CBSE MATHEMATICS • SELECTED CENTRAL & STATE COMPETITIVE EXAMS
AIMSSchool of MathematicsAim through AIMS

GANITA PRAKASH · GRADE 8 · PART II · CHAPTER 2

Chapter Summary

Think first. Use AI purposefully. Verify always.

Concept Map

Theorem / Formula Bank

  • Right triangle: a2+b2=c2, with c opposite the established right angle.
  • Converse: for a triangle with longest side c, a2+b2=c2 establishes the right angle.
  • Isosceles right triangle: c=a2. An arbitrary isosceles triangle does not satisfy this rule.
  • Square diagonal: d=a2. Midpoint-square area is a2/2.
  • 1.414<2<1.415; finite decimals are approximations.
  • Primitive triple: positive integer sides satisfying the theorem, with gcd 1.
  • Odd-square generator: n=(m2+1)2, 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.

Visual Recall

Move the pieces; preserve the area

WXV

Two cuts separate W and X from the five-sided piece V. Total area: 32 + 62 = 45.

All four new sides are congruent hypotenuses. The two acute angles of a right triangle are complementary, so each new corner is 90°. This establishes a square before using its area formula.

All eleven mathematical labs →

Triple Bank

Primitive with all entries at most 20: (3,4,5), (5,12,13), (8,15,17). Other triples within that bound: (6,8,10), (9,12,15), (12,16,20). Further examples: (7,24,25), (12,35,37), (15,36,39).

Important Connections

Congruence and area preservation establish the dissection. The theorem connects area to length, then supports exact radicals, integer triples and applications.

On a unit lattice, every square area has the form p²+q² for integers p,q not both zero. The odd-square generator produces primitive triples but does not produce them all.

Common Mistakes

  • Doubling the side makes four times the area, not twice.
  • Halving the side leaves one quarter of the area.
  • The formula needs an isosceles right triangle, not just any isosceles triangle.
  • Write √2 ≈ 1.414, never √2 = 1.414. Repeating decimals such as 1/3 are rational.
  • Equal total area alone does not prove the new boundary is a square; check its sides and angles.
  • Never use the theorem for a triangle merely because its drawing looks right-angled.
  • An integer right-triangle triple need not be primitive.
  • (8,15,17) is primitive but not generated: neither leg is 16, one less than its hypotenuse.
  • Use half-diagonals in a rhombus. Distinguish a square’s area from its side length.
  • Square the lengths before adding or subtracting.
  • The positive square root gives the length; negative lengths are not valid.

Mini Examples

DOUBLE — Area of the square constructed on its diagonal

Given

A square with side 3 units.

To Find

Area of the square constructed on its diagonal

Property

Four congruent half-square triangles form the new square.

Solution

  1. The original area is 32 = 9 square units.
  2. The new square contains twice as many congruent triangles.

Result

The new area is 18 square units.

HALVE — Area of its midpoint square

Given

An outer square of side 8 units.

To Find

Area of its midpoint square

Property

Subtract the four congruent corner triangles.

Solution

  1. Each corner triangle has area (4×4)2 = 8.
  2. Outer area = 64; removed area = 4 × 8 = 32.

Result

Midpoint-square area = 32 square units.

ROOT — Length of its hypotenuse

Given

An isosceles right triangle with equal legs 12.

To Find

Length of its hypotenuse

Property

c2 = 2a2 from square doubling.

Solution

  1. c2 = 2 × 122 = 288.
  2. c = 288 = 122.
  3. 162 < 288 < 172, so 16 < c < 17.

Result

The hypotenuse is exactly 122 units.

BOUND — Three-decimal lower and upper bounds for √2

Given

Positive candidates 1.414 and 1.415.

To Find

Three-decimal lower and upper bounds for 2

Property

Order of squares of positive numbers.

Solution

  1. 1.4142 = 1.999396 < 2.
  2. 1.4152 = 2.002225 > 2.

Result

1.414 < 2 < 1.415.

COMBINE — Area of the square made by the source dissection

Given

Squares with sides 3 and 4 units.

To Find

Area of the square made by the source dissection

Property

Rearrangement preserves area; the boundary is a square by congruence and complementary angles.

Solution

  1. The two original areas are 9 and 16.
  2. The central piece stays fixed; the two right triangles are rigidly moved.

Result

The combined square has area 25 square units.

THEOREM — Length of the other leg b

Given

A right triangle with hypotenuse 17 cm and one leg 8 cm.

To Find

Length of the other leg b

Property

82 + b2 = 172.

Solution

  1. b2 = 289 − 64 = 225.
  2. b = 225 = 15; length is positive.

Result

The other leg is 15 cm.

TRIPLES — Its primitive triple and scale factor

Given

The triple (15,36,39).

To Find

Its primitive triple and scale factor

Property

Divide every entry by their greatest common divisor.

Solution

  1. gcd(15,36,39) = 3.
  2. Dividing gives (5,12,13).
  3. 52 + 122 = 132 and gcd(5,12,13) = 1.

Result

Primitive triple (5,12,13), scale factor 3.

GENERATE — The triple generated from this odd square

Given

Odd square 49 = 72.

To Find

The triple generated from this odd square

Property

2n−1 = 49 and (n−1)2 + 49 = n2.

Solution

  1. n = 25 and n−1 = 24.
  2. 242 + 72 = 576 + 49 = 625 = 252.

Result

The primitive triple is (24,7,25).

APPLY — Length of each rhombus side

Given

A rhombus has diagonals 24 and 70 units.

To Find

Length of each rhombus side

Property

Rhombus diagonals bisect each other at right angles.

Solution

  1. Half-diagonals are 12 and 35.
  2. A side is the hypotenuse: s2 = 122 + 352 = 1369.
  3. s = 37.

Result

Each side is 37 units.

5-Question Quick Check

Progress: 05 checked correctly

Q1. A square has area 18. Find the area of the square on its diagonal.
Attempts: 0
Q2. Which statement is correct?
Attempts: 0
Q3. A right triangle has hypotenuse 17 and leg 8. Find the other leg.
Attempts: 0
Q4. Why does the generator miss (8,15,17)?
Attempts: 0
Q5. Rhombus diagonals are 24 and 70. Find its side.
Attempts: 0