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The Pythagorean Theorem: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A triangle named by its corners

Triangle $JKL$ has a right angle at $K$ . Side $JK$ is $12$ cm long and side $KL$ is $35$ cm long.

Which side is the hypotenuse, and how long is it?
1. Hint 1
  Look for the one side that does not touch the corner where the right angle is.
2. Hint 2
  The two sides that meet at $K$ are the legs. Square each of them and add the results.
3. Hint 3
  The total is $1369$ . Look for a whole number between $30$ and $40$ whose square is $1369$ .
Answer

The hypotenuse is $JL$ , and $JL = 37$ cm.

Full solution

The right angle is at $K$ , so the legs are $JK$ and $KL$ , the two sides that meet at $K$ .
The hypotenuse is the third side, $JL$ .

Add the squares of the legs.
$JL^2 = 12^2 + 35^2$ $JL^2 = 144 + 1225$ $JL^2 = 1369$

Since $36^2 = 1296$ and $37^2 = 1369$ , the square root of $1369$ is $37$ .
So $JL = 37$ cm, which is longer than both legs, as a hypotenuse must be.

Answer

The hypotenuse is $JL$ , and $JL = 37$ cm.

Key idea

Find the hypotenuse by its position, across from the right angle, before any numbers go into the rule.
Problem 2 Legs of $6$ and $9$

A right triangle has legs of $6$ cm and $9$ cm.

Find the exact length of its hypotenuse in simplest radical form. Then give the length to the nearest tenth of a centimeter.
1. Hint 1
  The square of the hypotenuse is the squares of the two legs added together.
2. Hint 2
  That sum is not a perfect square. Look for the largest perfect square that divides it evenly.
3. Hint 3
  To round, try $10.8$ , then the halfway value $10.85$ , and see which side of $117$ each square falls.
Answer

Exactly $3\sqrt{13}$ cm (the same as $\sqrt{117}$ cm), which is about $10.8$ cm.

Full solution

Add the squares of the legs.
$c^2 = 6^2 + 9^2$ $c^2 = 36 + 81 = 117$
So
$c = \sqrt{117}.$

The largest perfect square that divides $117$ is $9$ , because $117 = 9 \times 13$ and $13$ has no perfect-square factor other than $1$ .
$c = \sqrt{9} \times \sqrt{13}$ $c = 3\sqrt{13}$

To estimate, $10^2 = 100$ and $11^2 = 121$ , so $c$ is between $10$ and $11$ , and close to $11$ .
Next, $10.8^2 = 116.64$ and
$10.85^2 = 117.7225.$
The number $117$ lies between those two squares, so $c$ is between $10.8$ and $10.85$ and rounds to $10.8$ cm.

Answer

Exactly $3\sqrt{13}$ cm (the same as $\sqrt{117}$ cm), which is about $10.8$ cm.

Key idea

Keep the exact root as the answer, and round only when a decimal is asked for, marking it as approximate.
Problem 3 A long, thin right triangle

A right triangle has a hypotenuse of $61$ m and one leg of $11$ m. How long is the other leg?
1. Hint 1
  Decide which side the $61$ m is, and which kind of side is missing.
2. Hint 2
  The missing side is a leg. Take the square of the known leg away from the square of the hypotenuse.
3. Hint 3
  The result is a perfect square. Its root is a whole number between $50$ and $70$ .
Answer

$60$ m.

Full solution

The $61$ m side is the hypotenuse, so the missing side is a leg.
Its square is what remains after taking the known leg's square away from the hypotenuse's square.

$b^2 = 61^2 - 11^2$ $b^2 = 3721 - 121$ $b^2 = 3600$
Since $60 \times 60 = 3600$ , the missing leg is $b = 60$ .

Check by running the rule forward.
$11^2 + 60^2 = 121 + 3600$
That sum is $3721$ , which is $61^2$ , so the other leg is $60$ m.

Answer

$60$ m.

