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Sridharacharya’s Square Formula & Extension to Three Numbers

🎓 Class 8 Mathematics CBSE Theory Ch 6 — Linear Equations in One Variable ⏱ ~20 min
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This MCQ module is based on: Sridharacharya’s Square Formula & Extension to Three Numbers

This mathematics assessment will be based on: Sridharacharya’s Square Formula & Extension to Three Numbers
Targeting Class 8 level in Algebra, with Basic difficulty.

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Sridharacharya's Square Formula & Extension to Three Numbers

Sridharacharya (750 CE) gave an interesting method to quickly compute the squares of numbers using Identity 1C. Consider the following modified form of this identity:

\(a^2 \;=\; (a+b)(a-b) + b^2\)

Why is this identity true? (It is just Identity 1C rearranged.) Now, for example, \(31^2\) can be found by taking \(a=31\) and \(b=1\):

\(31^2 = (31+1)(31-1) + 1^2 = 32 \times 30 + 1 = 960 + 1 = 961\).

\(197^2\) can be found by taking \(a=197, b=3\): \(197^2 = (197+3)(197-3) + 3^2 = 200 \times 194 + 9 = 38800 + 9 = 38809\).

Figure it Out

1. Which is greater: \((a-b)^2\) or \((b-a)^2\)? Justify.
They are equal. \((b-a)^2 = (-(a-b))^2 = (a-b)^2\). A square is unchanged by a sign flip of the base.
2. Express 100 as the difference of two squares.
\(100 = 26^2 - 24^2\) since \((26-24)(26+24) = 2 \times 50 = 100\). Another: \(100 = 10^2 - 0^2\).
3. Find \(60^2, 72^2, 145^2, 1097^2\) and \(124^2\) using identities you have learned.
\(60^2 = 3600\); \(72^2 = (70+2)^2 = 4900+280+4 = 5184\); \(145^2 = (145+5)(145-5)+25 = 150\times 140 + 25 = 21025\); \(1097^2 = (1100)(1094)+9 = 1203409\); \(124^2 = (120+4)^2 = 14400+960+16 = 15376\).
4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? Also for fractions?
Yes — the identities \((a+b)^2+(a-b)^2 = 2(a^2+b^2)\) and \((a+b)(a-b) = a^2-b^2\) are true for every real number, including negatives and fractions, because they follow from the distributive property which itself works for all real numbers.

6.3 Mind the Mistake, Mend the Mistake

We have expanded and simplified some algebraic expressions below to their simplest forms. Check each of the simplifications and say if there is a mistake. If there is a mistake, try to explain what could have gone wrong. Then write the correct expression.

(1) \(-3p^2 -2p + 2p^2 = -3p^2 + 2p - 2p = -p^2 - 2p\). Is this correct?
Mistake: only \(-3p^2 + 2p^2\) should combine to \(-p^2\). Correct: \(-p^2 - 2p\).
(2) \(2x^{-1} + 3x - 10 = 2x - 1 + 3x - 10 = 5x - 11\). Correct?
Mistake: \(2x^{-1}\) is \(2/x\), not \(2x-1\). So the "simplification" is invalid.
(3) \(y + 22y - 23y = y + 22y - 23y = y\). Correct?
Mistake: \(y + 22y - 23y = 23y - 23y = 0\), not \(y\).
(4) \((5m + 6m^2)^2 = 25m^2 + 36m^4\). Correct?
Mistake: cross term missing. Correct: \((5m)^2 + 2(5m)(6m^2) + (6m^2)^2 = 25m^2 + 60m^3 + 36m^4\).
(5) \(6y^2 \cdot 4y^2 \cdot 6y^2 = 24y^2 \cdot 6y^2 = 144y^4\). Correct?
Mistake: the exponent addition was dropped. Correct: \(6\cdot 4\cdot 6 \cdot y^{2+2+2} = 144 y^6\).

6.4 This Way or That Way — All Ways Lead to the Bay

Observe the pattern in the figure: how many circles are there in Step 1, 2, 3…? How many total circles appear in Step \(k\)? Write an expression for the number of circles in Step \(k\). There are many different interpretations of this pattern.

Step 1 (3) Step 2 (8) Step 3 (15) …
Step-pattern of circles: 3, 8, 15, …, in general \(k(k+2) = (k+1)^2 - 1\)

Many Paths to the Same Expression

There are many ways of interpreting this pattern. Here are some possibilities:

  • Method 1: Step \(k\) forms a \((k+1) \times (k+1)\) square of dots with one corner removed: \((k+1)^2 - 1\).
  • Method 2: 1 + 2 + … + arranged symmetrically ⇒ \(k^2 + 2k\).
  • Method 3: \(3 + 5 + 7 + \ldots\) — a row of extras added each step.
  • Method 4: Sum of two triangles plus a middle strip \(= k(k+2)\).

