GANITA PRAKASH · GRADE 8 · PART II · CHAPTER 2
Study Materials
Think first. Use AI purposefully. Verify always.
- Right triangle: +=, with c opposite the established right angle.
- Converse: for a triangle with longest side c, += establishes the right angle.
- Isosceles right triangle: c=a. An arbitrary isosceles triangle does not satisfy this rule.
- Square diagonal: d=a. Midpoint-square area is /2.
- 1.414<<1.415; finite decimals are approximations.
- Primitive triple: positive integer sides satisfying the theorem, with gcd 1.
- Odd-square generator: n=, with odd m≥3, gives (n−1,m,n).
For the converse, construct a right triangle with legs a,b. Its hypotenuse has length c by the theorem. SSS makes it congruent to the given triangle, establishing the right angle.
DOUBLE — Double a square using its diagonal
Four congruent isosceles right triangles fit around their right-angle vertices to form a square. One original square contains two of these triangles. The new square therefore has twice the area, and each new side is an original diagonal.
New area = 2 × original area
Cut each of two equal squares into two triangles. Place the four right-angle corners at one centre. Their hypotenuses form the boundary; each outer corner is 45° + 45° = 90°.
Doubling the side makes four times the area, not twice.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 34
Worked example →HALVE — Reverse the construction
The four removed corner triangles are congruent right triangles with legs equal to half the original side. Their combined area is half the original square. The midpoint figure has four equal sides and right angles, so it is a square.
Midpoint-square area = original area ÷ 2
Mark all four midpoints and join consecutive ones. Each corner triangle is isosceles right. At each midpoint the inner angle is 180° − 45° − 45° = 90°; equal hypotenuses give equal sides.
Halving the side leaves one quarter of the area.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 36
Worked example →ROOT — Discover
The diagonal is a positive length whose square is 2. We call that length . More generally, an isosceles right triangle with equal legs a has hypotenuse a.
=
Use the geometric doubling result, then the relation between a square’s area and its positive side length. Keep the radical exact.
The formula needs an isosceles right triangle, not just any isosceles triangle.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 37
Worked example →BOUND — Locate a number that is not a fraction
Squaring positive numbers preserves their order. Successive decimal bounds locate without making a finite decimal exact. Irrationality requires a proof: a non-terminating decimal alone need not be irrational.
1.414 < < 1.415
Compare squares with 2. For irrationality suppose = for positive integers. Then = . Prime 2 occurs an even number of times in but an odd number in , a contradiction.
Write ≈ 1.414, never = 1.414. Repeating decimals such as are rational.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 39
Worked example →COMBINE — Combine two different squares
The source construction keeps the five-sided piece V and relocates two congruent right triangles W and X to T and U. The resulting outer boundary has four equal hypotenuses and four right angles. No area is added or lost.
Area(T + U + V) = Area(W + X + V)
Mark a width-a strip at the right of the larger square. Draw its diagonal, then the second cut from the top-left corner of the smaller square to the strip’s bottom-left corner. Move W and X without stretching. Check the complementary acute angles at each new corner.
Equal total area alone does not prove the new boundary is a square; check its sides and angles.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 44
Worked example →THEOREM — Connect the three squares
For a right triangle with perpendicular legs a and b and hypotenuse c, + = . The hypotenuse is opposite the right angle. Conversely, if a triangle’s longest side c satisfies this equation, the opposite angle is right.
+ =
Establish the right angle and identify the hypotenuse before substituting. Add squared legs to find c; subtract the known leg’s square from to find the other leg. Take the positive square root.
Never use the theorem for a triangle merely because its drawing looks right-angled.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 46
Worked example →TRIPLES — Integer sides and primitive triples
A triple consists of positive integers a, b, c satisfying + = . It is primitive when gcd(a,b,c) = 1. Multiplication by a positive integer preserves the equation; scaling a primitive triple by k > 1 makes it non-primitive.
(ka) + (kb) =
Check positivity and integrality, put the largest number last, verify the squared equality, then compute the common divisor. Divide a non-primitive triple by its gcd to obtain a primitive one.
An integer right-triangle triple need not be primitive.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 49
Worked example →GENERATE — Turn an odd square into a triple
Since (n−1) + (2n−1) = , an odd square = 2n−1 yields (n−1,m,n). Use odd integers m ≥ 3 so every side is positive. These generated triples are primitive because n−1 and n are consecutive.
(n − 1) + (2n − 1) =
Choose an odd m ≥ 3. Calculate n = . Use legs n−1 and m with hypotenuse n. Verify the equality and do not claim the method generates every primitive triple.
(8,15,17) is primitive but not generated: neither leg is 16, one less than its hypotenuse.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 50
Worked example →APPLY — Model, construct and justify
Translate the situation into a right triangle before using the theorem. For the lotus model, a straight stem of unchanged length pivots at a fixed root; its initial position is perpendicular to level water. Rhombi, square diagonals and grid squares also reveal useful right triangles.
+ = (x + 1)
Draw and label the model, identify known quantities and the target, state assumptions and solve. Preserve radicals and check the result in the original equation.
Use half-diagonals in a rhombus. Distinguish a square’s area from its side length.
Textbook connection: Ganita Prakash · Grade 8 · Part II · Chapter 2 — The Baudhāyana–Pythagoras Theorem · printed page 52
Worked example →