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Using the Pythagorean theorem

A ramp, a television, a kite on its string. Nobody handed you a right triangle, so where is it hiding, and what can it tell you?

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When you reach for it

You want a length that nobody measured: how far out a ramp starts, how tall a screen is, how long a brace must be, how high a kite is flying. The Pythagorean theorem can find it, but only after the length has been seen as one side of a right triangle whose other two sides are known, or can be worked out.

So this skill is mostly about seeing. Right angles hide in ordinary things: a wall standing on level ground, every corner of anything rectangular, a pole standing straight up, the line straight down from something in the air. Find the right angle and the reason it is there, and the theorem does the rest.

The procedure

Part 1: one right triangle in a real figure

Find the right angle first, and know why it is there. The theorem is true only for right triangles, so the right angle is the one thing you may not guess. In a real problem it comes from how things are built or placed: vertical meets level at a right angle, so a wall on flat ground makes one, and so does a mast. Every corner of a rectangle is one. A corner that merely looks square in a sketch is not a reason.

Draw just the triangle and label it. Sketch the triangle on its own, with the known lengths on it and a letter on the unknown. Often one of its sides is not drawn in the real thing at all: the diagonal of a screen is not a line on the screen.

Name the hypotenuse before writing the equation. The hypotenuse is the side across from the right angle, and it is always the longest side. In real figures it is usually the slanting thing: the ramp, the diagonal, the wire from a high point down to the ground, the kite string. If the unknown is the hypotenuse, add the squares of the two legs. If it is a leg, take the square of the known leg away from the square of the hypotenuse:

c = \sqrt{a^2 + b^2} \qquad \text{or} \qquad a = \sqrt{c^2 - b^2}

Both come straight from a^2 + b^2 = c^2, the second by taking b^2 from both sides. Only the positive root is a length.

Keep it exact, then round once, sensibly. \displaystyle \sqrt{15} is an exact length; it just has no tidy decimal. Keep the exact value through the working and round only at the end, to a precision that fits the measurements that went in: a ramp measured to the nearest meter does not start 3.872983 meters out. When a rounded value stands in for an irrational one, say so, with \approx or the word “about”. To see that a root is sensible, place it between two whole numbers first.

Ask whether the answer is believable. The hypotenuse must be longer than each leg, and shorter than the two legs added together. A screen whose diagonal is shorter than its width means a slip somewhere.

Part 2: two steps, and no right angle

When the triangle gives only part of the answer. Often the side the theorem finds is not what the question asks for. It gives a kite’s height above the hand, and the question wants the height above the ground; it gives a path’s length, and the question wants how much shorter that path is. After the theorem, ask what is still missing: an addition, a subtraction, an area, a comparison.

When the right angle has to be made. The front of a tent or the end of a roof is often an isosceles triangle, with no right angle at any corner. A vertical line from its top to the middle of its base meets the base at a right angle and cuts it into two right triangles. Then the leg along the bottom is half the base, not the whole of it.

When there is no right angle at all, stop. The equation a^2 + b^2 = c^2 will take any two numbers and produce a third, so the arithmetic never warns you. If the angle between two sides is not a right angle, that third number is simply not the third side. Example 7 shows what happens.

Worked examples

Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.

Example 1

A ramp up to a loading dock

A ramp 4 meters long runs from level ground up to the edge of a loading dock 1 meter high. The front wall of the dock is vertical. How far out from the dock does the ramp meet the ground?

1 x 4
Figure 1The dock’s front wall meets the level ground at a right angle, so the ramp is the hypotenuse.
Solution

The dock’s front wall is vertical and the ground is level, so they meet at a right angle, and the wall, the ground and the ramp make a right triangle (Figure 1). The ramp is across from that right angle, so it is the hypotenuse, 4. The legs are the 1 meter up the wall and the unknown distance x along the ground.

