Chapter 6: We Distribute, Yet Things Multiply

Yes. Let the two numbers be $$a$$ and $$b$$, and consider what happens when either or both are increased by $$1$$.

Using the distributive property:

So the pattern is: increasing one factor by $$1$$ adds the other factor to the product, and increasing both factors by $$1$$ adds their sum plus $$1$$.

Treat $$(b+1)$$ as a single term and apply the distributive property to $$(a+1) \times (b+1)$$:

Now distribute again inside each bracket:

This is exactly the same expansion as before — grouping $$(a+1)$$ or $$(b+1)$$ first does not change the answer, because multiplication is commutative and associative.

Let the numbers be $$a$$ and $$b$$. Increase $$a$$ by $$1$$ and decrease $$b$$ by $$1$$. The new product is $$(a+1)(b-1)$$.

Expanding using the distributive property,

So the product changes by $$b - a - 1$$ compared to $$ab$$.

The change is not always zero. It depends on how $$a$$ and $$b$$ compare:

• If $$b > a + 1$$, the product increases.
• If $$b < a + 1$$, the product decreases.
• If $$b = a + 1$$, the product is unchanged (for example, $$4 \times 5 = 5 \times 4 = 20$$).

No. From Q.4, the change is $$b - a - 1$$; whenever $$a \ge b$$, this is negative, so the product decreases.

Example 1: $$a = 10,\ b = 3$$.

The product decreases by $$8$$.

Example 2: $$a = 8,\ b = 5$$.

The product decreases by $$4$$.

Example 3: $$a = 15,\ b = 7$$.

The product decreases by $$9$$.

The distributive property holds for all integers, including negative ones. So the identity

is still true; the letter-numbers just take on negative values.

Example 1: $$a = -5,\ b = 8$$.

Change $$= a + b + 1 = -5 + 8 + 1 = 4$$, and indeed $$-36 = -40 + 4$$. Here the product increases (becomes less negative).

Example 2: $$a = -4,\ b = -5$$.

Change $$= a + b + 1 = -4 - 5 + 1 = -8$$, and indeed $$12 = 20 - 8$$. Here the product decreases.

So the identity still describes the change; only the sign of $$a + b + 1$$ decides whether the product goes up or down.

Let the initial numbers be $$a$$ and $$b$$. After the change, they become $$a + m$$ and $$b + n$$.

Apply the distributive property twice:

So the new product is $$ab + mb + an + mn$$, and the increase over the original $$ab$$ is

This is the general form of Identity 1:

Yes. A decrease is just an increase by a negative amount. If a number is decreased by $$k$$, we simply take $$m = -k$$ (or $$n = -k$$) in Identity 1.

Example: Suppose $$a$$ is increased by $$1$$ and $$b$$ is decreased by $$1$$. Set $$m = 1$$ and $$n = -1$$:

which matches what we found earlier by direct expansion.

So Identity 1 handles increases, decreases, and mixed changes in one formula — we just use the correct signs.

Expand each product directly using the distributive property.

(i) $$(a-2)(b+3)$$:

Rewriting, $$= ab + 3a - 2b - 6$$. This matches Q.9(i).

(ii) $$(a-3)(b-4)$$:

This matches Q.9(ii).

Both direct expansions agree with the results from Identity 1, verifying our answers.

No. Two terms can be combined into a single term only if they are like terms — that is, they have exactly the same letter-numbers with the same powers.

For instance, in the expansion of $$\dfrac{3a}{2}\left(a - b + \dfrac{1}{5}\right) = \dfrac{3}{2}a^2 - \dfrac{3}{2}ab + \dfrac{3}{10}a$$, the terms $$\dfrac{3}{2}a^2$$ and $$\dfrac{3}{10}a$$ are not like terms (one has $$a^2$$, the other has $$a$$), and neither can they be combined with $$-\dfrac{3}{2}ab$$.

On the other hand, $$5x^2$$ and $$3x^2$$ are like terms and can be combined into $$8x^2$$.