When carried out correctly, all methods lead to the same answer: \(k^2 + 2k\). The expression \(k^2 + 2k\) gives the number of circles in Step \(k\) of this pattern.

Big Idea
In mathematics, there are often multiple ways of looking at a pattern, and different ways of approaching and solving the same problem. Finding such ways often requires a great deal of creativity and imagination! While one or two of the ways might be your favourite(s), it can be amusing and enriching to explore other ways as well.

Consider the pattern made of square tiles in the picture below

1 2 3
Square-tile pattern: Step 1 = 1, Step 2 = 8, Step 3 = 16 → number = \((2n-1)^2 - (2n-3)^2\) = \(8(n-1)\) for n≥2

How many square tiles are there in each figure? Step 1: 1; Step 2: 8; Step 3: 16.
Step 4: 24. Step 10: \(8 \times 9 = 72\).
General expression (for \(n \geq 2\)): \((2n-1)^2 - (2n-3)^2 = 8(n-1)\) tiles.

Find the area of the (interior) shaded region in the figure below (all three rectangles have the same dimensions)

side m side m m
Three different methods of computing shaded area — all give \(m^2 - 3mn\)

Tadang's method: Big square area = \(m^2\). Subtract three rectangles = \(3 \times (m \times n) = 3mn\). Shaded area = \(m^2 - 3mn = m(m-3n)\).

Yusuf's method: Shaded region = square of side \((m-3n)\) plus 3-strips-adjusted \(= (m-3n)m\). By expanding: \((m-3n)m = m^2 - 3mn\). Same answer.

All three rectangles same dimensions — L-shape area

Find the area of the region of the figure with slanting lines in the figure (Fig 1).

Anusha's method: Required area = Area(ABCD) − Area(EFGH) = \(xy - x'y' = xy - x^2\) (using given). Area of EFGH = \(x^2\). So required area = \(xy - x^2\).

Vaishnavi's method: \(QS = y + x + y = x + 2y\). Area of \(PQRS = x(x+2y)\). Required area = \(x(x+2y) - (\text{area of three rectangles}) = x(x+2y) - 3xy\).

Aditya's method: Required area \(=2 \times \frac{(x-y)}{2}\cdot (\text{KM}) = (x-y)\cdot x \cdot ?\) — by expanding, verify all three expressions are equivalent. If \(x=8, y=3\), find the area of the shaded region.

Dashed-region puzzle (p = 6, r = 3.5, s = 9)

Write an expression for the area of the dashed region and substitute the values to get the numerical area. The dashed region is a rectangle of width \(s-r\) and length \(p-r\), adjusted for the overlap: Area = \(p\cdot s - r\cdot s - r\cdot (p - r) = ps - rs - rp + r^2\). Plug in: \(6\cdot 9 - 3.5\cdot 9 - 3.5\cdot 6 + 3.5^2 = 54 - 31.5 - 21 + 12.25 = 13.75\) sq. units.

Fast Multiplications Using the Distributive Property

The distributive property can be used to come up with quick methods of multiplication when certain types of numbers are multiplied.

When one of the numbers is 11, 101, 1001, …

The following multiplications to find the product of a number with 11 in a single step. \(3874 \times 11\). Let us take the first multiplication: \(3874 \times 11 = 3874 \times 10 + 3874 = 38740 + 3874\).

Notice that the digits are being added. That is, the number that has \(d\) in the thousands place, \(c\) in the hundreds place, \(b\) in the tens place and \(a\) in the units place. This becomes:

PlaceThHTU
\(dcba\times 10\)dcba · 0
\(dcba \times 1\)dcb · a
Sumdc+db+ca+b · a

This can be used to obtain the product in one line.

Step 1: write 4 on the right. Step 2: 4+7=11, write 1 carry 1. Step 3: 7+8+1=16, write 6 carry 1. Step 4: 8+3+1=12, write 2 carry 1. Step 5: 3+1=4. Result: \(3874 \times 11 = 42614\).

Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line

Evaluate (i) \(94 \times 11\), (ii) \(456 \times 11\), (iii) \(5279 \times 11\), (iv) \(4791256 \times 11\).

  • \(94 \times 11 = 1034\)
  • \(456 \times 11 = 5016\)
  • \(5279 \times 11 = 58069\)
  • \(4791256 \times 11 = 52703816\)

Can we come up with a similar rule for multiplying a number by 101?

