Math 8 · 8.G.B.6 · 8.G.B.7

The Pythagorean theorem

You know two sides of a right triangle. Is the third side already decided, and if so, how long is it?

You need

Area of trianglesMath 6 Square and cube rootsGrade 8

Unlocks

Converse of the Pythagorean theoremMath 8 Hypotenuse leg congruenceGeometry Pythagorean triplesMath 8 The distance formulaMath 8 Using the Pythagorean theoremMath 8 Special right trianglesGeometry Chord theoremsGeometry Surface area of pyramids and conesGeometry Proving the Pythagorean theorem by rearrangementMath 8 Opposite, adjacent and hypotenuseGeometry Right triangle trigonometryGrades 9 to 10 Arc lengthCalculus 2 +9 more

Where you are 18 in Triangles

Geometry11

Triangles18

▴ 9 above Hinge theorem Isosceles and equilateral triangles

The Pythagorean theorem

Converse of the Pythagorean theorem Pythagorean triples ▾ Show all 18 (4 more)

Why this exists

Tell someone that a triangle has sides of 6 and 8 and ask for the third side, and they cannot answer. The two sides could meet at any angle, and the third side could be anything from just over 2 to just under 14.

Now add one fact: the two given sides meet at a right angle. Suddenly there is no freedom left. The triangle is pinned down, and the third side is exactly 10. The Pythagorean theorem is the rule that says how to get that number from the other two, for every right triangle there is.

That turns a length you cannot reach into one you can calculate: the diagonal of a screen, the slant of a roof, the straight-line distance across a field you would rather not walk. Any time a right angle hides inside a figure, two lengths you know give you the third. Finding that hidden right angle is half the work, and it has its own part of the results below, “Using the Pythagorean theorem”.

Definition

A right triangle is a triangle with one right angle, a 90^\circ angle, usually marked with a small square in its corner. Its sides have their own names.

The two sides that meet at the right angle are the legs.
The side opposite the right angle is the hypotenuse. It never touches the square corner.

Figure 1The legs meet at the right angle. The hypotenuse sits opposite it.

It is the custom to call the legs a and b and the hypotenuse c, as in Figure 1. Which leg is a and which is b does not matter. Which side is c matters a great deal.

The Pythagorean theorem. In a right triangle, the square of the hypotenuse equals the sum of the squares of the legs:

a^2 + b^2 = c^2

In words: leg squared plus leg squared equals hypotenuse squared.

Intuition

“Squared” is not just a word here. The number a^2 is the area of a square whose side is a. So the theorem is really about three squares: build a square outward on each side of a right triangle, and the two squares on the legs together hold exactly as much area as the square on the hypotenuse.

Figure 2The two squares on the legs, 9 + 16, hold exactly as much as the square on the hypotenuse, 25.

Figure 2 shows it for the triangle with legs 3 and 4. The squares on the legs have areas 9 and 16. The tilted square on the hypotenuse has area 25, which is 9 + 16. Its side is therefore 5, because 5^2 = 25.

That is also why the hypotenuse is always the longest side. Its square holds the areas of both of the other squares, so it must be bigger than either one.

Results

Pythagorean theorem. If a triangle has a right angle, with legs a and b and hypotenuse c, then

a^2 + b^2 = c^2

To find the hypotenuse, add the squares of the legs and take the square root:

c = \sqrt{a^2 + b^2}

To find a leg, take the square of the other leg away from the square of the hypotenuse, then take the square root:

a = \sqrt{c^2 - b^2}

Only the positive square root counts, because a length cannot be negative.

Using the Pythagorean theorem

You need a distance nobody measured, and no triangle was handed to you. Where does the right triangle come from?

Someone asks how high a ladder reaches, how long a brace must be, how far apart two points are, or how much walking a shortcut saves. Nobody measured that length, and it is not a side of a triangle you were handed. But it is very often one side of a right triangle whose other sides you do know, or can get.

Reach for the theorem when the length you want sits across from a right angle, or beside one, in a figure that has one: the corner of a rectangle, the place where a wall meets the floor, where north meets east, where the two axes of a grid cross, where a height meets its base. The theorem links the three sides of a right triangle. This part of the lesson is the other half of the job: finding the triangle in the first place.

Finding the triangle and using it

The theorem, a^2 + b^2 = c^2, needs three things before it will give anything: a triangle, a right angle in it, and two of its sides. The whole skill is getting those three out of a real situation.

Find the right angle. This comes first because without it the theorem is false, so it is the one thing you may not assume. In real problems the right angle is seldom drawn. It is stated in words, or it follows from how things are placed: a vertical wall on level ground, the corner of a rectangle or a square, the directions north and east, a horizontal line and a vertical one on a grid, a height dropped to a base. A corner that only looks square in a sketch proves nothing.

Draw the triangle, adding a segment if you need one. Sometimes the triangle is already in the picture, like a ladder, a wall and the ground. More often the length you want is one side and you must supply the rest: the diagonal of a rectangle, the horizontal and vertical steps between two points, a height dropped to a base. Adding a segment to a figure changes nothing about the figure. It only lets you see a triangle that was already implied. Mark every known length on it, and put a letter on the one you want.

