opposite top corner. Find it with two right triangles. First, the floor is a rectangle, so its diagonal d satisfies d² = l² + w². Second, the floor diagonal, the vertical edge at its far end and the space diagonal D form a right triangle, because a vertical edge is perpendicular to every segment on the floor. So D² = d² + h² = l² + w² + h². Model: a box 12 cm by 4 cm by 3 cm. Floor diagonal: d² = 144 + 16 = 160, so d = √160 ≈ 12.65 cm. Space diagonal: D² = 160 + 9 = 169, so D = 13 cm exactly. Keep d² = 160 exact in the second step; using the rounded 12.65² = 160.02 would introduce a small error for no reason. A straight rod 13 cm long is the longest that fits inside this box.
Cones and pyramids. A right cone, one whose apex sits directly above the center of its circular base (the only kind used in this unit), with radius r and perpendicular height h has slant height s, the distance from the apex to the edge of the base, with s² = r² + h², because the height meets the base at its center at a right angle and the radius runs from that center to the edge; for a cone whose apex is off-center the three lengths do not form one right triangle. A cone of radius 6 cm and slant height 10 cm therefore has height h² = 100 − 36 = 64, h = 8 cm. This link is needed in the volume lesson, where the formula wants the height, not the slant height.
Misconceptions, and why they are wrong. Adding the legs (5 + 12 = 17) instead of their squares ignores that the theorem is about areas of squares on the sides; the true hypotenuse is 13, and 17 would make the straight path longer than the bent path. Forgetting the square root reports c² instead of c: a diagonal of 169 cm for a 5 cm by 12 cm rectangle is impossible because a diagonal cannot exceed the two sides added together. Using the wrong side as the hypotenuse, or applying the theorem to a triangle with no right angle, produces numbers with no meaning.
Worked leg with a real-world check: A 20 m guy wire is anchored 12 m from the base of a vertical pole. Height: h² = 20² − 12² = 400 − 144 = 256, so h = 16 m. Check: 16² + 12² = 256 + 144 = 400 = 20², so the three lengths form a right triangle and the answer is consistent. The answer is in meters because every length used was in meters.
Worked irrational diagonal: A rectangular park is 20 m by 12 m. A straight path across it from corner to corner has length p with p² = 400 + 144 = 544. Since 23² = 529 and 24² = 576, p is between 23 and 24; 23.3² = 542.89 and 23.4² = 547.56, so p ≈ 23.3 m, and a calculator gives 23.32 m. Exact form: √544 = √(16 × 34) = 4√34 m. Walking along two sides instead is 20 + 12 = 32 m, so the diagonal path saves about 8.7 m.
Worked three-dimensional diagonal: A shipping box is 6 cm by 8 cm by 24 cm. Floor diagonal: d² = 36 + 64 = 100, d = 10 cm. Space diagonal: D² = 100 + 24² = 100 + 576 = 676, D = 26 cm. A 25 cm rod fits lying along the space diagonal; a 27 cm rod does not fit straight anywhere in the box. For a cube of edge 5 cm the same two steps give a face diagonal √50 ≈ 7.07 cm and a space diagonal √75 = 5√3 ≈ 8.66 cm, both irrational.
Situation-to-triangle record: For each situation, draw the right triangle, mark the right angle, label the hypotenuse and the known sides with units, and state whether you will add or subtract squares: (1) a 10 ft ladder with its foot 6 ft from a wall, find the height reached; (2) a ramp rising 1.5 m over a horizontal run of 2 m, find the ramp length; (3) a box 2 cm by 3 cm by 6 cm, find the space diagonal. Solve all three (8 ft; 2.5 m; 7 cm) and show a check for each by substituting back.
Independent performance: Write and solve three original problems: one where the hypotenuse is unknown, one where a leg is unknown, and one in three dimensions. Success criteria: each problem names a real situation with units; a labeled diagram shows the right angle and the hypotenuse; the solution shows the squares, the add-or-subtract step and the square root; any irrational answer is given in exact form and as a decimal rounded to a stated precision with the reason for that precision; and each answer is checked by substitution or by comparing with the sum of the other two sides.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
A right triangle has legs 8 cm and 15 cm. Find the hypotenuse and explain each step.
The hypotenuse is opposite the right angle, so add the squares of the legs: 8² + 15² = 64 + 225 = 289 cm². Take the square root: c = 17 cm, because 17² = 289. The answer is in centimeters because the root undoes the squaring.
A right triangle has hypotenuse 25 m and one leg 7 m. Find the other leg and explain why you subtract.
The unknown is a leg, so its square is the hypotenuse squared minus the known leg squared: b² = 625 − 49 = 576, b = 24 m. Adding instead would give a leg longer than the hypotenuse, which is impossible since the hypotenuse is the longest side.
A right triangle has legs 2 and 6. Give the hypotenuse in exact form and as a decimal, and explain how you know the decimal is reasonable.
c² = 4 + 36 = 40, so c = √40 = √(4 × 10) = 2√10. Since 36 < 40 < 49, c is between 6 and 7, and 6.5² = 42.25 > 40 puts it nearer 6; 6.3² = 39.69 and 6.4² = 40.96 put it between 6.3 and 6.4, and the midpoint test 6.35² = 40.3225 > 40 puts it below 6.35, so c ≈ 6.3 to the nearest tenth; a calculator gives 6.32. The decimal is reasonable because it is longer than the longer leg, 6, and shorter than the two legs together, 8.
A 13 ft ladder leans against a vertical wall with its foot 5 ft from the wall on level ground. How high up the wall does it reach?
The wall and ground meet at a right angle, so the ladder is the hypotenuse. Height squared is 13² − 5² = 169 − 25 = 144, so the ladder reaches 12 ft up the wall. Check: 12² + 5² = 144 + 25 = 169 = 13².
A box is 2 cm by 6 cm by 9 cm. Find the length of its space diagonal.
Floor diagonal squared (using the 2 cm and 6 cm edges): 2² + 6² = 4 + 36 = 40. Space diagonal squared: 40 + 9² = 40 + 81 = 121, so the space diagonal is 11 cm. The floor diagonal and the vertical edge form a right angle, which is why the second application of the theorem is allowed. Keeping 40 exact rather than rounding √40 avoids error, and any choice of which two edges form the floor gives the same 121.
A right cone (apex directly above the center of the base) has radius 20 cm and slant height 29 cm. Find its perpendicular height.
The height meets the center of the base at a right angle and the radius runs from that center to the edge, so height, radius and slant height form a right triangle with the slant height as hypotenuse. h² = 29² − 20² = 841 − 400 = 441, so h = 21 cm.
A student says the hypotenuse of a right triangle with legs 6 and 8 is 6 + 8 = 14. Explain the error and give the correct value.
The theorem relates the squares of the sides, not the sides themselves: c² = 6² + 8² = 36 + 64 = 100, so c = 10. A hypotenuse of 14 would make the straight path from one end to the other equal to the bent path along both legs, but a straight path is always shorter.