Multiply 3874 by 101: Let us take a 4-digit number \(dcba\), that is, the number that has \(d\) in the thousands place, \(c\) in the hundreds place, \(b\) in the tens place and \(a\) in the units place. This becomes \(dcba \times 101 = dcba \times 100 + dcba \times 1\). Write \(dcba\) shifted two places to the left, and add \(dcba\):

\(3874 \times 101 = 387400 + 3874 = 391274\).

What could be a general rule to multiply a number by 101? Extend this rule for multiplying by 1001, 10001, … Evaluate: \(1111 \times 101, 263831 \times 1001, 11111 \times 1001, 9734 \times 99\) and \(23478 \times 999\).
• \(1111\times 101 = 111100 + 1111 = 112211\)
• \(263831 \times 1001 = 263831263831 - \text{no, carefully: } 263831000 + 263831 = 264094831\)
• \(11111 \times 1001 = 11122111\)
• \(9734 \times 99 = 9734\times 100 - 9734 = 973400 - 9734 = 963666\)
• \(23478 \times 999 = 23478000 - 23478 = 23454522\).
A Pinch of History
The distributive property of multiplication over addition was implicitly in the calculations of mathematicians in many ancient civilisations, particularly in ancient Egypt, Mesopotamia, Greece, China, and India. For example, the mathematicians Euclid (in geometric form) and Aryabhata (in algebraic form) used the distributive property to multiply long numbers extensively in their mathematical and scientific works. The first explicit statement of the distributive property was given by Brahmagupta in his work Brāhmasphuṭasiddhānta (Verse 12.55), who referred to the use of the property for multiplication as khanda-gunana (multiplication by parts). His verse states, "The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added". That is, if there are two parts, then using letter symbols this is equivalent to the rule \(a \cdot (c+b) = ac + bc\). In the next verse (Verse 12.56), Brahmagupta further describes a method for doing fast multiplication using this distributive property, which we explore further in the next section. Such methods of applying the distributive property to easily multiply two numbers were discussed extensively in the ancient mathematical works of Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE). In his work Brāhmasphuṭasiddhānta (Verse 12.56), Brahmagupta refers to such methods for fast multiplication using the distributive property as iṣṭa-guṇana.

Figure it Out

1. Compute the following products using the suggested identity.
(i) \(46^2\) using Identity 1A for \((a+b)^2\)
(ii) \(397 \times 403\) using Identity 1C for \((a+b)(a-b)\)
(iii) \(91^2\) using Identity 1B for \((a-b)^2\)
(iv) \(43 \times 45\) using Identity 1C for \((a+b)(a-b)\).
(i) \((40+6)^2 = 1600 + 480 + 36 = 2116\).
(ii) \((400-3)(400+3) = 160000 - 9 = 159991\).
(iii) \((90+1)^2 = 8100 + 180 + 1 = 8281\) — or \((100-9)^2 = 10000-1800+81=8281\).
(iv) Use \((44-1)(44+1) = 44^2 - 1 = 1936 - 1 = 1935\).
2. Use a suitable identity or the distributive property to find each of the following products.
(i) \((p-1)(p+11)\)
(ii) \((3a-9b)(3a+9b)\)
(iii) \((-2y+5)(5+y)\)
(iv) \((6x+5y)^2\)
(v) \((4-\frac{1}{3}c)^2\)
(vi) \((7p)(3x)(p+2)\)
(i) \(p^2 + 10p - 11\).
(ii) \(9a^2 - 81b^2\).
(iii) \((5-2y)(5+y) = 25 + 5y - 10y - 2y^2 = 25 - 5y - 2y^2\).
(iv) \(36x^2 + 60xy + 25y^2\).
(v) \(16 - \frac{8c}{3} + \frac{c^2}{9}\).
(vi) \(21px(p+2) = 21p^2 x + 42px\).

Identify the appropriate algebraic expression

3. For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number: \(2 + x^2\) / \(2(x^2)\) / \(x^2 + 2\) / \(2 x^2\) / etc.
(ii) The sum of the squares of two consecutive numbers: \(m^2 + n^2\), \((m+1)^2 + n^2\), \(m^2 + (n+1)^2\), \(m^2 + (m+1)^2\), \(m^2 + (m-1)^2\), \((2m+1)^2 + (2m+2)^2\).

Activity: February Diagonal Products (Math Talk)
L4 Analyse
Materials: A calendar page of February
Predict: In every 2-by-2 square on a calendar, how do the two diagonal products compare?
  1. Consider any 2-by-2 square of numbers in a calendar month, e.g. \(\begin{pmatrix} 4 & 5 \\ 11 & 12 \end{pmatrix}\).
  2. Find both diagonal products: \(4 \times 12 = 48\); \(5 \times 11 = 55\).
  3. Difference = \(55 - 48 = 7\). Try another square — you will always get 7.
  4. Explain using algebra: label the top-left \(a\); then the square is \(\begin{pmatrix} a & a+1 \\ a+7 & a+8 \end{pmatrix}\). Diagonal products: \(a(a+8) = a^2+8a\) and \((a+1)(a+7) = a^2+8a+7\). Difference = 7 for every square!

