Baudhayana Pythagoras Theorem Class 8

Quick answer

The Baudhāyana–Pythagoras theorem states that in a right triangle with legs a and b and hypotenuse c, a²+b²=c². Class 8 Ganita Prakash develops this using square areas, constructions, √2, integer triples and real-world applications.

Key points

The diagonal of a square of side s is s√2 and produces a double-area square.
Joining side midpoints creates a square of half the original area.
For equal right-triangle legs a, the hypotenuse is a√2.
The right-triangle theorem is a²+b²=c².
Use c=√(a²+b²) for hypotenuse and √(c²−a²) for a missing leg.
Integer triples satisfy a²+b²=c²; primitive triples have HCF 1.
Scaling a triple by a positive integer produces another triple.
The theorem applies only to right triangles; units and the longest side matter.

Use these notes alongside your current textbook. Check important formulas, worked answers and syllabus coverage with your teacher or official learning material.

The Baudhāyana–Pythagoras Theorem — NCERT Class 8 Mathematics Chapter 9

Direct answer: In a right triangle with perpendicular legs a and b and hypotenuse c, a²+b²=c². The theorem connects lengths to areas of squares and helps find diagonals, unknown sides and integer right-triangle triples.

How can you double the area of a square without guessing its new side? Baudhāyana’s geometric construction shows that the diagonal of the original square becomes the side of a square with exactly twice the area.

Textbook: Ganita Prakash Part II, Chapter 2, pages 33–54. Sections 2.1–2.7 and four Figure it Out sets are explained below.

2.1 Doubling a Square

A square of side s has area s². Doubling its side gives area (2s)²=4s², not 2s². The diagonal of the original square has length s√2, so the square built on it has area 2s². An area dissection proves the result by dividing the original into two congruent triangles and the new square into four.

2.2 Halving a Square

Join the midpoints of consecutive sides of a square of side s. The four corner triangles are congruent, each of area s²/8. Their combined area is s²/2, so the central tilted square has half the original area. Its side is s/√2, not s/2; halving the side would quarter the area.

2.3 Isosceles right triangles and √2

An isosceles right triangle has two equal legs a and hypotenuse a√2. The diagonal of a unit square is √2≈1.41421356. Since 1.414²=1.999396 and 1.415²=2.002225, 1.414<√2<1.415. √2 is irrational: if √2=m/n in lowest terms, m²=2n² implies m even and then n even, contradicting lowest terms.

2.4 Combining squares of different sizes

For a right triangle with legs a and b and hypotenuse c, the square on c has the combined areas of the squares on a and b. One proof arranges four congruent right triangles in a square of side a+b: (a+b)²=4(ab/2)+c², hence c²=a²+b². This works for all positive side lengths, not only integers.

How to find an unknown side

If the two perpendicular legs are known, c=√(a²+b²). If the hypotenuse and one leg are known, the other leg is √(c²−a²). Check that the hypotenuse is longest and use the same units for all lengths. Example: legs 6 and 8 give c=√100=10.

2.5 Baudhāyana triples and primitive triples

Positive integers (a,b,c) with a²+b²=c² form a Baudhāyana triple. Examples: (3,4,5), (5,12,13), (8,15,17). Scaling all three by positive integer k preserves the property. A primitive triple has HCF(a,b,c)=1. Every non-primitive triple is a scaled primitive triple.

Generate triples from odd squares

Since (n−1)²+(2n−1)=n², whenever 2n−1=m² for odd positive m≥3, (m,(m²−1)/2,(m²+1)/2) is a triple. For m=5 this gives (5,12,13). This method generates primitive triples, but not every primitive triple.

2.6 Fermat's Last Theorem

Fermat's Last Theorem says xⁿ+yⁿ=zⁿ has no solutions in positive integers for integer n>2. It is distinct from the square-power case n=2, which has many triples. The theorem was proved by Andrew Wiles, with the completed proof published in 1995 after a 1994 correction.

2.7 Applications

Find diagonals of rectangles and squares, test right-angle measurements, calculate sloping distances, find rhombus side lengths from diagonals, and determine the height of equilateral triangles. In the lotus problem from Līlāvatī, if a lotus rises 1 unit above water and touches the water 3 units away, let depth x; x²+3²=(x+1)² gives x=4.

