The Baudhayana-Pythagoras Theorem · Lesson 2 of 9
Hypotenuse of an Isosceles Right Triangle
“Discover the exact length √2, estimate square roots, and calculate sides of isosceles right triangles.”
• Identify the hypotenuse regardless of a triangle’s orientation. • Explain why a unit square’s diagonal has length √2. • Find decimal bounds and distinguish exact from approximate lengths. • Explain the main idea behind the irrationality of √2. • Use c² = 2a² in both directions.
Draw a square whose sides are each one unit long, then draw a line from one corner to the opposite corner. The diagonal is longer than a side, but how long is it exactly? A ruler may suggest a decimal close to 1.4. A more careful ruler might give more digits. Neither measurement, however, explains why the length has that value or whether those digits give the exact answer.
You already know something more dependable than a ruler reading: a square built on this diagonal has twice the area of the unit square. We will use that area fact to name the diagonal’s exact length, then learn how to place it between familiar decimal numbers.
Finally, we will use the same reasoning for larger and smaller isosceles right triangles. The unit triangle is the starting example, not the only triangle to which the idea applies.
Identifying the Hypotenuse
A right triangle has one angle of 90°. The side directly opposite that angle is called its hypotenuse. It does not touch the right-angle vertex. The other two sides meet at the right angle and are perpendicular to one another.
Do not identify the hypotenuse by asking which side is sloping on the page. If you rotate the drawing, a different side may look horizontal or vertical, but the right angle remains opposite the same side. Find the angle first; then find the side across from it.
The side opposite the right angle in a right-angled triangle.
The Triangle with Two Unit Sides
A diagonal divides a unit square into two congruent right triangles. In each triangle the perpendicular sides are 1 unit and 1 unit. Because two sides are equal, each is an isosceles right triangle.
Let the diagonal length be c units. The square built on that diagonal has area c² square units. From the doubling construction, this area is 2 × 1² = 2. Therefore c² = 2. Naming this number does not require rounding: the positive number whose square is 2 is written √2.
So the hypotenuse is exactly √2 units. The symbol is not an unfinished calculation. Just as the fraction 1/3 names an exact number, √2 names an exact number even when its decimal expansion does not end.
For a positive number N, √N is the positive number whose square is N. For example, √9 = 3 because 3² = 9.
Decimal Representation of √2
To locate √2, compare the squares of familiar positive numbers with 2. Since 1² = 1 and 2² = 4, the required number lies between 1 and 2. The lower bound is a number below the required value; the upper bound is a number above it.
Next try tenths. Since 1.4² = 1.96 and 1.5² = 2.25, the required number lies between 1.4 and 1.5. We can continue with hundredths and thousandths. This works because a larger positive side length gives a larger square area.
Notice the direction of the inequalities. If a decimal’s square is slightly less than 2, the decimal itself is slightly less than √2. If its square is slightly greater than 2, it is an upper bound. Comparing squares avoids guessing from the digits of the radical.
| Lower candidate and its square | Upper candidate and its square | Conclusion |
|---|---|---|
| 1² = 1 | 2² = 4 | 1 < √2 < 2 |
| 1.4² = 1.96 | 1.5² = 2.25 | 1.4 < √2 < 1.5 |
| 1.41² = 1.9881 | 1.42² = 2.0164 | 1.41 < √2 < 1.42 |
| 1.414² = 1.999396 | 1.415² = 2.002225 | 1.414 < √2 < 1.415 |
Problem
Is 1.41 the exact value of √2, or an approximation?
- 1.Square the proposed number: 1.41 × 1.41 = 1.9881.
- 2.The result is not 2, so 1.41 is not the exact square root.
- 3.Because 1.9881 < 2, the number 1.41 is below √2.
- 4.Use √2 ≈ 1.41 for an approximation to two decimal places, but keep √2 when an exact answer is required.
Why a Terminating Decimal Cannot Equal √2
Suppose there were an exact terminating decimal for √2. Remove any unnecessary zeros at its end. Its last digit would then be one of 1, 2, 3, 4, 5, 6, 7, 8, or 9. On squaring, the last digit comes from the square of that last digit.
The possible last digits of those squares are 1, 4, 9, 6, 5, 6, 9, 4, and 1. None is zero. If the original number has d decimal places, its square has 2d decimal places with a nonzero final digit. It cannot be the whole number 2.
Thus √2 has a non-terminating decimal expansion. Its opening digits are 1.41421356… . No finite list of these digits is exactly √2. More digits can improve an approximation without ever making a finite decimal exact.
Can √2 Be Written as a Fraction?
Non-terminating does not automatically mean “not a fraction”. For example, 1/3 = 0.333… continues forever. We need a separate argument to show that √2 cannot be written as m/n with positive integers m and n.
Suppose √2 = m/n. Squaring both sides gives 2 = m²/n², and multiplying by n² gives 2n² = m². Now recall a property of perfect squares: every prime factor occurs an even number of times, because squaring doubles the number of each prime factor.