The result is constant because the two products differ by \((a+1)(a+7) - a(a+8) = 7\), independent of \(a\). This is Identity 1 in action.

Verify which of the following statements are true

  1. \((k+1)(k+2) - (k+1)(k+3)\) is always 2.
  2. \((2p+1)(2q+3)\) is an odd number.
  3. Squares of even numbers are multiples of 4 and squares of odd numbers are 1 more than a multiple of 4.
  4. \((4k+2)^2 - (4k-2)^2\) is \(16k\).

Number of letters pattern

A number has a remainder of 3 when divided by 7, and another number has a remainder of 1 when divided by 7. What is the remainder when the sum is divided by 7? What about when the difference (first minus second) is divided by 7?

Competency-Based Questions

Scenario: A tile-layer needs to cover a square courtyard of side \((a+b+c)\) metres using exactly three kinds of tiles: large \(a^2, b^2, c^2\) squares plus rectangular fillers.
Q1. Applying the distributive property, \((a+b+c)^2\) expands to
L3 Apply
  • (a) \(a^2+b^2+c^2\)
  • (b) \(a^2+b^2+c^2+ab+bc+ca\)
  • (c) \(a^2+b^2+c^2+2ab+2bc+2ca\)
  • (d) \(a^2+b^2+c^2+3abc\)
(c). Treat \((a+b+c)\) as \(((a+b)+c)\) and apply Identity 1A twice.
Q2. Analyse the pattern \(3874 \times 11\) by "adding adjacent digits with carry". Explain why the rule works, using place-value.
L4 Analyse
\(N \times 11 = N \times 10 + N\). When we add \(N\) shifted one place left to itself, each column becomes the sum of two adjacent digits of \(N\) — exactly the "add the neighbours" rule, with carries handled normally.
Q3. Evaluate which shortcut is faster for \(99 \times 99\): Identity 1B \((100-1)^2\) or Identity 1C \((99)(99) = (100-1)^2\)? Justify.
L5 Evaluate
Identity 1B is faster here: \((100-1)^2 = 10000 - 200 + 1 = 9801\) — one subtraction and one addition. Identity 1C applies to products of the form \((a+b)(a-b)\), not to a repeated factor, so it does not shortcut \(99^2\) directly unless rewritten as \(99^2 = 98 \times 100 + 1\).
Q4. Create: design a "calendar trick" for a 3-by-3 square in which you predict the sum of the four corners using only the middle number. Prove the trick algebraically.
L6 Create
Let the middle number be \(m\). In a 3-by-3 block the corners are \(m-8, m-6, m+6, m+8\). Sum = \(4m\). Trick: "The sum of the 4 corners of any 3-by-3 calendar block is 4 times the middle number."

Assertion–Reason Questions

Assertion (A): \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\).
Reason (R): Grouping \((a+b+c) = (a+b)+c\) and applying \((x+y)^2 = x^2+2xy+y^2\) gives the identity.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — The expansion flows directly from the grouping.
Assertion (A): \(3874 \times 11 = 42614\).
Reason (R): Multiplication by 11 equals adding the number to itself shifted one place to the left.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — \(3874 \times 11 = 38740 + 3874 = 42614\). R explains the shortcut.
Assertion (A): Two different-looking algebraic expressions can never give the same numerical value for every input.
Reason (R): Equal algebraic expressions are identical letter-by-letter.
(a) Both true.
(b) A true, R false.
(c) A false, R true.
(d) A false, R false.
Answer: (d) — Both false. Expressions like \((a+b)^2\) and \(a^2+2ab+b^2\) look different yet agree on every input. Equality of expressions is about values, not letters.

Frequently Asked Questions — Linear Equations in One Variable

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Part 2 — Factorisation, (a+b+c)² & Multiplication Shortcuts | Class 8 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 6: Linear Equations in One Variable. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

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To solve problems on Part 2 — Factorisation, (a+b+c)² & Multiplication Shortcuts | Class 8 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

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Part 2 — Factorisation, (a+b+c)² & Multiplication Shortcuts | Class 8 Maths Ch 6 | MyAiSchool is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

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Common mistakes in Part 2 — Factorisation, (a+b+c)² & Multiplication Shortcuts | Class 8 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

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End-of-chapter NCERT exercises for Part 2 — Factorisation, (a+b+c)² & Multiplication Shortcuts | Class 8 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.

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