CBSE MATHEMATICS • SELECTED CENTRAL & STATE COMPETITIVE EXAMS
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GANITA PRAKASH · GRADE 8 · PART II · CHAPTER 2

Source Investigations

Think first. Use AI purposefully. Verify always.

Source application figures

The six right triangles on printed page 53 are redrawn with the same data and right-angle relationships. They are numbered here in reading order.

Page 53 — figure 1

79?
Check the missing side

72 + 92 = 130; missing length = 130.

Page 53 — figure 2

410?
Check the missing side

42 + 102 = 116; missing length = 229.

Page 53 — figure 3

40?41
Check the missing side

412 − 402 = 81; missing length = 9.

Page 53 — figure 4

10?√200
Check the missing side

200 − 102 = 100; missing length = 10.

Page 53 — figure 5

10√150?
Check the missing side

102 + 150 = 250; missing length = 510.

Page 53 — figure 6

27?45
Check the missing side

452 − 272 = 1296; missing length = 36.

Rhombus half-diagonals

1235

Schematic, not to scale. Diagonals bisect at right angles; each side is √(12²+35²)=37.

Additional source checks

Printed page 39 — equal legs 3,4,6,8,9

Hypotenuses: 32, 42, 62, 82, 92. One-decimal bounds: (4.2,4.3), (5.6,5.7), (8.4,8.5), (11.3,11.4), (12.7,12.8). Square the endpoints to verify each strict bound.

Printed pages 40 and 47 — missing lengths

Isosceles right hypotenuse 10 gives each leg 52. Page 47 Q4 answers: 74, 413, 12, 193, 58/2. For the final case, 1.52+3.52=14.5=584.

Construct areas three and five times an original square

Combine a square of area A with its doubled square of area 2A to obtain 3A. Double the doubled square to obtain 4A, then combine with A to obtain 5A. The required sides are 3A and 5A.

Why is the hypotenuse longest?

Positive legs give c2=a2+b2>a2 and c2>b2. Taking positive roots gives c>a and c>b.

Integer rectangles and primitive reduction

Rectangles 3×4, 5×12, 8×15, 7×24 and 12×35 have diagonals 5,13,17,25,37 respectively. Divide every entry of any non-primitive triple by its gcd: dividing the squared equality by that gcd squared yields a primitive triple.

A thread through mathematical history

The textbook places square constructions in Baudhāyana’s Śulba-Sūtra (c. 800 BCE) and connects the theorem’s familiar name with Pythagoras (c. 500 BCE). It attributes the prime-factor argument for √2 to Euclid’s Elements (c. 300 BCE).

It traces Fermat’s seventeenth-century question about higher powers to Andrew Wiles’s 1994 proof. For positive integers x,y,z and integer n>2, xⁿ+yⁿ=zⁿ has no solution. The lotus application comes from Bhāskarāchārya’s Līlāvatī. These notes follow the textbook’s historical framing.

Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 53