# Pythagorean Theorem Word Problems

TR Smith is a product designer and former teacher who uses math in her work every day.

Statue of Pythagoras in Samos, Greece

The Pythagorean Theorem is one of the oldest and most useful mathematical and geometric formulas. Although it was named after the Greek mathematician Pythagoras because he gave the first recorded proof of the equation, the formula for the hypotenuse of a right triangle has been known to other earlier civilizations. The theorem states that if the perpendicular legs of a right triangle have lengths of A and B, then the hypotenuse length C is related to A and B by the equation A^2 + B^ 2 = C^2, or equivalently C = sqrt(A^2 + B^2).

The following set of word problems tests your ability to apply the Pythagorean Theorem in more challenging scenarios than typical math drills. Word problems are a great way to synthesize different mathematical reasoning skills, diagram drawing, and logical thinking.

## (1) Crossing Ladders Problem

Two buildings are 15 feet apart. A 17-foot ladder is leaning against the right building with its base situated at the base of the left building. A 25-foot ladder is leaning against the left building with its base situated at the base of the right building. How high is the point where the ladders cross?

SOLUTION: This problem is solved using both the Pythagorean Theorem and properties of similar triangles. The diagram below shows how the ladders are arranged between the two buildings. The length or distance that we want to solve for is shown in green with question mark.

Each ladder is the hypotenuse of a right triangle. Using the Pythagorean Theorem, we can see that the longer ladder reaches a height of sqrt(25^2 - 15^2) = 20 feet, and the shorter ladder reaches a height of sqrt(17^2 - 15^2) = 8 feet. In the following diagram, we can also see that the two ladders form two similar triangles, colored in yellow.

The crossing point of the two ladders creates four line segments. Using properties of similar triangles we can obtain the ratio equalities to find the lengths of these segments..

y/8 = (17-y)/20

x/8 = (25-x)/20

Solving for x and y we get x = 50/7 and y = 34/7, as well as 25-x = 125/7 and 17-y = 85/7. With the lengths of these segments, we can now use similar triangles once more to find the height of the bottom triangle.

The small pink triangle has a hypotenuse of 50/7 and a height of H. This triangle is similar to the large right triangle on the right, which has a hypotenuse of 25 and a height of 20. This gives us the ratio equation

h/(50/7) = 20/25

which gives us h = 40/7 feet, or approximately 5 feet and 8.5 inches.

## (2) Find All Three Sides

A flower garden in the shape of a right triangle has a perimeter of 20 meters and an area of 15 square meters. What are the lengths of the three sides of the garden?

SOLUTION: We want to find a right triangle with legs A and B and a hypotenuse C such that A + B + C = 20 and AB/2 = 15. Combined with the Pythagorean Theorem equation A^2 + B^2 = C^2, we have a system of 3 equations in 3 variables (though it is not a linear system), so we have a chance at finding a unique solution. (Note: With n non-linear equations in n variables it is not always possible to find a unique solution. Sometimes there is no solution, a finite set of solutions, or infinitely many solutions.)

We start by transforming the perimeter equation 20 = A + B + C into 20 - A - B = C. Squaring both sides of this equation gives us 400 + A^2 + B^2 - 40A - 40B + 2AB = C^2. Now we can use this relation to replace C^2 in Pythagoras's equation:

A^2 + B^2 = 400 + A^2 + B^2 - 40A - 40B + 2AB

0 = 400 - 40A - 40B + 2AB

The area equation tells us that AB/2 = 15, which implies that AB = 30 and 2AB = 60. Replacing the AB term in the equation above with a number gives us

0 = 400 - 40A - 40B + 60

40(A + B) = 460

A + B = 11.5

Since A + B = 11.5 and the perimeter is 20, this tells us that C = 8.5. Now we need to find A and B. Using the fact that AB = 30, we can replace A with 30/B. Plugging this into the equation above gives us

30/B + B = 11.5
30 + B^2 = 11.5B
B^2 - 11.5B + 30 = 0
B = [11.5 ± sqrt(132.25 - 120)]/2
B = [11.5 ± 3.5]/2
B = 4 or 7.5

If B = 4 then A = 7.5, and if B = 7.5 then A = 4. Thus, the three sides of the right triangle are 4, 7.5, and 8.5.