Key idea

For a missing leg, subtract the known leg's square from the hypotenuse's square.
Problem 4 Hypotenuse $8$ , leg $4$

A right triangle has a hypotenuse of $8$ in and one leg of $4$ in.

Find the other leg exactly, in simplest radical form, and then to the nearest tenth of an inch.
1. Hint 1
  The missing side is a leg, so the squares are subtracted rather than added.
2. Hint 2
  After subtracting, pull the largest perfect-square factor out from under the root.
3. Hint 3
  For the decimal, trap the number under the root between two perfect squares first, then try decimals.
Answer

Exactly $4\sqrt{3}$ in (the same as $\sqrt{48}$ in), which is about $6.9$ in.

Full solution

The $8$ in side is the hypotenuse, so subtract the square of the known leg.
$b^2 = 8^2 - 4^2$ $b^2 = 64 - 16 = 48$
So
$b = \sqrt{48}.$

The largest perfect square dividing $48$ is $16$ , because $48 = 16 \times 3$ .
$b = \sqrt{16} \times \sqrt{3}$ $b = 4\sqrt{3}$

For the decimal, $36 < 48 < 49$ , so $b$ is between $6$ and $7$ , and close to $7$ .
Next, $6.9^2 = 47.61$ and
$6.95^2 = 48.3025.$
The number $48$ lies between those squares, so $b$ rounds to $6.9$ in.

Answer

Exactly $4\sqrt{3}$ in (the same as $\sqrt{48}$ in), which is about $6.9$ in.

Key idea

A missing leg can be irrational, and its exact form comes from the same simplifying you use for any square root.
Problem 5 The wire to the top of the pole

A straight wire runs from the top of a pole $4.5$ m tall to a stake in the ground $2.8$ m from the foot of the pole. The pole stands at a right angle to the level ground.

How long is the wire?

A wire from the top of a pole to a stake. The pole meets the level ground at a right angle.

Text description of this figure

Level ground runs along the bottom of the picture. On the right, a pole rises straight up from the ground and is labeled 4.5 m. On the ground to the left is a stake, and the ground distance from the stake to the foot of the pole is labeled 2.8 m. A straight wire joins the stake to the top of the pole and is labeled with a question mark. A small square at the foot of the pole marks the right angle between the pole and the ground.
1. Hint 1
  Find the right angle in the picture, and decide which length sits across from it.
2. Hint 2
  The wire is across from the right angle. Square the two other lengths and add them.
3. Hint 3
  The sum is $28.09$ . Its square root is a number with one decimal place.
Answer

$5.3$ m.

Full solution

The pole and the ground meet at a right angle, so the pole and the ground distance are the legs.
The wire, across from the right angle, is the hypotenuse.

Square the legs and add.
$w^2 = 4.5^2 + 2.8^2$ $w^2 = 20.25 + 7.84$ $w^2 = 28.09$

Since $5^2 = 25$ and $6^2 = 36$ , the wire is between $5$ m and $6$ m long.
Trying $5.3$ gives $5.3 \times 5.3 = 28.09$ exactly, so the wire is $5.3$ m long.

Answer

$5.3$ m.

Key idea

In a measurement problem, find the right angle first, because it tells you which length is the hypotenuse.
Problem 6 Is the corner square?

A builder wants a square corner. She marks a point $33$ cm along one edge and a point $56$ cm along the other edge, both measured from the corner. The two marks turn out to be $66$ cm apart.

Is the corner a right angle? Explain how you know, and say how far apart the marks would be if it were.
1. Hint 1
  Suppose for a moment that the corner is a right angle. Which side of a right triangle would the gap between the marks be?
2. Hint 2
  Work out the gap that a right angle would force, using the two marked lengths as the legs.
3. Hint 3
  Compare the gap a right angle forces with the $66$ cm the builder actually measured.
Answer

No, the corner is not a right angle. A right angle would put the marks exactly $65$ cm apart, and they are $66$ cm apart.