Important formulas and comparison table

Situation	Formula	Example
Right triangle	a²+b²=c²	3²+4²=5²
Hypotenuse	c=√(a²+b²)	5,12→13
Missing leg	b=√(c²−a²)	c=17,a=8→15
Isosceles right triangle	c=a√2	a=5→5√2
Square diagonal	d=s√2	s=4→4√2
Rectangle diagonal	d=√(l²+w²)	6×8→10
Rhombus side	s=√((d₁/2)²+(d₂/2)²)	24,70→37

Original competency practice and common mistakes

Original MCQ: Which is a right-triangle triple? (A) 2,3,4 (B) 8,15,17 (C) 5,6,7; answer B. HOTS: A 10 m ladder reaches 8 m up a wall; its foot is √(100−64)=6 m from the wall. Common errors: applying the theorem to non-right triangles, using the wrong side as hypotenuse, forgetting the square root, adding side lengths instead of their squares, and confusing area with length.

Learning checklist and flashcards

Can you explain why doubling a square's side quadruples its area? Can you derive the hypotenuse for equal legs? Can you identify primitive triples and scale them? Can you solve a real-world right-triangle problem and label units? Flashcard: longest side of right triangle? Hypotenuse. Flashcard: diagonal of unit square? √2. Flashcard: area of square built on hypotenuse? Sum of areas of squares on legs.

NCERT Figure it Out — page 39

Question wording is paraphrased; the numbered order and numerical values follow the official textbook. Any answer requiring the printed diagram is explicitly identified.

Question 1. Using pieces cut from two equal square papers, make a square with twice the area of either original square.
Step-by-step solution: If each original square has side s, the combined area is 2s². Arrange the four congruent triangular pieces around a square so that the outside boundary is a square; its side equals s√2. Check: (s√2)²=2s².
Final answer: A square of side s√2 and area 2s²; use the printed cut-out diagram for the physical arrangement.
Question 2. For isosceles right triangles with equal legs 3, 4, 6, 8 and 9 units, find hypotenuses and one-decimal bounds.
Step-by-step solution: For equal legs a, c²=2a², so c=a√2. Using √2≈1.41421356: 3√2≈4.2426, 4√2≈5.6569, 6√2≈8.4853, 8√2≈11.3137, 9√2≈12.7279.
Final answer: 3→4.2<c<4.3; 4→5.6<c<5.7; 6→8.4<c<8.5; 8→11.3<c<11.4; 9→12.7<c<12.8.
Question 3. An isosceles right triangle has hypotenuse 10 units. Find both equal legs.
Step-by-step solution: Let each leg be a. Then 2a²=10²=100; a²=50; a=√50=5√2.
Final answer: Each leg is 5√2 units, approximately 7.071 units.

NCERT Figure it Out — page 47

Question wording is paraphrased; the numbered order and numerical values follow the official textbook. Any answer requiring the printed diagram is explicitly identified.

Question 1. A right triangle has perpendicular sides 5 cm and 12 cm. Find the hypotenuse.
Step-by-step solution: c²=5²+12²=25+144=169. Taking the positive square root, c=13.
Final answer: 13 cm.
Question 2. A right triangle has one leg 8 cm and hypotenuse 17 cm. Find its other leg.
Step-by-step solution: b²=17²−8²=289−64=225, so b=15.
Final answer: 15 cm.
Question 3. Construct squares of area three times and five times a given square.
Step-by-step solution: For original side s, construct a right triangle with legs s and s√2 to obtain hypotenuse s√3, the side of a square with area 3s². For five times, construct a right triangle with legs s and 2s to obtain hypotenuse s√5; its square has area 5s².
Final answer: Construct squares of sides s√3 and s√5.
Question 4. Find missing side lengths: (i) a=5,b=7; (ii) a=8,b=12; (iii) a=9,c=15; (iv) a=7,b=12; (v) a=1.5,b=3.5.
Step-by-step solution: (i) c=√(25+49)=√74. (ii) c=√(64+144)=√208=4√13. (iii) b=√(225−81)=12. (iv) c=√(49+144)=√193. (v) c=√(2.25+12.25)=√14.5.
Final answer: (i) √74≈8.60; (ii) 4√13≈14.42; (iii) 12; (iv) √193≈13.89; (v) √14.5≈3.81 (in given units).

NCERT Figure it Out — page 50

Question wording is paraphrased; the numbered order and numerical values follow the official textbook. Any answer requiring the printed diagram is explicitly identified.

Question 1. Generate five more integer right-triangle triples using consecutive-square differences.
Step-by-step solution: Use (n−1)²+(2n−1)=n². Choose odd squares 2n−1=m² for m=7,9,11,13,15. Then n=(m²+1)/2, giving triples (m,(m²−1)/2,(m²+1)/2).
Final answer: (7,24,25), (9,40,41), (11,60,61), (13,84,85), (15,112,113).
Question 2. Can this odd-square method produce non-primitive triples?
Step-by-step solution: For odd m=9, the method yields (9,40,41), whose HCF is 1; for any odd m, gcd(m,(m²−1)/2)=1 because a common divisor of m and (m²−1)/2 would divide 1. Therefore the method always produces primitive triples.
Final answer: No; all triples produced by this construction are primitive.
Question 3. Are there primitive right-triangle triples not produced by this method?
Step-by-step solution: The method always produces a triple in which the two largest sides differ by 1. But (8,15,17) is primitive (HCF 1) and its two largest sides differ by 2.
Final answer: Yes; (8,15,17) is an example.