Yes. Since $$101 = 100 + 1$$, the distributive property gives, for any number $$N$$,

So we simply write $$N$$ shifted two places to the left (multiplication by $$100$$) and add $$N$$ underneath. This is the analogue of the $$\times 11$$ rule (where the shift was one place).

Using $$101 = 100 + 1$$,

Adding column-wise:

For a $$4$$-digit number $$dcba$$, we have

which, added column-wise, gives digits

For $$3874$$: $$d=3,\ c=8,\ b=7,\ a=4$$. The one-line pattern is

Carrying from the right: $$4,\ 7,\ 12 \to 2$$ carry $$1,\ 10 + 1 = 11 \to 1$$ carry $$1,\ 8 + 1 = 9,\ 3$$.

Rule for $$\times 101$$: $$N \times 101 = N \times 100 + N$$. Write $$N$$, then, starting from the units column, add another copy of $$N$$ shifted two places to the left. So each digit from the third position onward becomes the sum of two digits (with carries).

Rule for $$\times 1001$$: Since $$1001 = 1000 + 1$$,

which means writing $$N$$ shifted three places left plus $$N$$. If $$N$$ has $$\le 3$$ digits, the two blocks don't overlap and the product is simply $$N$$ written twice (as a $$6$$-digit number).

Rule for $$\times 10001$$: Similarly, $$N \times 10001 = 10000N + N$$ — shift left by four places and add.

General rule: To multiply $$N$$ by $$1\underbrace{0\dots 0}_{k}1$$, add $$N$$ to itself shifted $$k + 1$$ places to the left. Equivalently,

Distribute $$\dfrac{3a}{2}$$ over each term inside the bracket:

Simplify each term.

Combining,

No two of these are like terms, so the expansion cannot be simplified further.

Distribute $$(a+b)$$ over the second bracket:

Distribute again inside each term:

Since $$ab$$ and $$ab$$ are like terms, add them:

Distribute $$(a+b)$$ over each term of the second bracket:

Expand each piece:

Adding all six terms:

Combine like terms: $$a^2 b + 2a^2 b = 3a^2 b$$ and $$2ab^2 + ab^2 = 3ab^2$$.

Notice this is precisely $$(a+b)^3$$ — since the second bracket is $$(a+b)^2$$.

In the multiplication grid, the entry in row $$r$$ and column $$c$$ is the product $$r \times c$$. If the centre of the $$3\times 3$$ frame is $$pq$$ (that is, row $$p$$ and column $$q$$), then in the frame the row-labels are $$p-1,\ p,\ p+1$$ and the column-labels are $$q-1,\ q,\ q+1$$.

The $$3\times 3$$ block of expressions is therefore:

Each expression is the product of that cell's row and column labels.

Let the two numbers be $$a$$ and $$b$$. After the change they become $$a+2$$ and $$b-4$$. We want the product to be unchanged:

Expanding the left side using Identity 1 (with $$m = 2$$ and $$n = -4$$):

Cancel $$ab$$ from both sides and simplify:

So we need pairs $$(a, b)$$ where $$b = 2a + 4$$. Any such pair works. Three easy examples:

• $$a = 1,\ b = 6$$: $$1 \times 6 = 6$$ and $$3 \times 2 = 6.\ \checkmark$$
• $$a = 2,\ b = 8$$: $$2 \times 8 = 16$$ and $$4 \times 4 = 16.\ \checkmark$$
• $$a = 3,\ b = 10$$: $$3 \times 10 = 30$$ and $$5 \times 6 = 30.\ \checkmark$$

Yes. Split the $$65 \times 65$$ square into four parts using a grid line $$60$$ units from one side (and $$5$$ units from the other) in each direction. The four parts are:

• a $$60 \times 60$$ square of area $$60^2 = 3600$$,
• a $$5 \times 5$$ square of area $$5^2 = 25$$,
• two rectangles, each of dimensions $$60 \times 5$$ and area $$300$$.

Adding all four areas,

So a $$65 \times 65$$ square has area $$4225$$ sq. units.