Decide which side is the hypotenuse before writing anything. It is the side opposite the right angle. The equation is not symmetric: the hypotenuse is the one that stands alone, squared, on one side. A ladder, a diagonal and the straight line between two points are all hypotenuses, because each one closes off two sides that meet at a right angle.

Write the equation, solve for the square, then take the root. Put the known lengths into a^2 + b^2 = c^2 and leave the unknown squared until the last line. Then take the positive square root, because a length is never negative. When the answer is not a whole number, keep it exact, such as \displaystyle \sqrt{52} = 2\sqrt{13} (see simplifying radicals), and round only at the very end, so that no rounding error gets squared along the way.

Answer the question that was asked. The side you found is often not what the question wants. It may want the area, the perimeter, the distance saved, or how far something moved. Then check that the answer makes sense: the hypotenuse must be the longest side of its triangle, and the units must agree.

When one triangle is not enough. There are three common cases, and each has an example below. A side found in one triangle can be a known side in a second triangle, as with two triangles that share a side (Example 9) or the diagonal of a box (Example 11). Two triangles can share a length, such as the ladder that keeps its length while it slides (Example 10) or the height that two halves of a triangle stand on (Example 13), and then each triangle gives its own equation. And when the unknown appears in two sides of the same triangle, the theorem gives an equation in that unknown to solve (Example 12).

Why it is true

Take any right triangle with legs a and b and hypotenuse c, and make four copies of it. Now build two squares, each with side a + b. They are the same size, so they have the same area.

Figure 3Same big square, same four triangles. What is left over must be the same area.

In the left square, put one copy of the triangle in each corner, its right angle fitting the square’s corner, so that each side of the big square is split into a piece a and a piece b. The space left in the middle has four sides, each a hypotenuse c. Its corners are right angles too: at each one, the two acute angles of the triangle sit beside it along a straight edge, and they add to 90^\circ because the three angles of the triangle add to 180^\circ. That leaves 180^\circ - 90^\circ = 90^\circ for the corner. So the middle is a square, with area c^2.

In the right square, put the same four triangles together in pairs to make two rectangles, each a by b. What is left is one square of side a and one square of side b, with total area a^2 + b^2.

Both big squares have the same area, and each has the same four triangles taken out of it. So what is left over must be the same area in both:

c^2 = a^2 + b^2

Nothing depended on the particular lengths, so this works for every right triangle. This rearrangement proof has a lesson of its own, proving the theorem by rearrangement. A different proof, using similar triangles, comes later in proving it with similarity. Euclid’s version is Proposition 47 of Book I of the Elements.

Worked examples

Example 1

Finding the hypotenuse

A right triangle has legs 8 and 6, as in Figure 4. Find the hypotenuse c.

Figure 4Two legs given. Find the hypotenuse.

Solution

The two given sides meet at the right angle, so they are the legs, and the unknown side is the hypotenuse. Put the legs in for a and b:

\begin{aligned} 8^2 + 6^2 &= c^2 \\ 64 + 36 &= c^2 \\ 100 &= c^2 \end{aligned}

The number whose square is 100 is 10, so c = 10. The negative root, -10, is thrown away, because a side cannot have negative length.

Check. The hypotenuse should be the longest side, and 10 is longer than 8 and 6.

Example 2

Finding a leg

In Figure 5 the triangle rests on its longest side. Find the side marked b.

Figure 5The hypotenuse is the side opposite the square corner, wherever it lies.

Solution

First decide which side is the hypotenuse. The right angle is at the top, and the side opposite it is the one along the bottom, so the hypotenuse is 17. The sides 8 and b are the legs. The theorem says

\begin{aligned} 8^2 + b^2 &= 17^2 \\ 64 + b^2 &= 289 \\ b^2 &= 225 \end{aligned}

Taking 64 from both sides left b^2 on its own. Since 15^2 = 225, b = 15.

Check. 64 + 225 = 289, and 17^2 = 289. Here the unknown was a leg, so it was found by subtracting. Putting b in the place of c would have given 8^2 + 17^2 = 353 and a wrong answer longer than the hypotenuse.

Example 3

When the answer is not a whole number

A right triangle has legs 3 and 7. Find the length of the hypotenuse, exactly and to the nearest hundredth.

Solution

\begin{aligned} 3^2 + 7^2 &= c^2 \\ 9 + 49 &= c^2 \\ 58 &= c^2 \end{aligned}

No whole number squares to 58, so the exact answer is written with a square root:

c = \sqrt{58}

To see about how long that is, notice 7^2 = 49 and 8^2 = 64. Since 58 lies between 49 and 64, \displaystyle \sqrt{58} lies between 7 and 8, a bit nearer 8. A calculator gives

c = \sqrt{58} \approx 7.62

Most right triangles look like this one. Whole-number sides on all three, like 6, 8, 10, are the exception, and they have a lesson of their own: Pythagorean triples.