NCERT Figure it Out — pages 52–54

Question wording is paraphrased; the numbered order and numerical values follow the official textbook. Any answer requiring the printed diagram is explicitly identified.

Question 1. Find the diagonal of a square with side 5 cm.
Step-by-step solution: A diagonal divides the square into two right triangles with legs 5 cm each. d²=5²+5²=50.
Final answer: 5√2 cm, approximately 7.07 cm.
Question 2. Find the missing side lengths in the right-triangle diagrams.
Step-by-step solution: The printed figure lists several sets of side labels (including 7,9,4,10,40,41,27,45,200,10,150,10), but the text layer does not reliably associate each number with its triangle or indicate which is the hypotenuse. For each diagram, identify the side opposite 90°, then use c=√(a²+b²) or missing leg=√(c²−known leg²).
Final answer: Requires diagram-by-diagram verification; no fabricated assignment of side labels.
Question 3. Find the side of a rhombus with diagonals 24 and 70 units.
Step-by-step solution: Rhombus diagonals bisect at 90°, giving half-diagonals 12 and 35. Side²=12²+35²=144+1225=1369.
Final answer: 37 units.
Question 4. Is the hypotenuse the longest side of a right triangle? Justify.
Step-by-step solution: For positive legs a,b, c²=a²+b²>a² and c²>b². All lengths are positive, so c>a and c>b.
Final answer: Yes, always.
Question 5. True or false: Every integer right-triangle triple is primitive or a scaled primitive triple.
Step-by-step solution: Take HCF d of the three positive integer side lengths. Dividing each by d produces integer lengths with HCF 1. Their squares still satisfy the theorem, so the original is d times a primitive triple.
Final answer: True.
Question 6. Give five rectangles with integer sides and integer diagonals.
Step-by-step solution: Use five integer triples where the largest entry is the diagonal. Examples 3²+4²=5²; 5²+12²=13²; 8²+15²=17²; 7²+24²=25²; 9²+12²=15².
Final answer: Rectangles 3×4 (diagonal 5), 5×12 (13), 8×15 (17), 7×24 (25), 9×12 (15).
Question 7. Construct a square with area equal to the difference between squares of sides 5 and 7.
Step-by-step solution: Required area=7²−5²=49−25=24. Construct a right triangle with hypotenuse 7 and one leg 5; the remaining leg has length √24=2√6. Construct a square on that leg.
Final answer: Square side 2√6 units, area 24 square units.
Question 8. On a square dot grid, can you form squares of areas 2, 3, 4 and 5? Which integer areas are possible?
Step-by-step solution: A grid-square side vector (p,q) between lattice points has squared length p²+q²; rotating by 90° gives another integer-coordinate side, so area=p²+q². For 2 choose (1,1); for 4 choose (2,0); for 5 choose (1,2). Area 3 cannot be written as a sum of two integer squares.
Final answer: 2 yes; 3 no; 4 yes; 5 yes. In general, possible areas are positive integers expressible as p²+q² with integers p,q.
Question 9. Find the area of an equilateral triangle with side 6 units.
Step-by-step solution: The altitude bisects the base into two segments of 3. Height²=6²−3²=27, so height=3√3. Area=(1/2)×6×3√3.
Final answer: 9√3 square units, approximately 15.59 square units.

Further textbook activities

Paper folding: Join the side midpoints to obtain a half-area square. Paper rearrangement: Combine two equal or unequal squares by rearranging pieces, preserving area. Lotus puzzle: x²+3²=(x+1)², so depth x=4 units. Find the Colours: The last page presents a logic puzzle with three wrongly labelled boxes. It is a separate reasoning task, not an application of the theorem; with only the labels and permission to open one box, opening the box labelled RED reveals its actual colour but does not necessarily distinguish the two possible assignments of the other boxes unless additional information or constraints are supplied.

Official reference and accuracy note

Verified against the official NCERT Ganita Prakash Part II Chapter 2 PDF. The four Figure it Out sets comprise 3 questions (page 39–40), 4 questions (page 47), 3 questions (page 50) and 9 questions (pages 52–54). The second question of the final set depends on printed figures and is not assigned unverified numerical labels. Historical chronology is attributed to the textbook where appropriate, and is not independently adjudicated.