Full solution

If the corner were a right angle, the two marked lengths would be the legs of a right triangle, and the gap between the marks would be its hypotenuse.

$g^2 = 33^2 + 56^2$ $g^2 = 1089 + 3136$ $g^2 = 4225$
Since $65 \times 65 = 4225$ , a right angle would make the gap exactly $65$ cm.

The builder measured $66$ cm, not $65$ cm.
A right angle at the corner forces a gap of $65$ cm, so this corner cannot be a right angle.

Answer

No, the corner is not a right angle. A right angle would put the marks exactly $65$ cm apart, and they are $66$ cm apart.

Key idea

A right angle forces the length of the third side, so any other measurement shows the corner is not square.
Problem 7 A triangle with two equal sides

A triangle has two sides of $53$ cm and a base of $56$ cm. The height drawn from the top corner meets the base at its midpoint, at a right angle.

Find the height and the area of the triangle.
1. Hint 1
  The height splits the triangle into two right triangles. Name the hypotenuse of one of them.
2. Hint 2
  One leg of each small triangle is half of the base, and the height is the other leg.
3. Hint 3
  Once you know the height, the area is half of the base times the height.
Answer

The height is $45$ cm, and the area is $1260 \text{ cm}^2$ .

Full solution

The height meets the base at its midpoint at a right angle, so it cuts the triangle into two right triangles.
Each has a hypotenuse of $53$ cm and a leg of half the base, which is $28$ cm.
The height is the other leg.

$h^2 = 53^2 - 28^2$ $h^2 = 2809 - 784$ $h^2 = 2025$
Since $45 \times 45 = 2025$ , the height is $45$ cm.

The area of a triangle is half the base times the height.
$A = \tfrac12 \times 56 \times 45$ $A = 1260$
So the area is $1260 \text{ cm}^2$ .

Answer

The height is $45$ cm, and the area is $1260 \text{ cm}^2$ .

Key idea

A height can cut a shape into right triangles, and then the theorem supplies the length an area formula needs.
Problem 8 Checking a classmate's leg

Leo is finding the missing leg of a right triangle whose hypotenuse is $85$ and whose other leg is $13$ . He writes $b^2 = 85^2 + 13^2 = 7394$ , so $b$ is about $86$ .

Explain how you can tell that Leo's answer is wrong without redoing his arithmetic. Then find the correct leg.
1. Hint 1
  Compare the size of Leo's answer with the size of the hypotenuse.
2. Hint 2
  In a right triangle the hypotenuse is the longest side. Which operation on the squares gives a leg?
3. Hint 3
  Subtract the square of $13$ from the square of $85$ , then look for a whole-number root.
Answer

A leg of about $86$ would be longer than the hypotenuse of $85$ , which no leg can be. The correct leg is $84$ .

Full solution

The hypotenuse is always the longest side of a right triangle.
Leo's leg of about $86$ is longer than the hypotenuse of $85$ , so it cannot be right.
He added the squares, which is the method for a hypotenuse, not for a leg.

For a leg, subtract the squares.
$b^2 = 85^2 - 13^2$ $b^2 = 7225 - 169$ $b^2 = 7056$

Since $80^2 = 6400$ and $85^2 = 7225$ , the root is between $80$ and $85$ .
Trying $84$ gives $84 \times 84 = 7056$ , so $b = 84$ .

Check by running the rule forward.
$13^2 + 84^2 = 169 + 7056$
That sum is $7225$ , which is $85^2$ .

Answer

A leg of about $86$ would be longer than the hypotenuse of $85$ , which no leg can be. The correct leg is $84$ .

Key idea

A leg must come out shorter than the hypotenuse, which gives a quick check on any answer.
Problem 9 Two ways to fill one square

A right triangle has legs of $4$ cm and $7$ cm. Four copies of it are placed inside a square with sides of $11$ cm, in two different ways.

In the first way, the triangles sit in the four corners and leave a tilted square in the middle whose sides are hypotenuses. In the second way, the triangles pair up into two rectangles and leave two squares.