Example 4

Two equal legs

A right triangle has two legs of equal length and a hypotenuse of 10. How long is each leg?

Solution

Call each leg x. Both legs go into the theorem, and the hypotenuse is 10:

\begin{aligned} x^2 + x^2 &= 10^2 \\ 2x^2 &= 100 \\ x^2 &= 50 \\ x &= \sqrt{50} \approx 7.07 \end{aligned}

The two equal squares were collected into 2x^2, then both sides were divided by 2.

Check. \displaystyle \sqrt{50} is between 7 and 8, since 49 < 50 < 64, and 2 \times 50 = 100. Each leg is a little over seven tenths of the hypotenuse, a fact that comes back as one of the special right triangles.

Example 5

A ladder against a wall

A ladder 25 feet long leans against a vertical wall on level ground, with its foot 7 feet from the wall. How high up the wall does it reach?

Figure 6The wall and the ground meet at a right angle, so the ladder is the hypotenuse.

Solution

The wall is vertical and the ground is level, so they meet at a right angle. The ladder, the wall and the ground make a right triangle. The ladder is across from the right angle, so it is the hypotenuse. The legs are the 7 feet along the ground and the unknown height h:

\begin{aligned} 7^2 + h^2 &= 25^2 \\ 49 + h^2 &= 625 \\ h^2 &= 576 \\ h &= 24 \end{aligned}

The 49 was taken from both sides to leave h^2 alone, and 24^2 = 576. The ladder reaches 24 feet up the wall.

Check. 49 + 576 = 625 = 25^2, and the ladder, at 25, is longer than both the height and the distance from the wall.

Example 6

A shortcut across a field

A rectangular field is 80 meters long and 60 meters wide. Walking along two sides takes you from one corner to the opposite one. How long is the straight path across the field, and how much shorter is it?

Figure 7The diagonal closes the two sides of the field into a right triangle.

Solution

Every corner of a rectangle is a right angle, so the diagonal closes the two sides into a right triangle with the diagonal d as its hypotenuse:

\begin{aligned} 80^2 + 60^2 &= d^2 \\ 6400 + 3600 &= d^2 \\ 10000 &= d^2 \\ d &= 100 \end{aligned}

The straight path is 100 meters. Going round the two sides is 80 + 60 = 140 meters, so the shortcut saves 140 - 100 = 40 meters.

The question asked for two things, and the theorem supplied only the first. The subtraction is the second, which is the step that is easy to forget. The same fact, that the diagonal is shorter than the two sides together, is true for every rectangle; Exercise 43 asks why.

Example 7

The distance between two points on a grid

Find the distance between A(1, 2) and B(7, 6).

Figure 8The slanted segment from A to B is the hypotenuse of a triangle whose legs run along the grid.

Solution

The segment AB is slanted, so no grid line measures it. Make a right triangle from grid lines. Go from A across to the point (7, 2), straight below B, and then straight up to B. A horizontal line meets a vertical line at a right angle, so the corner at (7, 2) is a right angle, and AB is the hypotenuse.

The legs are found by subtracting coordinates. The horizontal leg is 7 - 1 = 6, and the vertical leg is 6 - 2 = 4:

\begin{aligned} AB^2 &= 6^2 + 4^2 = 36 + 16 = 52 \\ AB &= \sqrt{52} = 2\sqrt{13} \approx 7.21 \end{aligned}

Since 52 = 4 \cdot 13 and 4 is a perfect square, \displaystyle \sqrt{52} = \sqrt{4}\,\sqrt{13} = 2\sqrt{13}. Doing exactly this with the letters (x_1, y_1) and (x_2, y_2) in place of numbers gives the distance formula. You never need to memorize it, because the triangle is always there to rebuild it from.

Example 8

The height of an isosceles triangle

An isosceles triangle has two sides of 15 and a base of 18. You are given that its height from the top vertex meets the base at a right angle, at the midpoint of the base, as in Figure 9. (That this always happens in an isosceles triangle is proved later, in hypotenuse leg congruence and in proving the Base Angles Theorem; here it is part of what you are told.) Find the height h and the area of the triangle.

Figure 9The height cuts the isosceles triangle into two right triangles.

Solution

The height splits the triangle into two right triangles. Look at the left one. Its right angle is where the height meets the base, so its hypotenuse is the slanted side, 15. Its legs are h and, because the height lands at the given midpoint, half the base, 18 \div 2 = 9:

\begin{aligned} 9^2 + h^2 &= 15^2 \\ 81 + h^2 &= 225 \\ h^2 &= 144 \\ h &= 12 \end{aligned}

Now the area of a triangle is half the base times the height:

\text{area} = \tfrac{1}{2} \times 18 \times 12 = 108

The theorem found a length that was not given, so that another formula could be used. That is the most common way it gets used.

Example 9

Two right triangles sharing a side

In Figure 10, \overline{AD} is perpendicular to \overline{BC}. AB = 15, BD = 9 and DC = 16. Find AD and AC.