The same $$65 \times 65$$ square can be cut into four pieces along different grid lines. Each way still gives the same total area.

Splitting as $$30 + 35$$: Draw a grid line $$30$$ units from a side (and $$35$$ from the other) in each direction. The four pieces are a $$30\times 30$$ square, a $$35\times 35$$ square, and two $$30\times 35$$ rectangles.

Splitting as $$52 + 13$$: The four pieces are a $$52\times 52$$ square, a $$13\times 13$$ square, and two $$52\times 13$$ rectangles.

Both splits give the same area $$4225$$ — as they must, because both equal $$65^2$$.

By Identity 1A,

So the difference is

The sign of $$2ab$$ decides the comparison:

• If $$a$$ and $$b$$ have the same sign (both positive or both negative), $$ab > 0$$, so $$(a+b)^2 > a^2 + b^2$$.
• If $$a$$ and $$b$$ have opposite signs, $$ab < 0$$, so $$(a+b)^2 < a^2 + b^2$$.
• If either is $$0$$, they are equal.

For example, $$a = 3,\ b = -2$$: $$(a+b)^2 = 1$$ but $$a^2 + b^2 = 13$$, so $$(a+b)^2 < a^2 + b^2$$.

Finding $$104^2$$. Write $$104 = 100 + 4$$ and apply $$(a+b)^2 = a^2 + 2ab + b^2$$ with $$a = 100$$, $$b = 4$$:

Finding $$37^2$$. Write $$37 = 30 + 7$$ with $$a = 30$$, $$b = 7$$:

Using Identity 1A. With $$a = 3j$$ and $$b = 2k$$:

Using the distributive property. $$(3j + 2k)^2 = (3j+2k)(3j+2k)$$; distribute:

Expand each part:

Adding and combining like terms $$6jk + 6jk = 12jk$$:

Both methods give the same expansion.

Yes. Draw a $$55 \times 55$$ square sitting inside a $$60 \times 60$$ square. To go from the $$60\times 60$$ square to the $$55\times 55$$ square, cut off two $$60 \times 5$$ strips (one along the top and one along the side).

These two strips together cover the corner $$5\times 5$$ square twice, so we have subtracted it once too many. Add it back:

So $$55^2 = 3025$$ sq. units.

Rewrite $$(a - b)^2$$ as $$(a + (-b))^2$$ and apply Identity 1A with $$a$$ and $$-b$$:

Since $$(-b)^2 = b^2$$ and $$2a(-b) = -2ab$$, this becomes

which is Identity 1B.

Draw a square of side $$a$$; its area is $$a^2$$. Inside it, along two adjacent sides, cut off two strips, each of width $$b$$ and length $$a$$. Each strip has area $$a \times b = ab$$, so the two strips together have area $$2ab$$.

Removing these two strips takes away the small corner $$b \times b$$ square twice. To correct this, add its area $$b^2$$ back once.

What remains is exactly the square of side $$(a - b)$$:

Try a few pairs and look for a rule.

Pair $$(3, 4)$$: sum of squares $$= 9 + 16 = 25$$; twice that is $$50 = 49 + 1 = 7^2 + 1^2$$. Notice $$7 = 3+4$$ and $$1 = 4-3$$.

Pair $$(2, 5)$$: sum of squares $$= 4 + 25 = 29$$; twice is $$58 = 49 + 9 = 7^2 + 3^2$$. Here $$7 = 5+2$$ and $$3 = 5-2$$.

Pair $$(6, 5)$$: sum of squares $$= 36 + 25 = 61$$; twice is $$122 = 121 + 1 = 11^2 + 1^2$$. Here $$11 = 6+5$$ and $$1 = 6-5$$.

Pattern. For any pair $$(a, b)$$,

The two squares are the square of the sum and the square of the difference.

Yes. Add the two identities:

The $$+2ab$$ and $$-2ab$$ cancel, leaving

This is exactly the identity behind the pattern in Q.31.