The unknown is a leg, so the square of the known leg comes off the square of the hypotenuse:

\begin{aligned} 1^2 + x^2 &= 4^2 \\ x^2 &= 16 - 1 = 15 \\ x &= \sqrt{15} \end{aligned}

No whole number squares to 15. Since 3^2 = 9 and 4^2 = 16, \displaystyle \sqrt{15} lies between 3 and 4, very close to 4. A calculator gives 3.8729\ldots, so the ramp meets the ground about 3.9 meters out from the dock.

Check. About 3.9 is less than the ramp’s 4, as a leg must be. The dock is only 1 meter high, so the ramp lies nearly flat, and a run just under its length is what to expect.

Example 2

A brace across a goal

The back of a practice goal is a rectangular frame 240 centimeters wide and 150 centimeters high. To keep the frame square, a straight brace is fixed across it from one corner to the opposite corner. About how long is the brace?

240 150 d
Figure 2The brace is not one of the rectangle’s sides. Drawing it in shows the right triangle.
Solution

The brace is not one of the frame’s sides, so draw it in: the diagonal d from one corner to the opposite corner (Figure 2). Every corner of a rectangle is a right angle, so the brace, the bottom bar and an upright make a right triangle, with its right angle at the corner the brace skips. The brace is across from that corner, so the unknown is the hypotenuse, and the squares are added:

\begin{aligned} d^2 &= 240^2 + 150^2 \\ &= 57600 + 22500 = 80100 \\ d &= \sqrt{80100} \end{aligned}

Since 283^2 = 80089 and 284^2 = 80656, the root is just over 283. A calculator gives 283.02\ldots, so the brace is about 283 centimeters long.

Check. 283 is longer than the long side, 240, and shorter than the two sides together, 240 + 150 = 390.

Example 3

What size is a television?

A television is sold as a “55-inch” set. That number is the length of the screen’s diagonal, from one corner to the opposite one. The screen is 48 inches wide. About how tall is it?

48 h 55
Figure 3The size a screen is sold by is its diagonal, the hypotenuse.
Solution

The screen is a rectangle, so its bottom edge and its side meet at a right angle, and the diagonal closes them into a right triangle (Figure 3). The diagonal is across from that corner, so the hypotenuse is the 55, not the 48. The legs are the width, 48, and the height h. A leg is wanted, so subtract:

\begin{aligned} 48^2 + h^2 &= 55^2 \\ h^2 &= 3025 - 2304 = 721 \\ h &= \sqrt{721} \approx 26.85 \end{aligned}

The screen is about 26.9 inches tall, a little under 27, since 27^2 = 729.

Had the 55 gone in as a leg, the working would have given \displaystyle \sqrt{48^2 + 55^2} = \sqrt{5329} = 73 inches, a height longer than the diagonal, which no rectangle has. Naming the hypotenuse first is what prevents that.

Example 4

Can he stand up in the tent?

The front of a tent is an isosceles triangle. Its two sloping sides are each 2.4 meters, and its base along the level ground is 3.2 meters. A pole stands straight up from the middle of the base to the peak. Prosper is 1.8 meters tall. Can he stand up straight under the peak?

2.4 2.4 1.6 1.6 h
Figure 4The tent has no right angle at its corners. The pole makes two.
Solution

No corner of the tent’s front is a right angle, so the theorem cannot be used on the whole triangle. The pole makes one: it stands straight up on level ground, so it meets the base at a right angle and cuts the front into two right triangles (Figure 4). Take the right-hand one. Its hypotenuse is the sloping side, 2.4, across from the right angle at the foot of the pole. Its legs are the pole h and half the base, 3.2 \div 2 = 1.6, because the pole stands at the middle.

\begin{aligned} 1.6^2 + h^2 &= 2.4^2 \\ 2.56 + h^2 &= 5.76 \\ h^2 &= 3.2 \\ h &= \sqrt{3.2} \approx 1.789 \end{aligned}

The peak is about 1.79 meters up, so Prosper, at 1.80 meters, cannot quite stand straight. He is about a centimeter too tall.

