Revision and important questions: The Baudhayana-Pythagoras Theorem

The Baudhayana-Pythagoras Theorem18 min readLesson 5 of 6

5.1 Formula box Baudhayana-Pythagoras theorem: in a right triangle with legs a and b and hypotenuse c, a² + b² = c². Hypotenuse from legs: c = \sqrt(a² + b²). Leg from the other leg and the hypotenuse: a = \sqrt(c² - b²). Isosceles right triangle: c² = 2a², so c = a\sqrt2. Diagonal of a square of side s: d = s\sqrt2. Diagonal of a rectangle with sides l and b: d = \sqrt(l² + b²). Rhombus with diagonals p and q: side = \sqrt((p/2)² + (q/2)²). Equilateral triangle of side s: height = (\sqrt3/2)s and area = (\sqrt3/4)s². Triple scaling: if (a, b, c) is a triple and k a positive integer, then (ka, kb, kc) is a triple. Odd-square identity: (n - 1)² + (2n - 1) = n², which gives a triple when 2n - 1 is an odd square. Square root bounds: square the decimals just below and just above to trap \sqrt2, \sqrt3 and other irrational lengths.

a^{2} + b^{2} = c^{2}, c = a^{2} + b^{2}, a = c^{2} - b^{2}

The theorem in three forms

d_{square} = s 2, area_{equilateral} = \frac{3}{4} s^{2}, side_{rhombus} = (\frac{p}{2})^{2} + (\frac{q}{2})^{2}

Standard results that follow from the theorem

Exam

Exam habit. Label the hypotenuse before substituting. The theorem only works when the side opposite the right angle is treated as c. If the question gives the hypotenuse and asks for a leg, the equation must be rearranged, not applied with the given numbers in the wrong places.

5.2 Common mistakes Adding when the hypotenuse is known. If c = 13 and a = 5, the missing leg is \sqrt(169 - 25) = 12, not \sqrt(169 + 25). Writing \sqrt(a² + b²) as a + b. The square root does not distribute over a sum: \sqrt(9 + 16) = 5, while 3 + 4 = 7. Doubling the side to double the area. Doubling the side multiplies the area by 4. The square of double area is built on the diagonal. Calling every integer triple primitive. (6, 8, 10) is not primitive; it is 2 \times (3, 4, 5). Treating the odd-square method as complete. It always gives a triple with two consecutive sides, so triples such as (8, 15, 17) are missed. Forgetting the unit. A length is in cm, m or units, and an area is in square units. Writing 13 instead of 13 cm loses a mark. Using the grid-area shortcut for non-lattice squares. A grid square of area p² + q² exists only when p and q are whole steps; that is why area 3 fails.

5.3 Important exam-style questions Question 1. A right triangle has legs 9 cm and 12 cm. Find its hypotenuse. Answer. c² = 81 + 144 = 225, so c = 15 cm. This is the (3, 4, 5) triple scaled by 3.

Question 2. A right triangle has hypotenuse 25 cm and one leg 7 cm. Find the other leg. Answer. The missing leg squared is 25² - 7² = 625 - 49 = 576, so the leg is \sqrt576 = 24 cm. This is the (7, 24, 25) triple.

Question 3. Show that (6, 8, 10) is a Baudhayana triple and say whether it is primitive. Answer. 6² + 8² = 36 + 64 = 100 = 10², so it is a triple. It is not primitive, because all three numbers share the factor 2; dividing by 2 gives the primitive triple (3, 4, 5).

Question 4. Find the diagonal of a square of side 7 cm. Answer. The diagonal divides the square into two right triangles with legs 7 and 7, so d² = 49 + 49 = 98 and d = 7\sqrt2 cm. Since 9.89² = 97.8121 and 9.9² = 98.01, the value is about 9.90 cm.

Question 5. A rectangular field is 24 m long and 10 m wide. Find the length of its diagonal. Answer. The diagonal is the hypotenuse of a right triangle with legs 24 and 10: d² = 576 + 100 = 676, so d = 26 m. This is the (5, 12, 13) triple scaled by 2.

Question 6. A lotus stem stands 2 units above the water. When swayed by the wind, its tip touches the surface 4 units from the base. Find the depth of the water. Answer. Let the depth be x. The stem length is x + 2, and the tip, the base and the new touch point form a right triangle. Then 4² + x² = (x + 2)². Expanding gives 16 + x² = x² + 4x + 4, so 12 = 4x and x = 3. The water is 3 units deep.

Question 7. Find the height and area of an equilateral triangle with side 8 cm. Answer. The altitude bisects the base into two 4 cm segments. Then h² + 4² = 8², so h² = 64 - 16 = 48 and h = 4\sqrt3 cm. The area is (1/2) \times 8 \times 4\sqrt3 = 16\sqrt3 square cm, about 27.71 square cm.

Question 8. A rhombus has diagonals 30 cm and 40 cm. Find its side. Answer. The diagonals bisect each other at right angles into 15 cm and 20 cm. The side is the hypotenuse of the right triangle with legs 15 and 20: side² = 225 + 400 = 625, so side = 25 cm.

Question 9. If (a, b, c) is a Baudhayana triple, prove that (ka, kb, kc) is a triple for any positive integer k. Answer. Since (a, b, c) is a triple, a² + b² = c². Multiply both sides by k²: k²a² + k²b² = k²c². The left side is (ka)² + (kb)² and the right side is (kc)², so the equation holds for the scaled triple. Therefore (ka, kb, kc) is also a Baudhayana triple.

Question 10. On a square dot grid, can a square have area 6 or area 8? Explain. Answer. A grid square with side steps p and q has area p² + q². For area 8, take p = 2 and q = 2, giving 4 + 4 = 8, so area 8 is possible. For area 6, no two perfect squares add to 6 (the squares 0, 1, 4, 9 exceed or undercut it), so area 6 is impossible on a square grid.

I can state the theorem and identify the hypotenuse in any right triangle.
I can find an unknown leg by subtracting squares, not by adding them.
I can derive c = a√2 for an isosceles right triangle and find square diagonals.
I can list Baudhayana triples, recognise primitive triples and scale them.
I can prove that scaling a triple gives another triple.
I can generate a triple from an odd square using (n - 1) squared plus (2n - 1) equals n squared.
I can apply the theorem to the lotus problem, rhombus sides and equilateral triangle areas.