Figure 10Two right triangles sharing the side \overline{AD}.

Solution

The right angle at D belongs to two triangles at once, \triangle ABD and \triangle ADC.

Start with \triangle ABD, where two sides are known. Its hypotenuse is \overline{AB}, the side opposite the right angle at D, and its legs are \overline{BD} and \overline{AD}:

\begin{aligned} 9^2 + AD^2 &= 15^2 \\ 81 + AD^2 &= 225 \\ AD^2 &= 144 \\ AD &= 12 \end{aligned}

Now \triangle ADC has both legs known, AD = 12 and DC = 16, and its hypotenuse is \overline{AC}:

\begin{aligned} 12^2 + 16^2 &= AC^2 \\ 144 + 256 &= AC^2 \\ 400 &= AC^2 \\ AC &= 20 \end{aligned}

So AD = 12 and AC = 20. The side found in the first triangle was the missing piece for the second.

Example 10

A ladder that slides

A ladder 10 feet long stands with its top 8 feet up a wall. The top slides down 1 foot. How far does the foot of the ladder move out along the ground?

Figure 11The ladder does not change length, so both positions are right triangles with the same hypotenuse.

Solution

The ladder is the same length before and after, so it is the hypotenuse, 10, of a right triangle in both positions. The wall and ground meet at a right angle in both.

Before. The top is 8 up, so the foot is a distance f from the wall, where f^2 + 8^2 = 10^2. That gives f^2 = 36, so f = 6.

After. The top is 7 up. Call the new distance of the foot from the wall x:

\begin{aligned} x^2 + 7^2 &= 10^2 \\ x^2 &= 100 - 49 = 51 \\ x &= \sqrt{51} \approx 7.14 \end{aligned}

The foot moved from 6 to \displaystyle \sqrt{51}, so it moved

\sqrt{51} - 6 \approx 7.14 - 6 = 1.14 \text{ feet}

The top came down 1 foot and the foot moved out more than 1 foot. Both triangles were needed, and the ladder’s unchanged length is what tied them together.

Example 11

The longest rod in a box

A box is 12 centimeters long, 4 centimeters wide and 3 centimeters high. What is the longest straight rod that fits inside it?

Figure 12First the floor diagonal, then the diagonal of the whole box, which stands on it.

Solution

The longest rod runs from one bottom corner to the opposite top corner, the diagonal d of the box. In the box as drawn there is no right triangle with d as a side, so one has to be built in two steps.

Step one, across the floor. The floor is a 12 by 4 rectangle. Its diagonal f, from the corner to the far corner of the floor, is the hypotenuse of a right triangle with legs 12 and 4:

f^2 = 12^2 + 4^2 = 144 + 16 = 160

Step two, up to the ceiling. The vertical edge at the far end of that diagonal stands at a right angle to the floor, so it is at a right angle to every line in the floor through its foot, including f. The floor diagonal, that edge of length 3, and the rod d form a right triangle whose hypotenuse is d:

d^2 = f^2 + 3^2 = 160 + 9 = 169

So d = 13. The longest rod is 13 centimeters.

The number f itself was never needed, only f^2, and the theorem hands you f^2 directly. That is why the square roots can wait. The result is d^2 = 12^2 + 4^2 + 3^2, and the lesson coordinates and distance in three dimensions turns it into a formula for any two points.

Example 12

An unknown in two sides

The shorter side of a rectangle is 9 meters. Its diagonal is 1 meter longer than its longer side. Find the sides and the diagonal.

Solution

Let the longer side be x meters, so the diagonal is x + 1. The diagonal is the hypotenuse of the right triangle formed by the two sides at a corner:

9^2 + x^2 = (x + 1)^2

The unknown is now on both sides of the equation, so expand the right side first, using (x + 1)^2 = x^2 + 2x + 1:

\begin{aligned} 81 + x^2 &= x^2 + 2x + 1 \\ 81 &= 2x + 1 \\ 80 &= 2x \\ x &= 40 \end{aligned}

The x^2 was on both sides, so it canceled, and what is left is a linear equation. The longer side is 40 meters and the diagonal is 41 meters.

Check. 9^2 + 40^2 = 81 + 1600 = 1681 = 41^2.

The canceling is not a coincidence of these numbers. Whenever the diagonal is given as the longer side plus a fixed amount, both sides of the equation start with x^2, and it drops out. When the unknown sits in both legs instead, as in Exercises 33 and 36, the two x^2 terms add rather than cancel, and a quadratic is left to solve.

Example 13

Making a right triangle where none was given

A triangle has sides 13, 14 and 15. Find its area.

Figure 13No right angle was given. The height makes two.

Solution

The area is half the base times the height. The base can be the side 14, but no height is given, and the triangle as given has no right angle to work with. Make one: drop the height h from the top vertex straight down to the base. A height meets its base at a right angle, so it cuts the triangle into two right triangles, and h is a leg of both.

