The Baudhayana-Pythagoras Theorem · Lesson 4 of 9
Using Baudhāyana’s Theorem
“Find unknown sides, estimate non-integer lengths, and check calculations using the right-angle condition.”
• Choose the correct hypotenuse before substituting values. • Find an unknown hypotenuse by adding square areas. • Find an unknown perpendicular side by subtracting square areas. • Handle exact square roots, decimal lengths, and units. • Check answers and explain common errors.
A triangle is drawn on a sheet of paper, and two of its side lengths are labelled. You would like to find the missing length without measuring the drawing. Perhaps the drawing has been enlarged, or perhaps it is only a rough sketch. The numbers and geometric information can still determine the answer, but first you must look for the detail that makes the calculation possible: a right angle.
The area proof has given you a dependable relationship among the sides of a right triangle. Using it well involves more than remembering three letters. You must decide which side is the hypotenuse, what is unknown, and whether the unknown square area must be found by addition or subtraction.
A good solution explains each decision. By the end of this lesson you should be able to work with whole-number and decimal lengths and check whether an answer is sensible.
Recognising When the Theorem Applies
Look for a right-angle mark, a statement that the triangle is right-angled, or information that establishes perpendicular lines. A shape that merely looks right-angled is not sufficient evidence. A sketch can be misleading.
Identify the vertex of the right angle. The side opposite it is the hypotenuse; call its length c. The other two lengths may be called a and b in either order because a² + b² and b² + a² are equal.
All measurements must be in the same unit before you substitute them. If one side is 30 cm and another is 0.4 m, convert 0.4 m to 40 cm, or convert 30 cm to 0.3 m. Mixing the bare numbers 30 and 0.4 would compare different units as though they were identical.
Finding the Hypotenuse
When both perpendicular sides are known, you know the areas of the two smaller squares. Add these areas to get the area of the hypotenuse square. Then take the positive square root to recover its side length.
For example, with perpendicular sides 3 cm and 4 cm, the areas are 9 cm² and 16 cm². Their sum is 25 cm². The hypotenuse is the side of a square with that area, so it measures 5 cm. Notice the two distinct stages: finding c² and finding c.
Problem
Find the hypotenuse.
- 1.The unknown is opposite the right angle, so call it c.
- 2.Write c² = 5² + 12².
- 3.Calculate the squares separately: c² = 25 + 144 = 169.
- 4.Take the positive square root: c = 13 cm.
- 5.Check: 13 is longer than both 5 and 12, and 13² equals 169.
Draw perpendicular segments of 5 cm and 12 cm from a common endpoint. Join their other endpoints to make a triangle and measure its hypotenuse. Compare your measurement with 13 cm. Explain why a slightly different ruler reading may reflect drawing or measuring error, while the mathematical result for the stated lengths is exact.
Finding a Missing Perpendicular Side
Now suppose the hypotenuse and one other side are given. The hypotenuse square already represents the total area. Remove the area of the known smaller square to find the other smaller square’s area.
Starting from a² + b² = c², subtract a² from both sides. This gives b² = c² − a². Take a square root only after carrying out the subtraction. You are subtracting areas before finding a length.
Always subtract the smaller side’s square from the hypotenuse’s square. A negative result signals a problem: perhaps the hypotenuse was identified incorrectly or the supplied lengths are inconsistent with the stated triangle.
Problem
Find the other perpendicular side.
- 1.The known hypotenuse is 17 cm, so 8² + b² = 17².
- 2.Evaluate: 64 + b² = 289.
- 3.Subtract 64 from both sides: b² = 225.
- 4.Take the positive square root: b = 15 cm.
- 5.Check: 64 + 225 = 289, and 17 remains the longest side.
Problem
A right triangle has hypotenuse 15 units and one perpendicular side 9 units. Find the remaining side.
- 1.Write 9² + b² = 15².
- 2.Calculate b² = 225 − 81 = 144.
- 3.Therefore b = 12 units.
- 4.Subtracting the lengths directly would give 15 − 9 = 6, but 9² + 6² = 117, not 225. The theorem requires squares.
Answers That Are Not Whole Numbers
The square root at the final step will not always be an integer. This is a normal outcome, not a sign that the method failed. A right triangle with sides 5 and 7 has c² = 25 + 49 = 74, so c = √74.
For an exact answer, keep the radical. If an approximate size is useful, compare nearby squares. Since 8² = 64 and 9² = 81, √74 lies between 8 and 9. Testing tenths gives narrower bounds.
Do not round intermediate square areas unnecessarily. Carry exact values through the calculation and round only the final length if requested. This prevents small early errors from affecting later steps.
Problem
Find the hypotenuse for perpendicular sides 5 cm and 7 cm, with bounds to one decimal place.
- 1.Calculate c² = 25 + 49 = 74.