Find the area left uncovered in each arrangement, explain why the two areas must be equal, and give the exact length of the hypotenuse.
1. Hint 1
  Start with the area of the big square, and then the area of one triangle.
2. Hint 2
  Each triangle is half of a rectangle that measures $4$ cm by $7$ cm.
3. Hint 3
  In the second arrangement, the two squares left uncovered have the two legs as their sides.
Answer

Both arrangements leave $65 \text{ cm}^2$ , since each is the same $121 \text{ cm}^2$ square minus the same four triangles. The hypotenuse is exactly $\sqrt{65}$ cm.

Full solution

The big square has area
$11^2 = 121 \text{ cm}^2.$
Each triangle is half of a $4$ by $7$ rectangle, so it covers $14 \text{ cm}^2$ , and the four triangles cover $56 \text{ cm}^2$ together.

In the first arrangement, the tilted square is all that the triangles leave.
$121 - 56 = 65$
In the second arrangement, the uncovered squares have sides $4$ cm and $7$ cm.
$16 + 49 = 65$

The two uncovered areas must be equal, because each is the same $121 \text{ cm}^2$ square with the same $56 \text{ cm}^2$ of triangles taken away.
The tilted square is the square on the hypotenuse, so the hypotenuse squared is $65$ .

The number $65 = 5 \times 13$ has no perfect-square factor other than $1$ , so the hypotenuse is exactly $\sqrt{65}$ cm, a little more than $8$ cm.

Answer

Both arrangements leave $65 \text{ cm}^2$ , since each is the same $121 \text{ cm}^2$ square minus the same four triangles. The hypotenuse is exactly $\sqrt{65}$ cm.

Key idea

The same square minus the same four triangles leaves the same area, which is why the square on the hypotenuse equals the two squares on the legs.
Problem 10 The shortcut across

Explain why the hypotenuse of any right triangle is shorter than its two legs added together.

Then check your explanation on a right triangle whose legs are both $5$ cm, giving the hypotenuse exactly and to the nearest tenth of a centimeter.
1. Hint 1
  Compare squares instead of lengths: a bigger square always has a longer side.
2. Hint 2
  Picture a square whose side is the two legs laid end to end, like the big square in the lesson's second arrangement.
3. Hint 3
  That big square holds the two squares on the legs and two rectangles as well. Compare its area with the square on the hypotenuse.
Answer

A square with side equal to the two legs together holds both leg squares plus two rectangles, so it is bigger than the square on the hypotenuse. For legs of $5$ cm, the hypotenuse is exactly $5\sqrt{2}$ cm, about $7.1$ cm.

Full solution

Lay the two legs end to end and build a square on that combined length.
Cut it as in the second arrangement: it splits into the square on one leg, the square on the other leg, and two rectangles with the legs as their sides.

So that square has more area than the two leg squares alone.
The two leg squares together are exactly the square on the hypotenuse, so the square on the two legs together is bigger than the square on the hypotenuse.

A bigger square has a longer side.
So the two legs together are longer than the hypotenuse, which means the hypotenuse is shorter than the two legs added together.

For legs of $5$ cm, add the squares.
$c^2 = 25 + 25 = 50$
Since $50 = 25 \times 2$ , the hypotenuse is $c = 5\sqrt{2}$ cm.

Next, $7.05^2 = 49.7025$ and $7.15^2 = 51.1225$ , so $c$ is between them and rounds to $7.1$ cm.
That is less than $5 + 5 = 10$ cm, as the explanation says it must be.

Answer

A square with side equal to the two legs together holds both leg squares plus two rectangles, so it is bigger than the square on the hypotenuse. For legs of $5$ cm, the hypotenuse is exactly $5\sqrt{2}$ cm, about $7.1$ cm.

Key idea

The square on the two legs together holds the hypotenuse's square plus two rectangles, so walking across is always shorter than walking along both legs.

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