Say the foot of the height is a distance x from the left end of the base. Then it is 14 - x from the right end. The side 13 is the hypotenuse of the left triangle, and the side 15 is the hypotenuse of the right one:

\begin{aligned} x^2 + h^2 &= 13^2 = 169 \\ (14 - x)^2 + h^2 &= 15^2 = 225 \end{aligned}

Two equations share the unknown h^2, and taking one from the other removes it:

\begin{aligned} (14 - x)^2 - x^2 &= 225 - 169 \\ 196 - 28x + x^2 - x^2 &= 56 \\ 196 - 28x &= 56 \\ x &= 5 \end{aligned}

Since 0 < 5 < 14, the foot does lie on the base, as the drawing assumed. Now go back to the left triangle:

h^2 = 169 - 5^2 = 144, \qquad h = 12

Then the area is

\tfrac{1}{2} \times 14 \times 12 = 84

Check. The right triangle on the right has legs 14 - 5 = 9 and 12, and 9^2 + 12^2 = 81 + 144 = 225 = 15^2.

The triangle was not a right triangle, but it was made of two, and that is all the theorem needs. Unlike Example 8, nothing here said where the height lands; the two equations found it.

Boundaries

The triangle must have a right angle, and it has to be real. For any other triangle the equation is false. A triangle with all three sides 2 has 2^2 + 2^2 = 8, but the third side squared is 4. When the angle between two sides is wider than a right angle, the third side comes out longer than \displaystyle \sqrt{a^2 + b^2}; when it is narrower, shorter. The law of cosines measures by how much, and it is what replaces this theorem for triangles that are not right. In a real problem, a sketch that looks square is not evidence: the right angle must be given, or follow from something given, such as vertical and horizontal, the corner of a rectangle, perpendicular lines or a height.

c must be the side opposite the right angle. The equation is not the same with the letters moved around. Before writing anything down, find the square corner and take the side across from it as c, however the triangle is turned. The numbers alone never say which side that is: two sides of 9 and 12 give a third side of 15 if they are the legs, and \displaystyle \sqrt{63} if 12 is the hypotenuse. The figure has to decide. Exercise 45 asks for both.

It gives c^2, not c. The last step is always a square root, and only the positive root is a length. Often the root is not a whole number, and the exact answer stays as \displaystyle \sqrt{\;}.

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

The figure must be flat. The theorem describes triangles in a plane. A ladder against a wall is close enough, and so is a field or a map of a town. Over very long distances on the Earth’s curved surface, walking north and then east does not make a flat right triangle, and the answer from the theorem drifts away from the real distance.

Real things are not perfect lines. A ladder bends a little and a wall is not perfectly vertical. The answer is exactly as good as the measurements that went in, so giving 1.1414\ldots feet when the ladder was measured to the nearest foot claims more than is known.

Two unknowns need two facts. One equation from one triangle can find one unknown. When a problem has more, as in the 13, 14, 15 triangle, you need a second triangle or a second stated relation, and then two equations.

It does not run backwards by itself. The theorem says: right angle, therefore a^2 + b^2 = c^2. Going the other way, from the equation to a right angle, is a separate theorem with its own proof, the converse of the Pythagorean theorem.

Where this goes next

Put the right triangle on a grid, with its legs along the axes, and the work of Example 7 becomes the distance formula between two points. Take a third coordinate and use the theorem twice, as in the box, and you have distance in three dimensions. A right triangle made of the height, a radius and the slant height gives the height of a cone, and one like it gives the height of a pyramid (Exercises 29 and 30), which is what the volume of pyramids and cones needs.

Turned around, it tests whether a corner is square: the converse says that if the squares add up, the angle is right, which is why a corner marked out with sides 3, 4 and 5 is a square corner. The whole-number cases, 3, 4, 5 and 5, 12, 13 and 8, 15, 17, are Pythagorean triples. Two right triangles turn up so often that their side ratios are worth memorizing: special right triangles.

In geometry it proves that two right triangles with matching hypotenuses and one matching leg are congruent, hypotenuse leg congruence. And it sits under all of right triangle trigonometry, which picks up exactly where this lesson stops: once the sides of a right triangle are tied together, the angles can be tied to them too, and a problem that gives an angle instead of a second side, which this theorem cannot reach, becomes solvable.

Exercises

Naming the parts

In \triangle PQR the right angle is at Q. Name the hypotenuse and the two legs.
A right triangle has sides 9, 40 and 41. Which side is the hypotenuse? Show that the three lengths fit the Pythagorean theorem.
A right triangle has legs m and n and hypotenuse k. Write the Pythagorean theorem for it, then rewrite it to give m^2 on its own.

Finding the hypotenuse

The legs of a right triangle are given. Find the hypotenuse. Give an exact answer, and a decimal to the nearest hundredth when it is not a whole number.

5 and 12
15 and 20
1 and 3
1.5 and 2

Finding a leg

The hypotenuse and one leg of a right triangle are given. Find the other leg, exactly, and to the nearest hundredth when it is not a whole number.