Here the rounding decides the answer. To the nearest tenth, 1.789 is 1.8, which would say the peak is exactly his height. Squares avoid rounding altogether: h^2 = 3.2, while 1.8^2 = 3.24. Since 3.2 < 3.24, the peak is lower than 1.8 meters.

Example 5

How high is the kite?

Leontyne flies a kite on 60 meters of string, pulled tight so that it is straight. She holds the string 1.4 meters above level ground, and the kite is directly above a point 38 meters from her feet. How high is the kite above the ground?

1.4 60 38 v
Figure 5The right angle is level with her hand, not on the ground. The hand height is drawn larger than scale.
Solution

The right angle is not at Leontyne’s feet. It is at the point level with her hand and directly under the kite, where a horizontal line from her hand meets the vertical line down from the kite (Figure 5). Her hand is straight above her feet, so that horizontal line is also 38 meters long.

In that triangle the string is across from the right angle, so it is the hypotenuse, 60. The legs are the 38 across and the height v of the kite above her hand:

\begin{aligned} 38^2 + v^2 &= 60^2 \\ v^2 &= 3600 - 1444 = 2156 \\ v &= \sqrt{2156} \approx 46.43 \end{aligned}

That is the height above her hand. The question asked for the height above the ground, so the second step adds the 1.4 meters between her hand and the ground:

\sqrt{2156} + 1.4 \approx 46.43 + 1.4 = 47.83

The kite is about 48 meters up. A real string sags, so the straight line from hand to kite is a little shorter than 60 meters and the kite a little lower: one more reason not to give it to the centimeter.

Example 6

A shortcut to the fountain

A park is a rectangle 240 meters by 130 meters. Ottilie comes in at the gate on one corner, A, and heads for the fountain F at the middle of the far long side. The paths run along the edges, but on a dry day she cuts straight across the grass. How much shorter is her walk?

240 130 120 s A F D
Figure 6The path to the fountain is the hypotenuse of a triangle with legs 130 and 120, not the park’s diagonal.
Solution

The straight path from A to F is not the park’s diagonal, because F is only halfway along the far side. Find the triangle it does belong to (Figure 6). By the edges, she goes up the short side from A to the corner D, 130 meters, and then along the far side to the fountain, half of 240, which is 120 meters. The corner D is a corner of the rectangle, so it is a right angle, and the straight path s is across from it: s is the hypotenuse.

\begin{aligned} s^2 &= 130^2 + 120^2 \\ &= 16900 + 14400 = 31300 \\ s &= \sqrt{31300} \approx 176.9 \end{aligned}

That says how long the shortcut is. The question asked how much shorter it is, which is a subtraction. The walk by the edges is 130 + 120 = 250 meters, so the shortcut saves

250 - \sqrt{31300} \approx 250 - 176.9 = 73.1

about 73 meters.

Example 7

No right angle, no theorem

Mattias is edging a triangular flower bed. Two of its sides are 6 meters and 9 meters, and on his sketch the corner between them looks square. He works out the third side as \displaystyle \sqrt{6^2 + 9^2} = \sqrt{117} \approx 10.8 meters and buys that much edging. When he measures the bed, the third side is 11.5 meters. What went wrong?

6 9
\displaystyle \sqrt{117}
 if square 6 9 11.5 as measured
Figure 7Same two sides, a wider corner, a longer third side. The theorem only describes the left one.
Solution

Nothing said the corner was a right angle; only the sketch looked that way. The theorem promises a^2 + b^2 = c^2 only when the angle between a and b is a right angle, so his \displaystyle \sqrt{117} was the third side of a triangle he did not have.

