Applications of the theorem

The theorem is most useful when a problem hides a right triangle inside it. Spotting that right triangle is half the challenge , applying the formula is the other half.

Concept

Ladders. A leaning ladder forms a right triangle with the wall and the floor. Knowing any two of (length of ladder, distance from wall, height reached) lets you find the third.

TV/screen sizes. A TV is sold by its diagonal length. If a $43$-inch TV has aspect ratio $16 : 9$, the actual width and height come from the diagonal using Pythagoras. Width $:$ height $:$ diagonal $= 16 : 9 : \sqrt{16^2 + 9^2} = 16 : 9 : \sqrt{337}$.

Screen-size formula. For a screen of diagonal $d$ and aspect ratio $w : h$, the dimensions are

$\text{width} = \frac{wd}{\sqrt{w^2 + h^2}}, \quad \text{height} = \frac{hd}{\sqrt{w^2 + h^2}}.$

For $43$-inch, $16:9$: width $\approx 37.5$ inches, height $\approx 21.1$ inches.

Ramps. A ramp of length $L$ rising height $h$ over horizontal run $b$ satisfies $L^2 = h^2 + b^2$.

Distance between two points. On a flat map (Cartesian plane), the straight-line distance between $(x_1, y_1)$ and $(x_2, y_2)$ is

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

This is just Pythagoras applied to the right triangle whose legs are the horizontal and vertical differences.

Shortest-path arguments. Sometimes the shortest path between two points avoids an obstacle by going through the square root of a sum of squares , exactly the kind of expression the theorem produces.

Indirect measurement. Cannot reach the top of a tree? Walk to a known distance from its base, measure the angle (or apply other tricks), and compute the height. Surveyors and astronomers have used variants of Pythagoras for centuries.

Worked examples

Example 1. A $13$-m ladder reaches $12$ m up a wall. How far is its base from the wall?

• $b = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ m.

Example 2. A TV has $50$-inch diagonal and aspect ratio $16:9$. Find its width and height.

• $\sqrt{16^2 + 9^2} = \sqrt{337} \approx 18.36$.
• Width: $50 \cdot 16 / 18.36 \approx 43.58$ inches.
• Height: $50 \cdot 9 / 18.36 \approx 24.51$ inches.

Example 3. A man walks $6$ km East then $8$ km North. Straight-line distance from start?

• $\sqrt{36 + 64} = \sqrt{100} = 10$ km.

Example 4. Distance between $(2, 3)$ and $(5, 7)$ in the Cartesian plane?

• Horizontal: $5 - 2 = 3$. Vertical: $7 - 3 = 4$.
• Distance: $\sqrt{9 + 16} = 5$.

Try it yourself

1. A ladder $10$ m long leans against a wall; its top reaches $8$ m up. Distance of foot from wall?
2. A TV has $32$-inch diagonal, aspect ratio $4:3$. Find width and height.
3. A ramp rises $1.5$ m over a $2$ m horizontal run. Find its length.
4. Find distance between $(1, 2)$ and $(7, 10)$.
5. A man cycles $9$ km North then $40$ km East. Direct distance home?
6. A rectangle's diagonal is $25$ cm and one side is $15$ cm. Find the other side.
7. A flag-pole casts a shadow of $12$ m on the ground. The distance from the tip of the shadow to the top of the pole is $13$ m. Find the height.
8. A square room has $5$ m sides. What is the length of the diagonal across the floor?

Activity / Insight

Estimate a tree. Stand at a known distance $d$ from the base of a tree. Use a string to mark the line from your eye level to the top of the tree (you'll need a partner). Measure that string. With a right triangle (eye-height level, distance to base, line to top), the height of the tree is roughly $\sqrt{\text{string}^2 - d^2} + \text{your eye-height}$. Try it on a small tree first. Surveyors do exactly this with better tools.