Hypotenuse 37, leg 12
Hypotenuse 26, leg 10
Hypotenuse 9, leg 4
Hypotenuse 6.5, leg 2.5

Finding the triangle in a situation

A 15-foot ladder has its foot 9 feet from a vertical wall. How high up the wall does it reach?
A television screen is 96 centimeters wide and 54 centimeters high. Find the length of its diagonal, exactly and to the nearest tenth of a centimeter.
A ramp is 8.5 meters long and rises 4 meters. How far along the ground does it run?
Two hikers leave the same point. One walks 7 kilometers north and the other walks 24 kilometers east. How far apart are they?
A square garden has a diagonal of 20 meters. How long is each side, exactly and to the nearest hundredth of a meter?
A cable runs from the top of a vertical mast to a point on level ground 9 meters from the foot of the mast. The cable is 41 meters long. How tall is the mast?

Right triangles inside other shapes

A rectangle is 20 by 21. How long is its diagonal?
A square has sides of 6. Find the length of its diagonal, exactly and to the nearest hundredth.
An isosceles triangle has two sides of 5 and a base of 8. You are given that its height meets the base at a right angle at the midpoint of the base (the fact proved in hypotenuse leg congruence). Find the height and the area.
A right triangle has two equal legs and a hypotenuse of 8. Find the length of each leg, exactly and to the nearest hundredth.
A right triangle has legs 3 and 4. A second right triangle has as its legs the hypotenuse of the first triangle and a side of 12. Find the hypotenuse of the second triangle.

Distance between points

Find the distance between (2, 1) and (8, 9).
Find the distance between (-3, 4) and (2, -1), exactly and to the nearest hundredth.
A triangle has vertices (-2, 1), (4, 1) and (4, 9). Find its perimeter.

Two steps

A 13-foot ladder stands with its top 12 feet up a wall. The top slides down 2 feet. How far does the foot of the ladder move out from where it started? Give the answer to the nearest hundredth of a foot.
A box is 2 centimeters by 3 centimeters by 6 centimeters. What is the longest straight rod that fits inside it?
A room is 12 meters long, 9 meters wide and 8 meters high. How far is it from a corner of the floor to the opposite corner of the ceiling?
A right circular cone, whose apex is directly above the center of its circular base, has a slant height of 13 centimeters and a base of radius 5 centimeters. Find its height, and then its volume, exactly.
A right square pyramid, whose apex is directly above the center of its square base, has a base of side 16 meters. Each triangular face has a height of 17 meters, measured from the middle of a base edge up to the apex. How tall is the pyramid, measured straight down from the apex to the center of the base?
An equilateral triangle has sides of 6. Find its height and its area, exactly. You may use that the height meets the base at its midpoint.
A triangle has sides 10, 17 and 21. Find the height onto the side of 21, and the area of the triangle.

An unknown in more than one place

The legs of a right triangle are x and 2x, and its hypotenuse is \displaystyle \sqrt{45}. Find x and the length of each leg.
The shorter side of a rectangle is 10 centimeters. Its diagonal is 2 centimeters longer than its longer side. Find the longer side and the diagonal.
One leg of a right triangle is 15, and its hypotenuse is 5 more than its other leg. Find the hypotenuse.
A 20-foot ladder reaches a height that is 4 feet more than the distance from its foot to the wall. Find both the distance and the height.
Two cars leave a crossroads at the same moment. One drives north at 60 kilometers per hour and the other drives east at 80 kilometers per hour, on straight roads. (a) How far apart are they after half an hour? (b) After how long are they 150 kilometers apart?

Reasoning

Use the Pythagorean theorem to explain why the hypotenuse is always the longest side of a right triangle.
A triangle has sides 4, 5 and 6. Explain, without drawing it, why it cannot be a right triangle.
Both legs of a right triangle are doubled, and the angle between them stays a right angle. What happens to the hypotenuse? Explain using the theorem.
In the proof in “Why it is true”, where exactly is the right angle used? Explain why the proof would fail for a triangle without one.
A box has edges a, b and c. Show that its longest diagonal has length \displaystyle \sqrt{a^2 + b^2 + c^2}, and say where each of the two right angles you used is.
Explain why the diagonal of a rectangle is always shorter than the two sides that run between the same corners added together.
A builder measures 6 feet along one board and 8 feet along another, from the corner where they meet, and finds the two marks are 10 feet apart. Which theorem lets her conclude the corner is square, and why does the Pythagorean theorem itself not do that?
A triangle is said to have a right angle and two sides of 9 and 12. Explain why the third side is not yet determined, and find both possible lengths.