In n², the number of factors 2 is even. Multiplying by one extra 2 makes that number odd in 2n². But m² must have an even number of factors 2. Equal positive integers cannot have different prime factorizations. This contradiction shows that the proposed fraction does not exist.
A number that cannot be written as a ratio of integers is called irrational. The reasoning here explains why √2 is irrational; it is not merely a conclusion drawn from seeing many decimal digits. You can follow the geometry and calculations even if you need time to revisit this factorisation argument.
For n = 6, n² = 36 = 2² × 3². Multiplying by 2 gives 72 = 2³ × 3². The exponent of 2 changes from even to odd, so 72 is not a perfect square. The argument above applies this same observation to every positive integer n.
General Solution
Now take an isosceles right triangle whose two equal perpendicular sides are a units long. Two copies form a square of side a, with area a². A square on its diagonal has area 2a². Since that diagonal is the triangle’s hypotenuse c, its square also has area c².
Therefore c² = 2a². To find c, take the positive square root. We can also write c = a√2: squaring a√2 gives a² × 2, exactly the required area. This relationship applies only when the two equal sides meet at a right angle.
Problem
Find the hypotenuse when both perpendicular sides are 3 units. Give an exact length and bounds to one decimal place.
- 1.Substitute a = 3: c² = 2 × 3² = 18.
- 2.The exact length is √18 = 3√2 units.
- 3.Compare 4.2² = 17.64 and 4.3² = 18.49.
- 4.Since 17.64 < 18 < 18.49, the length lies between 4.2 and 4.3 units.
Problem
Find and bound the hypotenuse when the equal sides are 12 units.
- 1.The square of the hypotenuse is 2 × 12² = 2 × 144 = 288.
- 2.Thus c = √288 units, or 12√2 units.
- 3.Since 16² = 256 and 17² = 289, we have 16 < √288 < 17.
- 4.The upper bound 17 is very close, but it is not exact because 17² is 289 rather than 288.
Working Backwards from the Hypotenuse
When the hypotenuse is known, remember that its square represents the combined area of two equal squares. Divide that area by two to get the area of one square. Only then take the square root to get its side.
Do not simply halve the hypotenuse. Halving a length and halving the square of a length are different operations. The area relationship tells you which quantity must be divided.
Problem
Find the two equal perpendicular sides.
- 1.Here c² = (√72)² = 72.
- 2.Use 72 = 2a², then divide by 2: a² = 36.
- 3.Take the positive square root: a = 6 units.
- 4.Check: 6² + 6² = 36 + 36 = 72, the square of the given hypotenuse.
Problem
Find the equal sides of an isosceles right triangle with hypotenuse 10 units.
- 1.Substitute c = 10 into c² = 2a²: 100 = 2a².
- 2.Divide by 2 to obtain a² = 50.
- 3.Therefore each side is √50 = 5√2 units, approximately 7.07 units.
- 4.Sides of 5 units would not work: their squares add to 50, not to 100.
Quiz
How do you identify the hypotenuse?
Which statement about √2 is correct?
Equal perpendicular sides are 4 units. What is the hypotenuse?
The hypotenuse is √98 units. What is each equal perpendicular side?
Why is non-termination alone insufficient to prove irrationality?
Practice Problems
- Find the hypotenuse for equal perpendicular sides of 6 units. Solution: 1. Use c² = 2 × 36 = 72. 2. Therefore c = √72 = 6√2 units. 3. Since 8.4² = 70.56 and 8.5² = 72.25, 8.4 < c < 8.5.
- Find decimal bounds for the hypotenuse when each equal side is 8 units. Solution: 1. Its square is 2 × 64 = 128. 2. Compare 11.3² = 127.69 and 11.4² = 129.96. 3. The hypotenuse lies between 11.3 and 11.4 units.
- An isosceles right triangle has hypotenuse √162 cm. Find its equal sides. Solution: 1. Square the hypotenuse to obtain 162 cm². 2. Divide by 2: a² = 81. 3. The equal sides are 9 cm each.
- A student writes √2 = 1.414 because a calculator shows those digits. Explain why the equality is not exact. Solution: 1. Compute 1.414² = 1.999396. 2. This is less than 2, so 1.414 is a lower approximation. 3. Write √2 ≈ 1.414, or leave the exact radical √2.
- An isosceles right triangle has equal sides 9 cm. Find its exact hypotenuse and one-decimal-place bounds; explain why doubling the equal side is incorrect. Solution: 1. Calculate c² = 2 × 81 = 162, hence c = 9√2 cm. 2. Since 12.7² = 161.29 and 12.8² = 163.84, 12.7 < c < 12.8. 3. Doubling 9 gives 18, but 18² = 324 rather than 162. The required doubling concerns square area, not length.
Key Takeaways
• The hypotenuse is opposite the right angle. • A unit square’s diagonal is exactly √2 units. • Bounds come from comparing squares of positive numbers. • A finite decimal approximation must not be written as the exact value of √2. • The prime-factorisation argument shows that √2 cannot be a fraction. • For equal perpendicular sides a, c² = 2a² and c = a√2. • When c is known, divide c² by two before taking the square root.