- 2.The exact answer is √74 cm.
- 3.Compare 8.6² = 73.96 and 8.7² = 75.69.
- 4.Therefore 8.6 < c < 8.7 cm. A decimal approximation may be given, but it does not replace the exact radical.
Problem
Find the hypotenuse for perpendicular sides 8 units and 12 units.
- 1.Calculate c² = 8² + 12² = 64 + 144 = 208.
- 2.The exact hypotenuse is √208 units.
- 3.Because 14² = 196 and 15² = 225, the answer lies between 14 and 15 units.
- 4.This also passes the basic size check: the hypotenuse is longer than 12 units.
Working with Decimal Lengths
The theorem describes geometric lengths, so it does not require whole numbers. Decimal and fractional side lengths follow the same steps. The care needed is arithmetic: square each entire decimal before adding or subtracting.
For 1.5, squaring means 1.5 × 1.5 = 2.25. It does not mean doubling 1.5 to obtain 3. Writing the multiplication explicitly is useful when a decimal square is unfamiliar.
Problem
Find the exact hypotenuse for sides 1.5 units and 3.5 units, and place it between consecutive tenths.
- 1.Calculate 1.5² = 2.25 and 3.5² = 12.25.
- 2.Add: c² = 14.5, so c = √14.5 units.
- 3.Since 3.8² = 14.44 and 3.9² = 15.21, the length lies between 3.8 and 3.9 units.
- 4.Check that the answer exceeds the longer perpendicular side, 3.5 units.
Checking the Result
The hypotenuse must be the longest side. From c² = a² + b² and b² > 0, we obtain c² > a². Since the lengths are positive, c > a. Similarly c > b. This is a reason, not just a pattern seen in a few examples.
Also substitute the answer back into the original equation. This catches an arithmetic error even when the resulting length seems plausible. If a radical answer is left exact, square the radical to verify it directly.
Keep units consistent with the quantity. A calculated square area such as c² = 169 cm² leads to a length c = 13 cm. Writing the final length as 13 cm² would attach an area unit to a distance.
Common Mistakes
| Incorrect step | Why it fails | Correct approach |
|---|---|---|
| c = a + b | The relationship adds square areas, not lengths. | Find c² = a² + b², then take √. |
| c² = 169, so c = 169 | The square root step is missing. | c = √169 = 13. |
| b² = a² − c² | This subtracts the larger square from the smaller. | b² = c² − a². |
| (a + b)² = a² + b² | Squaring a sum also gives a cross term. | Square the lengths separately. |
| Any triangle uses this equation | The right-angle condition is missing. | Establish a right angle first. |
Quiz
Which information is needed before applying the theorem to a triangle?
The perpendicular sides are 7 and 24. What is the hypotenuse?
A hypotenuse is 10 and one perpendicular side is 6. Which expression gives the other side?
If c² = 65 cm², which is an exact length for c?
Why must the hypotenuse exceed either perpendicular side?
Practice Problems
- Find the hypotenuse for perpendicular sides 9 cm and 12 cm. Solution: 1. c² = 81 + 144 = 225. 2. Therefore c = 15 cm. 3. Check: 15 exceeds both known sides.
- Find a perpendicular side when the other is 20 cm and the hypotenuse is 29 cm. Solution: 1. b² = 29² − 20² = 841 − 400 = 441. 2. Hence b = 21 cm. 3. Check: 400 + 441 = 841.
- Find the exact hypotenuse for perpendicular sides 7 and 12 units, and give whole-number bounds. Solution: 1. c² = 49 + 144 = 193. 2. Therefore c = √193 units. 3. Since 13² = 169 and 14² = 196, 13 < c < 14.
- A right triangle has perpendicular sides 30 cm and 0.4 m. Find its hypotenuse in centimetres. Solution: 1. Convert 0.4 m to 40 cm. 2. c² = 30² + 40² = 900 + 1600 = 2500 cm². 3. The hypotenuse is 50 cm.
- A student finds the missing perpendicular side using 13 − 5 = 8. The hypotenuse is 13 units and the other side is 5 units. Correct and check the solution. Solution: 1. Subtract square areas instead: b² = 13² − 5² = 169 − 25 = 144. 2. Thus b = 12 units. 3. Check: 5² + 12² = 25 + 144 = 169. 4. The proposed 8 would give 25 + 64 = 89, so it fails the theorem.
Key Takeaways
• Establish the right angle and locate the opposite side before substituting numbers. • Add the perpendicular side squares to find the hypotenuse square. • Subtract a known perpendicular side square from the hypotenuse square to find the other. • Take the positive square root to turn square area back into length. • Exact radicals are valid answers; bounds and decimals describe their approximate sizes. • Use one length unit throughout and verify the answer by substitution.