The measurement shows the corner is not square. If it were, the third side squared would be 6^2 + 9^2 = 117. In fact

11.5^2 = 132.25

which is more than 117, so the corner is not a right angle. The two sides open wider than a square corner, as on the right of Figure 7, and the third side comes out longer. (In high school geometry the hinge theorem makes this exact: with the same two sides, a wider angle between them gives a longer third side.)

The arithmetic gave no warning. The only guard is the reason for the right angle: Mattias needed to know the corner was square before using the theorem. Once a bed is built, the converse of the Pythagorean theorem tests a corner from the three measured sides: if the squares of the two shorter sides add up to the square of the longest, the corner is square.

Boundaries

There must be a right angle, and a reason for it. For any other triangle the equation is false. The reason has to come from the situation: vertical meeting level, the corner of a rectangle, a pole standing straight up, a line stated to be perpendicular. How a sketch looks is not a reason.

The hypotenuse is opposite the right angle and is the longest side. The theorem cannot find a leg from a “hypotenuse” that is shorter than the other leg: c^2 - b^2 would be negative, and no right triangle has those sides.

One flat triangle at a time. The theorem describes a triangle lying in a plane. A length that runs through a box from one corner to the opposite corner needs two right triangles, one after the other, which is the Pythagorean theorem in three dimensions.

All three lengths in one unit. Centimeters and meters cannot be squared and added until they are the same unit.

Real things are only nearly lines. A string sags, a wall leans a little, a frame is measured to the centimeter. An answer is no more precise than the measurements that went in, so round to match them. And round in the safe direction when it matters: wire that must reach a length is rounded up, since rounding down leaves it short.

It finds lengths, not angles. How steeply a ramp rises is an angle, and the theorem cannot give it. That needs right triangle trigonometry.

Where this goes next

Put the right triangle on a coordinate grid, with its legs along the grid lines, and the straight distance between two points comes from the same equation: that is the distance formula. Use the theorem twice, once across the floor of a box and once up to its far top corner, and you have the Pythagorean theorem in three dimensions.

When a problem gives an angle instead of a second side, a trigonometric ratio finds the side instead. The diagonal of a square has a shortcut of its own in special right triangles.

Exercises

Give each length exactly, then rounded to a sensible precision with the word “about”.

Part 1: one right triangle

  1. A ladder 6 meters long leans against a vertical wall on level ground and reaches 5.6 meters up the wall. How far is its foot from the wall?
  2. A computer monitor is 53 centimeters wide and 30 centimeters tall. How long is its diagonal?
  3. A phone screen has a diagonal of 15 centimeters and a width of 7 centimeters. How tall is it?
  4. A rope 9 meters long is pulled tight from the top of a vertical flagpole 7 meters tall to a peg in the level ground. How far is the peg from the foot of the pole?
  5. A rectangular gate is 3 meters wide and 1.2 meters tall. A wooden brace runs from one corner to the opposite corner. How long is the brace, to the nearest centimeter?
  6. A guy wire runs from a point 9 meters up a vertical mast to an anchor in the level ground 4 meters from the foot of the mast. The wire is sold in whole meters. How many meters must be bought?

Part 2: make the triangle, then finish the job

  1. The end wall of a house is topped by a triangle of roof. The two rafters are each 4.5 meters and the span between their feet is 7 meters, with the ridge directly above the middle of the span. How high is the ridge above the feet of the rafters?
  2. The front of a doghouse is an isosceles triangle with two sloping sides of 45 centimeters and a base of 30 centimeters. The line from the peak to the middle of the base meets it at a right angle. Find the height of the front and its area.
  3. Nyree flies a kite on 80 meters of straight string, holding it 1.5 meters above level ground. The kite is directly above a point 45 meters from her feet. How high is the kite above the ground?
  4. A kite is 40 meters above level ground, directly above a point 25 meters from Quilla-Mae’s feet. She holds the string 1.2 meters above the ground. How long is the string, if it is straight?
  5. Two guy wires hold a vertical mast. Both are anchored at the same point on the level ground, 6 meters from the foot of the mast. One is fixed 5 meters up the mast and the other 10 meters up. How much wire is used in all?