Answers

The hypotenuse is \overline{PR}, the side opposite the right angle at Q. The legs are \overline{PQ} and \overline{QR}, the two sides that meet at Q.
The hypotenuse is 41, the longest side. 9^2 + 40^2 = 81 + 1600 = 1681, and 41^2 = 1681.
m^2 + n^2 = k^2, so m^2 = k^2 - n^2, by taking n^2 from both sides.
13. 5^2 + 12^2 = 25 + 144 = 169, and \displaystyle \sqrt{169} = 13.
25. 15^2 + 20^2 = 225 + 400 = 625, and \displaystyle \sqrt{625} = 25.
\displaystyle \sqrt{10} \approx 3.16. 1^2 + 3^2 = 1 + 9 = 10, and 10 is not a perfect square.
2.5. 1.5^2 + 2^2 = 2.25 + 4 = 6.25, and 2.5^2 = 6.25.
35. 37^2 - 12^2 = 1369 - 144 = 1225, and \displaystyle \sqrt{1225} = 35.
24. 26^2 - 10^2 = 676 - 100 = 576, and \displaystyle \sqrt{576} = 24.
\displaystyle \sqrt{65} \approx 8.06. 9^2 - 4^2 = 81 - 16 = 65.
6. 6.5^2 - 2.5^2 = 42.25 - 6.25 = 36, and \displaystyle \sqrt{36} = 6.
12 feet. The ladder is the hypotenuse: 9^2 + h^2 = 15^2, so h^2 = 225 - 81 = 144 and h = 12.
\displaystyle 6\sqrt{337} \approx 110.1 centimeters. The diagonal is the hypotenuse: 96^2 + 54^2 = 9216 + 2916 = 12132 = 36 \cdot 337, so \displaystyle d = 6\sqrt{337}.
7.5 meters. The ramp is the hypotenuse: 8.5^2 - 4^2 = 72.25 - 16 = 56.25, and \displaystyle \sqrt{56.25} = 7.5.
25 kilometers. North and east are at right angles, so 7^2 + 24^2 = 49 + 576 = 625 and \displaystyle \sqrt{625} = 25.
\displaystyle 10\sqrt{2} \approx 14.14 meters. The diagonal is the hypotenuse of a triangle with two equal legs s: s^2 + s^2 = 400, so s^2 = 200 and \displaystyle s = \sqrt{200} = 10\sqrt{2}.
40 meters. The mast is vertical and the ground level, so the cable is the hypotenuse: 41^2 - 9^2 = 1681 - 81 = 1600, and \displaystyle \sqrt{1600} = 40.
29. The diagonal is the hypotenuse of a right triangle with legs 20 and 21: 400 + 441 = 841, and \displaystyle \sqrt{841} = 29.
\displaystyle \sqrt{72} \approx 8.49. The diagonal cuts the square into two right triangles with legs 6 and 6: 36 + 36 = 72.
Height 3, area 12. Each half is a right triangle with hypotenuse 5 and one leg 8 \div 2 = 4, so h^2 = 25 - 16 = 9 and h = 3. Area = \tfrac{1}{2} \times 8 \times 3 = 12.
\displaystyle \sqrt{32} \approx 5.66. x^2 + x^2 = 64, so 2x^2 = 64, x^2 = 32.
13. The first hypotenuse is \displaystyle \sqrt{9 + 16} = 5. The second triangle has legs 5 and 12, so its hypotenuse is \displaystyle \sqrt{25 + 144} = \sqrt{169} = 13.
10. The legs are 8 - 2 = 6 and 9 - 1 = 8, and 6^2 + 8^2 = 100.
\displaystyle 5\sqrt{2} \approx 7.07. The legs are 2 - (-3) = 5 and 4 - (-1) = 5, and 5^2 + 5^2 = 50, so the distance is \displaystyle \sqrt{50} = 5\sqrt{2}.
24. The first two vertices are 6 apart along a horizontal line, and the last two are 8 apart along a vertical line, so the triangle has a right angle at (4, 1). The third side is the hypotenuse, \displaystyle \sqrt{6^2 + 8^2} = 10. The perimeter is 6 + 8 + 10 = 24.
About 3.31 feet. At the start the top is 12 up, so the foot is \displaystyle \sqrt{13^2 - 12^2} = 5 from the wall. After the slide the top is 10 up, so the foot is \displaystyle \sqrt{13^2 - 10^2} = \sqrt{69} \approx 8.31 from the wall. It moved \displaystyle \sqrt{69} - 5 \approx 3.31.
7 centimeters. The floor diagonal has f^2 = 2^2 + 3^2 = 13. Then d^2 = 13 + 6^2 = 49, so d = 7.
17 meters. The floor diagonal has f^2 = 12^2 + 9^2 = 225. Then d^2 = 225 + 8^2 = 289, so d = 17.
The height is 12 centimeters and the volume is 100\pi cubic centimeters. The apex is directly above the center, so the height meets the base at its center, at a right angle to every radius. The height, a radius and the slant height make a right triangle with hypotenuse 13: h^2 = 13^2 - 5^2 = 144, so h = 12. Then V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi \cdot 25 \cdot 12 = 100\pi.
15 meters. The apex is directly above the center, so the height meets the base there at a right angle to every line in the base. From the center to the middle of a base edge is half the side, 8, and that line lies in the base, so it is at a right angle to the height. The slant height 17 is the hypotenuse: h^2 = 17^2 - 8^2 = 225, so h = 15.