## (3) Building a Ramp

Cheryl has several 18-foot long boards to build a ramp with a slanted length of 18 feet. If the horizontal length of the ramp has to be at least 7 feet long and at most 16 feet long, what is the range of the ramp's height?

SOLUTION: The highest the ramp can be is when the horizontal length is 7 feet, and the shortest the ramp can be is when the horizontal length is 16 feet. These heights are represented by x and y respectively in the diagram below.

Using the Pythagorean Theorem we have

18^2 = x^2 + 7^2
324 = x^2 + 49
275 = x^2
x = sqrt(275) ≈ 16.583

18^2 = y^2 + 16^2
324 = y^2 + 256
68 = y^2
y = sqrt(68) ≈ 8.246

Therefore, the height of the ramp can range from 8.246 feet to 16.583 feet.

## (4) Integer Right Triangles

Suppose a right triangle's area and perimeter are integers. Does this mean the side lengths always integers? Must the side lengths always rational numbers? Either prove that the side lengths must always be integer/rational or give some counter examples.

SOLUTION: Some right triangles with integer perimeter and area do have integer side lengths. For example, the primitive Pythagorean triples (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), etc. satisfy this property.

However, there are right triangles with rational but non-integer side lengths that also have integer perimeters and areas. For example, the triangle with sides 4, 7.5, and 8.5 is a right triangle since 4^2 + 7.5^2 = 8.5^2. It's perimeter and area are 20 and 15 respectively. (See problem above.)

And yet, it is also possible to find right triangles with irrational side lengths that still have integer area and perimeter. A few simple examples are

A = 4 + sqrt(2), B = 4 - sqrt(2), C = 6

A = 6 + sqrt(14), B = 6 - sqrt(14), C = 10

A = 7 + sqrt(23), B = 7 - sqrt(23), C = 12

In each set you can verify that

• A^2 + B^2 = C^2 (the triangles are right)
• AB/2 = integer (the triangles have integer area)
• A+B+C = integer (the triangles have integer perimeter)

Therefore, right triangles can have integer areas and perimeters without necessarily having integer or rational side lengths.

## (5) A Long Walk Around the Neighborhood

Starting from home, Kurt walks 0.5 km north, 0.7 km east, 0.4 km north, 0.2 km west, 0.6 km south, and 0.2 km west before he takes a break. How far from home is he when he takes a break?

SOLUTION: The net change in the east-west direction is 0.7 - 0.2 - 0.2 = 0.3 km east. The net change in the north-south direction is 0.5 + 0.5 - 0.6 = 0.4 km north. If you think of a right triangle with an east-west length of 0.3 and a north-south height of 0.4 km, the hypotenuse is 0.5 km, the distance between the starting and ending point. You can also solve this problem by drawing a diagram of Kurt's path:

In the diagram above, each square represents 0.1 km. The distance between where he started and where he finished is the hypotenuse of a triangle with legs 0.3 and 0.4. Since 0.4^2 + 0.3^2 = 0.25 = 0.5^2, Kurt ends up half a kilometer from his house.

## (6) Width of Diagonal Stripes

A 6-by-9 rectangle is divided into 6 diagonal stripes of equal width as shown in the diagram below. The stripes run parallel to the main diagonal of the rectangle. How wide are the stripes?

SOLUTION: If the 6-by-9 rectangle is divided into 6 stripes of equal width, then the segments along the long side of the rectangle are 3 units long and the segments along the short side are 2 units long. The stripe in the corner is a right triangle with legs of length 2 and 3 whose hypotenuse is sqrt(13), since 2^2 + 3^2 = 13. Notice that the altitude of this triangle is equal to the width of each stripe.

The altitude cuts the corner right triangle into two smaller right triangles that are similar. Since similar triangles have the same side length ratios, we can set up a ratio problem to solve for w. In the larger triangle, the ratio of the longer leg to the hypotenus is 3/sqrt(13). In the smaller green triangle, that same ratio is w/2. This gives us the easy-to-solve equation

3/sqrt(13) = w/2

Therefore, w = 6/sqrt(13), or w ≈ 1.6641.

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