9.3 The Hypotenuse of an Isosceles Right Triangle

In a right triangle the side facing the right angle carries the name hypotenuse. It is the slanted side, and, as we shall confirm, always the longest of the three.

Let us tackle the simplest case: an isosceles right triangle whose two equal (perpendicular) sides each measure 1.

Here is the key observation. A unit square consists of two such isosceles right triangles, joined along their hypotenuse. And we have just seen that the square raised on the diagonal (which is precisely the hypotenuse of those triangles) carries twice the area of the unit square.

So if c stands for the hypotenuse, the square on it has area c \times c = c^2, and

c^2 = 2 \times (\text{area of unit square}) = 2 \times 1 = 2.

Hence

c^2 = 2 \qquad\Longrightarrow\qquad c = \sqrt{2}.

The hypotenuse of a unit isosceles right triangle measures \sqrt{2} units. But what number is \sqrt{2}?

What is the value of \sqrt{2}?

Let us corner \sqrt{2} between bounds, tightening them step by step.

Is \sqrt{2} above or below 1? A square of side 1 has area 1; a square of side \sqrt{2} has area 2. Since 1^2 = 1 and (\sqrt2)^2 = 2, we get 1 < \sqrt{2}.

Is \sqrt{2} above or below 2? A square of side 2 has area 4; our square has area 2. Since (\sqrt2)^2 = 2 and 2^2 = 4, we get \sqrt{2} < 2. So far:

1 < \sqrt{2} < 2.

We name 1 a lower bound and 2 an upper bound. Now we squeeze inward by trying decimals:

Step What we test Conclusion
Tenths 1.4^2 = 1.96,   1.5^2 = 2.25 1.4 < \sqrt2 < 1.5
Hundredths 1.41^2 = 1.9881,   1.42^2 = 2.0164 1.41 < \sqrt2 < 1.42
Thousandths 1.414^2 = 1.999396,   1.415^2 = 2.002225 1.414 < \sqrt2 < 1.415

We could continue endlessly, yet the two bounds never lock onto a clean decimal. Why is that?

\sqrt{2} Never Terminates Imagine \sqrt2 were a terminating decimal such as 1.414\ldots4, ending in some non-zero final digit. Then its square would also finish with a non-zero digit somewhere after the decimal point, so the square could never land on exactly 2.000\ldots. That is a contradiction. Therefore the decimal form of \sqrt2 must run on forever, never ending and never repeating into a neat block: \sqrt{2} = 1.41421356\ldots

Can \sqrt{2} be written as a fraction?

Try This Could \sqrt{2} equal some fraction \dfrac{m}{n} built from counting numbers m, n? Suppose so. Then \sqrt2 = \frac{m}{n} \;\Rightarrow\; 2 = \frac{m^2}{n^2} \;\Rightarrow\; 2n^2 = m^2. Remember that in the prime factorisation of any perfect square, every prime shows up an even number of times. Now tally the prime 2 in 2n^2 = m^2. On the left, the lone extra factor of 2 pushes the count to odd; on the right, m^2 being a square keeps it even. No number can carry both an odd and an even count of the same prime, impossible! So \sqrt2 is not a fraction.

This graceful argument goes back to Euclid in his Elements (around 300 BCE). A number like \sqrt2, neither a terminating decimal nor a fraction, is what we will later call an irrational number.

A general formula

The very same area reasoning applies to any isosceles right triangle. Let each equal side measure a and the hypotenuse measure c. The square on the hypotenuse holds twice the area of the square on a side:

\boxed{\,c^2 = 2a^2\,} \qquad\text{equivalently}\qquad c = a\sqrt{2}.

This works in both directions, find c from a, or recover a from c.

Worked Example

Find the hypotenuse of an isosceles right triangle whose equal sides each measure 15.

Here a = 15, so c = \sqrt{2 \times 15^2} = \sqrt{2 \times 225} = \sqrt{450}. Since 21^2 = 441 and 22^2 = 484, the value \sqrt{450} sits between 21 and 22. (Closer in, \sqrt{450} \approx 21.21.)

Worked Example

If the hypotenuse of an isosceles right triangle is \sqrt{98}, find its other two sides.

Here c = \sqrt{98}, so c^2 = 98. Using c^2 = 2a^2: 98 = 2a^2 \;\Rightarrow\; a^2 = \frac{98}{2} = 49 \;\Rightarrow\; a = \sqrt{49} = 7. Each equal side measures 7.

Worked Example

A square floor tile has side 11 cm. How long is the diagonal, and between which two whole numbers does it lie?

The diagonal of a square is the hypotenuse of an isosceles right triangle with legs 11, so d = \sqrt{2 \times 11^2} = \sqrt{242}. Since 15^2 = 225 and 16^2 = 256, the diagonal \sqrt{242} lies between 15 and 16 cm (\approx 15.56 cm).

Figure it Out: Isosceles Right Triangles

Practice

  1. There is a second route to doubling a square: take two identical square papers, cut each along one diagonal into pieces 1, 2, 3, 4, and reassemble all the pieces into one square of double the area. Persuade yourself that it works.
  2. The two equal sides of an isosceles right triangle are given. Find the hypotenuse and give bounds with at least one digit after the decimal point:  (i) 2  (ii) 5  (iii) 7  (iv) 10  (v) 11.
  3. The hypotenuse of an isosceles right triangle is 14. What are its other two side lengths? (Hint: the square built from two such triangles has area 14^2 = 196.)
  1. Cutting each square along one diagonal yields four right triangles; gathering all four around a shared right-angle vertex builds a square whose side is the common hypotenuse, with double the total area. (The same total area as two squares, now pooled into one square.)
  2. Using c = a\sqrt2:
      1. a=2: c = \sqrt{8} \approx \mathbf{2.8}  (since 2.8^2 = 7.84, 2.9^2=8.41, so 2.8 < c < 2.9).
      1. a=5: c = \sqrt{50} \approx \mathbf{7.0}  (7.0 < c < 7.1).
      1. a=7: c = \sqrt{98} \approx \mathbf{9.8}  (9.8 < c < 9.9).
      1. a=10: c = \sqrt{200} \approx \mathbf{14.1}  (14.1 < c < 14.2).
      1. a=11: c = \sqrt{242} \approx \mathbf{15.5}  (15.5 < c < 15.6).
  3. From c^2 = 2a^2: \,196 = 2a^2 \Rightarrow a^2 = 98 \Rightarrow a = \sqrt{98} \approx \mathbf{9.90}. Each equal side is \sqrt{98} = 7\sqrt2 \approx 9.90.
Isosceles Right-Triangle Hypotenuse

Enter the length of an equal side a. We compute the hypotenuse using c = a√2.