Part 2: shortcuts

  1. Quilla-Mae crosses a rectangular park 180 meters by 120 meters from corner to opposite corner instead of walking along two sides. She walks at 1.4 meters per second. About how many seconds does the shortcut save?
  2. One park is a square 100 meters on a side. Another is a rectangle 150 meters by 50 meters. In both, walking along two sides from a corner to the opposite corner is 200 meters. Which park has the bigger saving from a straight path between those corners, and by about how much?

Part 2: is there a right angle?

  1. Say whether the Pythagorean theorem can be used as it stands, and if it can, find the length. (a) A ladder 5 meters long leans against a vertical wall on level ground, its foot 1.5 meters from the wall. How high does it reach? (b) The same ladder rests against a garden wall that leans back, away from the ladder, with its foot 1.5 meters from the bottom of the wall. How high does it reach? (c) A triangular sail has edges of 2.5 meters and 4 meters. How long is its third edge?
  2. Leontyne’s flower bed was meant to have a square corner between its sides of 2 meters and 3 meters. She measures the third side as 4 meters. Is the corner square? How long would the third side be if it were?

Reasoning

  1. Ottilie asks how high a ladder 2 meters long reaches when its foot is 2.5 meters from the wall. Explain why there is no answer, both from the picture and from the arithmetic.
  2. In Exercise 6 the answer was rounded up, while in Example 5 the kite’s height was rounded to the nearest meter. Explain why the two roundings go different ways.
  3. Prosper says the kite in Example 5 is \displaystyle \sqrt{60^2 - 38^2} \approx 46.4 meters high. What has he left out, and why does the right triangle in that example not reach down to the ground?
  4. A television is sold as a “50-inch” set. Explain why its width and its height must both be less than 50 inches.
  5. Without working out \displaystyle \sqrt{31300}, explain why the path from A to F in Example 6 must be longer than 130 meters and shorter than 250 meters.
Answers
  1. About 2.2 meters. The ladder is the hypotenuse: x^2 = 6^2 - 5.6^2 = 36 - 31.36 = 4.64, and \displaystyle x = \sqrt{4.64} \approx 2.15.
  2. About 61 centimeters. The diagonal is the hypotenuse: d^2 = 53^2 + 30^2 = 2809 + 900 = 3709, and \displaystyle d = \sqrt{3709} \approx 60.9.
  3. About 13.3 centimeters. The diagonal 15 is the hypotenuse: h^2 = 15^2 - 7^2 = 225 - 49 = 176, and \displaystyle h = \sqrt{176} = 4\sqrt{11} \approx 13.27.
  4. About 5.7 meters. The pole is vertical and the ground level, so the rope is the hypotenuse: x^2 = 9^2 - 7^2 = 81 - 49 = 32, and \displaystyle x = \sqrt{32} = 4\sqrt{2} \approx 5.66.
  5. About 3.23 meters. The brace is the hypotenuse: b^2 = 3^2 + 1.2^2 = 9 + 1.44 = 10.44, and \displaystyle b = \sqrt{10.44} \approx 3.231.
  6. Buy 10 meters. The wire is the hypotenuse: w^2 = 9^2 + 4^2 = 81 + 16 = 97, so \displaystyle w = \sqrt{97} \approx 9.85 meters. Anything less is too short, so round up.
  7. About 2.83 meters. The vertical line from the ridge meets the span at its middle at a right angle, so a rafter is the hypotenuse and half the span, 3.5, is a leg: h^2 = 4.5^2 - 3.5^2 = 20.25 - 12.25 = 8, and \displaystyle h = \sqrt{8} = 2\sqrt{2} \approx 2.83.