The height is \displaystyle 3\sqrt{3} and the area is \displaystyle 9\sqrt{3}. The height meets the base at its midpoint (the theorem shows why: the two right triangles share the leg h and both have hypotenuse 6, so each piece of the base is \displaystyle \sqrt{36 - h^2}, and the pieces are equal). So one leg is 3 and the hypotenuse is 6: h^2 = 6^2 - 3^2 = 27 and \displaystyle h = \sqrt{27} = 3\sqrt{3}. Area \displaystyle = \tfrac{1}{2} \cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3}.
The height is 8 and the area is 84. Let the foot be x from the end of the side 10. Then x^2 + h^2 = 100 and (21 - x)^2 + h^2 = 289. Subtract: 441 - 42x = 189, so x = 6. Then h^2 = 100 - 36 = 64 and h = 8. Since 0 < 6 < 21, the foot is on the base. Area = \tfrac{1}{2} \cdot 21 \cdot 8 = 84.
x = 3, and the legs are 3 and 6. x^2 + (2x)^2 = 45 gives x^2 + 4x^2 = 45, so 5x^2 = 45 and x^2 = 9. Only the positive root is a length. Check: 9 + 36 = 45.
The longer side is 24 centimeters and the diagonal is 26 centimeters. Let the longer side be x: 10^2 + x^2 = (x + 2)^2, so 100 = 4x + 4 and x = 24. Check: 100 + 576 = 676 = 26^2.
25. Let the other leg be x, so the hypotenuse is x + 5: 15^2 + x^2 = (x + 5)^2, so 225 = 10x + 25 and x = 20. The hypotenuse is 20 + 5 = 25.
The foot is 12 feet from the wall and the ladder reaches 16 feet. Let the distance be x, so the height is x + 4: x^2 + (x + 4)^2 = 400, which simplifies to x^2 + 4x = 192. Completing the square gives (x + 2)^2 = 196, so x + 2 = \pm 14, and x = 12 or x = -16. A distance is not negative, so x = 12, and the height is 16. Check: 144 + 256 = 400.
(a) 50 kilometers. In half an hour they have gone 30 and 40 kilometers along roads at right angles, and 30^2 + 40^2 = 2500. (b) 1.5 hours. After t hours they have gone 60t and 80t, so the distance is \displaystyle \sqrt{(60t)^2 + (80t)^2} = \sqrt{10000t^2} = 100t. Setting 100t = 150 gives t = 1.5.
c^2 = a^2 + b^2, and b^2 is more than 0, so c^2 is more than a^2, which makes c longer than a. In the same way a^2 is more than 0, so c is longer than b.
If it were a right triangle, its hypotenuse would be the longest side, 6, and the theorem would force 4^2 + 5^2 = 6^2. But 4^2 + 5^2 = 16 + 25 = 41, while 6^2 = 36. The equation fails, so the triangle has no right angle.
The hypotenuse doubles too. The new legs are 2a and 2b, and (2a)^2 + (2b)^2 = 4a^2 + 4b^2 = 4(a^2 + b^2) = 4c^2. The number whose square is 4c^2 is 2c.
In three places. Each triangle’s right angle fits into a corner of the big square. At each corner of the middle shape, the two acute angles of the triangle sit side by side along a straight edge, and they add to 90^\circ only because the third angle is 90^\circ; that leaves 90^\circ for the corner, so the middle shape is a square of area c^2. And two triangles fit together into an a by b rectangle only because each has a square corner. Without a right angle the triangles do not fill the corners of a square at all, so neither arrangement can be built.
Call the floor diagonal f. The floor is a rectangle a by b, with a right angle at its corner, so f^2 = a^2 + b^2. The vertical edge c stands at the end of f at a right angle to the floor, which is the second right angle. The box diagonal d is the hypotenuse of the triangle formed by f and c, so d^2 = f^2 + c^2 = a^2 + b^2 + c^2, and \displaystyle d = \sqrt{a^2 + b^2 + c^2}.
Call the sides a and b and the diagonal d. Then d^2 = a^2 + b^2, while (a + b)^2 = a^2 + 2ab + b^2. Because a and b are positive, 2ab > 0, so (a + b)^2 > d^2. Both a + b and d are positive, so a + b > d.
The converse of the Pythagorean theorem. She has found 6^2 + 8^2 = 100 = 10^2, and the converse says that if the squares of two sides add to the square of the third, the angle between the two is a right angle. The Pythagorean theorem itself starts from a right angle and ends at the equation. It says nothing about what an equation that holds implies, so it cannot be run backwards on its own.
A right angle does not say which side is the hypotenuse. If 9 and 12 are both legs, the third side is the hypotenuse, \displaystyle \sqrt{9^2 + 12^2} = \sqrt{225} = 15. If 12 is the hypotenuse and 9 is a leg, the third side is a leg, \displaystyle \sqrt{12^2 - 9^2} = \sqrt{63} = 3\sqrt{7} \approx 7.94. The case with 9 as the hypotenuse is impossible, since the hypotenuse is the longest side.