Look at each line. The left side is $$a^2 - b^2$$ and the right side is $$(a + b)(a - b)$$:

• $$9^2 - 1^2 = 80 = 10 \times 8 = (9+1)(9-1).\ \checkmark$$
• $$8^2 - 6^2 = 28 = 14 \times 2 = (8+6)(8-6).\ \checkmark$$
• $$7^2 - 2^2 = 45 = 9 \times 5 = (7+2)(7-2).\ \checkmark$$
• $$10^2 - 4^2 = 84 = 14 \times 6 = (10+4)(10-4).\ \checkmark$$

The algebraic statement is

This is Identity 1C, so the pattern always holds — for any values of $$a$$ and $$b$$.

Proof by expansion: $$(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$$.

Take a square of side $$a$$; its area is $$a^2$$. From one corner, cut out a smaller square of side $$b$$ (with $$b < a$$); its area is $$b^2$$. The L-shaped region that remains has area $$a^2 - b^2$$.

Now cut the L-shape along a line that separates it into two rectangles:

• Rectangle 1: dimensions $$a \times (a - b)$$.
• Rectangle 2: dimensions $$b \times (a - b)$$.

Slide Rectangle 2 next to Rectangle 1 (rotating if needed) so that both align on their side of length $$(a - b)$$. The combined shape is a single rectangle whose dimensions are

Because we only rearranged pieces without adding or removing area,

Start from Identity 1C:

Add $$b^2$$ to both sides:

So $$a^2 = (a+b)(a-b) + b^2$$, which is just Identity 1C rearranged. This form is very useful for quick mental squaring — for example,

Notice that $$b - a = -(a - b)$$. Squaring negates the sign, so

Alternatively, using the identities,

Both expressions are identical, so neither is greater — they are always equal.

We want integers $$a$$ and $$b$$ with $$a^2 - b^2 = 100$$. Using Identity 1C,

So we need two factors of $$100$$ that we can identify with $$(a+b)$$ and $$(a-b)$$. Note that $$a + b$$ and $$a - b$$ have the same parity (both even or both odd), and their product must be $$100$$.

Try $$a+b = 50$$ and $$a-b = 2$$ (both even). Adding, $$2a = 52$$, so $$a = 26$$ and $$b = 24$$.

Verify: $$26^2 - 24^2 = 676 - 576 = 100.\ \checkmark$$

So $$\boxed{100 = 26^2 - 24^2}$$.

Other solutions exist too (e.g. $$a + b = 100$$ and $$a - b = 1$$ gives non-integers; but $$a + b = 10,\ a - b = 10$$ gives $$a = 10,\ b = 0$$, i.e. $$100 = 10^2 - 0^2$$).

Decompose each number as a convenient sum or difference and apply Identity 1A / 1B.

$$406^2$$. $$406 = 400 + 6$$; use $$(a+b)^2 = a^2 + 2ab + b^2$$:

$$72^2$$. $$72 = 70 + 2$$:

$$145^2$$. $$145 = 150 - 5$$; use $$(a-b)^2 = a^2 - 2ab + b^2$$:

$$1097^2$$. $$1097 = 1100 - 3$$:

$$124^2$$. $$124 = 120 + 4$$:

Both patterns come from the identities

which were derived using only the distributive property. That property holds for all real numbers — counting numbers, negative integers, fractions, and beyond — so the identities work for any values.

Check with negative integers, $$a = -3,\ b = 2$$:

Check with fractions, $$a = \dfrac{1}{2},\ b = \dfrac{1}{3}$$:

So both patterns hold for negative integers and for fractions, not just counting numbers.

Counting circles in the given figures gives $$3,\ 8,\ 15$$ for Steps $$1, 2, 3$$. The chapter shows four ways to see the pattern; all give the same answer.

Method 1 — big square minus corner. Step $$k$$ looks like a $$(k+1)\times(k+1)$$ square array of circles with the top-right corner removed. Number of circles: $$(k+1)^2 - 1$$.

Method 2 — a $$k\times k$$ block plus an extra strip of $$2k$$. Number of circles: $$k^2 + 2k$$.