  8. The height is \displaystyle 30\sqrt{2} \approx 42.4 centimeters and the area is \displaystyle 450\sqrt{2} \approx 636 square centimeters. A sloping side is the hypotenuse and half the base, 15, is a leg: h^2 = 45^2 - 15^2 = 2025 - 225 = 1800, so \displaystyle h = \sqrt{1800} = 30\sqrt{2}. The area is \displaystyle \tfrac{1}{2} \times 30 \times 30\sqrt{2} = 450\sqrt{2}.
  9. About 67.6 meters. The right angle is level with her hand: v^2 = 80^2 - 45^2 = 6400 - 2025 = 4375, and \displaystyle v = \sqrt{4375} \approx 66.14 above her hand. Adding the 1.5 gives about 67.64.
  10. About 46.2 meters. Above her hand the kite is 40 - 1.2 = 38.8 meters, and the string is the hypotenuse: s^2 = 38.8^2 + 25^2 = 1505.44 + 625 = 2130.44, and s \approx 46.16.
  11. About 19.5 meters. The wires are hypotenuses of two right triangles that share the leg 6: \displaystyle \sqrt{6^2 + 5^2} = \sqrt{61} \approx 7.81 and \displaystyle \sqrt{6^2 + 10^2} = \sqrt{136} \approx 11.66. Together, \displaystyle \sqrt{61} + \sqrt{136} \approx 19.47.
  12. About 60 seconds. The diagonal is \displaystyle \sqrt{180^2 + 120^2} = \sqrt{46800} \approx 216.3, which saves \displaystyle 300 - \sqrt{46800} \approx 83.7 meters. At 1.4 meters per second that takes 83.7 \div 1.4 \approx 59.8 seconds, about a minute.
  13. The square park, by about 17 meters. Its diagonal is \displaystyle \sqrt{20000} \approx 141.4, saving \displaystyle 200 - \sqrt{20000} \approx 58.6 meters. The long park’s is \displaystyle \sqrt{25000} \approx 158.1, saving \displaystyle 200 - \sqrt{25000} \approx 41.9 meters. The difference is about 16.7 meters.
  14. (a) Yes, about 4.8 meters. Vertical meets level at a right angle and the ladder is the hypotenuse: h^2 = 5^2 - 1.5^2 = 25 - 2.25 = 22.75, and \displaystyle h = \sqrt{22.75} \approx 4.77. (b) No. A wall that leans does not meet the ground at a right angle, so the ladder, wall and ground do not make a right triangle. (c) No. Nothing says the angle between the two edges is a right angle; the third edge could be any length between 1.5 and 6.5 meters.
  15. No. If the corner were square, the third side would be \displaystyle \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.6 meters. The measured side is 4, and 4^2 = 16, not 13, so the corner is not square.
  16. The ladder is the hypotenuse, the longest side, but here the ground leg would be 2.5, longer than the ladder: a 2 meter ladder cannot even reach a wall 2.5 meters away. In the arithmetic, h^2 = 2^2 - 2.5^2 = 4 - 6.25 = -2.25, and no length has a negative square.
  17. The wire has to reach: 9.8 meters is shorter than the 9.85 needed, so rounding must go up. The kite’s height is only reported, and the string sags anyway, so the nearest whole meter is honest and neither direction is safer.
  18. He has left out the 1.4 meters between Leontyne’s hand and the ground. The string starts at her hand, so the horizontal leg is level with her hand, and \displaystyle \sqrt{60^2 - 38^2} is the height above her hand. The height above the ground is that plus 1.4, about 48 meters.
  19. A screen’s corner is a right angle, so the width, the height and the diagonal make a right triangle with the diagonal as hypotenuse. Its square is the sum of the other two squares, so it is bigger than either, and both the width and the height are less than 50 inches.
  20. The path is the hypotenuse of the right triangle ADF, so it is longer than each leg, in particular longer than 130. And a straight line from A to F is shorter than the bent route by the edges, 130 + 120 = 250.
kg1128 · updated 2026-10-09 · CC-BY-SA-4.0