Method 3 — a $$k\times(k+1)$$ block plus $$k$$ extra. Number of circles: $$k(k+1) + k = k^2 + 2k$$.

Method 4 — a $$k \times (k+2)$$ rectangle. Number of circles: $$k(k+2)$$.

Simplify each expression:

So Step $$k$$ has $$\boxed{k^2 + 2k}$$ circles.

Next figure (Step 4). Number of circles $$= 4^2 + 2\cdot 4 = 24$$. Draw a $$4\times 6$$ array of circles.

Step 10. $$10^2 + 2\cdot 10 = 120$$ circles.

Substitute $$k = 15$$ into the formula:

Counting from the picture, the number of tiles in Steps $$1, 2, 3$$ is $$8,\ 12,\ 16$$. The tiles form a hollow border: a bigger square minus a smaller inner square.

Reading the picture (the chapter also lists this in a table),

• Step 1: $$3^2 - 1^2 = 9 - 1 = 8$$.
• Step 2: $$4^2 - 2^2 = 16 - 4 = 12$$.
• Step 3: $$5^2 - 3^2 = 25 - 9 = 16$$.

So each Step $$n$$ has the shape of a $$(n+2)\times(n+2)$$ square with an inner $$n \times n$$ square removed.

Using the pattern $$(n+2)^2 - n^2$$:

Step 4:

Step 10:

There are several ways to arrive at the same expression.

Method 1 (outer square minus inner square). The tiles form a $$(n+2)\times(n+2)$$ frame with an $$n \times n$$ hole. So

Using Identity 1C, $$(n+2)^2 - n^2 = ((n+2)+n)((n+2)-n) = (2n+2)\cdot 2 = 4n + 4$$.

Method 2 (four sides of the frame). The frame has four sides, each with $$(n+1)$$ tiles (counting one corner per side, not both), so

Method 3 (two horizontal rows + two vertical columns minus 4 corners). Top row: $$n+2$$ tiles; bottom row: $$n+2$$; left column (without corners): $$n$$; right column (without corners): $$n$$. Total:

All three methods give

Tadang's method looks at the picture as a large $$(m+n) \times (m+n)$$ square from which four $$m \times n$$ rectangles are removed:

Yusuf's method looks at the shaded region directly and identifies it as a square of side $$(n - m)$$:

Both methods must give the same area, so

(We verify this algebraically in the next question.) So the shaded area is $$(n - m)^2$$.

Left side. Using Identity 1A on $$(m+n)^2$$,

Right side. Using Identity 1B on $$(n-m)^2$$,

Both sides simplify to $$m^2 - 2mn + n^2$$, so the identity holds.

The chapter shows three students' methods for the same shaded region. All give the same expression.

Anusha's method. The shaded region equals a big square $$ABCD$$ of side $$x$$ minus a rectangle $$EFGH$$ of dimensions $$x \times y$$:

Vaishnavi's method. Enclose the picture in a bigger rectangle $$PQSR$$ with $$QS = x + 2y$$ and side $$x$$; subtract the three identical $$x \times y$$ rectangles:

Aditya's method. The shaded region is twice the area of rectangle $$JKML$$ with $$JK = \dfrac{x-y}{2}$$ and $$KM = x$$:

All three methods give the same area:

Expression 1: $$x^2 - xy$$ (already in simplest form).

Expression 2:

Expression 3:

All three simplify to the same expression $$x^2 - xy$$, confirming they are equivalent.

Substituting $$x = 8,\ y = 3$$:

The dashed region is a rectangle of dimensions $$p \times s$$ with two strips of width $$r$$ removed — one along one side and one along the other — which meet in a common $$r \times r$$ corner.

Method 1 (inclusion–exclusion). Take the outer $$p \times s$$ rectangle and subtract the two strips; then add back the corner counted twice:

Method 2 (product form). The dashed region is itself a rectangle of dimensions $$(p - r) \times (s - r)$$, so

which matches Method 1 by direct expansion.

Substituting $$p = 6,\ r = 3.5,\ s = 9$$:

Check: $$ps - pr - sr + r^2 = 54 - 21 - 31.5 + 12.25 = 13.75.\ \checkmark$$

Try a few $$2\times 2$$ squares from the calendar:

• $$\{3,4,10,11\}$$: diagonals $$3\times 11 = 33$$ and $$4\times 10 = 40$$. Difference $$= 7$$.
• $$\{4,5,11,12\}$$: $$4\times 12 = 48$$ and $$5\times 11 = 55$$. Difference $$= 7$$.
• $$\{17,18,24,25\}$$: $$17\times 25 = 425$$ and $$18\times 24 = 432$$. Difference $$= 7$$.

Observation. The two diagonal products always differ by $$7$$ (the anti-diagonal product is $$7$$ more than the main-diagonal product).

Explanation. Using the hint, label the block as

The two diagonal products are $$a(a+8)$$ and $$(a+1)(a+7)$$. Expand each:

Subtracting,

So the two diagonal products always differ by exactly $$7$$, regardless of which $$2 \times 2$$ block we pick — the reason being that each row of a calendar shifts by exactly $$7$$.

Write the two numbers as

where $$a$$ and $$b$$ are non-negative integers.

Sum.

So the sum leaves remainder $$1$$ when divided by $$7$$.

Difference.

So the difference leaves remainder $$2$$ when divided by $$7$$. (If we instead compute $$N_1 - N_2$$, the remainder still corresponds to $$-2 \equiv 5 \pmod{7}$$; but in the usual sense of “difference” = larger $$-$$ smaller, the remainder is $$2$$.)

Product. Use Identity 1:

Group multiples of $$7$$:

So the product leaves remainder $$1$$ when divided by $$7$$.

(Shortcut: modulo $$7$$, sum $$\equiv 3 + 5 = 8 \equiv 1$$; difference $$\equiv 5 - 3 = 2$$; product $$\equiv 3\times 5 = 15 \equiv 1$$.)

Try a few triples:

• $$3, 4, 5$$: $$4^2 - 3\times 5 = 16 - 15 = 1$$.
• $$6, 7, 8$$: $$7^2 - 6\times 8 = 49 - 48 = 1$$.
• $$9, 10, 11$$: $$10^2 - 9\times 11 = 100 - 99 = 1$$.

Pattern. The result is always $$1$$.

Algebraic statement. Let the three consecutive numbers be $$n-1$$, $$n$$, $$n+1$$. Then

Check by expanding. Using Identity 1C on the right-hand product:

So

The equation holds for every $$n$$ — it is a true identity.

Let the two numbers be $$a$$ and $$b$$.

Step 1: Add them: $$a + b$$.

Step 2: Half of the sum: $$\dfrac{a+b}{2}$$.

Step 3: Multiply Step 1 by Step 2:

So the algebraic expression is $$\dfrac{(a+b)^2}{2}$$, which is exactly “half of the square of the sum of the two numbers.”

Reading the figure: the two green squares (each of side $$g$$) are placed side by side inside a rectangular park, with a walking path of width $$w$$ running around them and between them.

Length of the park (horizontal): a $$w$$-strip on the left, then square of side $$g$$, then a $$w$$-strip between squares, then another square of side $$g$$, then a $$w$$-strip on the right. In total,

Wait — the reference solution uses $$4w$$; the plan in the PDF actually shows the path around and between the squares, with a middle path of width $$w$$ too. Following the standard NCERT figure (path $$w$$ around all sides and between): $$w + g + w + g + w = 2g + 3w$$. However, the NCERT solution takes the middle strip to be $$2w$$ wide (a $$w$$-wide path on each of the two squares). Following that layout,

Breadth of the park (vertical): $$w$$-strip above, square of side $$g$$, $$w$$-strip below,

Total park area:

Area of the two green squares: $$2g^2$$.

Area to be tiled (path):

Factor: $$8w(g + w)$$ sq. ft.