CBSE Class 8 Mathematics We Distribute Yet Things Multiply Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Properties of multiplication.
(i) State the commutative, the associative and the distributive property of multiplication, giving one numerical example of each.
(ii) Using the distributive property twice, expand (a + m)(b + n) step by step.
(iii) Hence write by how much the product ab increases when both a and b are increased by 1, and check your answer for a = 23, b = 27.
Answer:
(i) Commutative: a × b = b × a. Example: 6 × 9 = 9 × 6 = 54.
Associative: (a × b) × c = a × (b × c). Example: (2 × 5) × 7 = 2 × (5 × 7) = 70.
Distributive: a(b + c) = ab + ac. Example: 7(10 + 3) = 70 + 21 = 91.
(ii) Treat (a + m) as one term and distribute it over b + n:
(a + m)(b + n) = (a + m)b + (a + m)n
= ab + mb + an + mn.
So the product is the sum of the products of each term of the first bracket with each term of the second.
(iii) Put m = 1 and n = 1: (a + 1)(b + 1) = ab + b + a + 1.
The product increases by a + b + 1.
Check: 23 + 27 + 1 = 51. Directly, 24 × 28 = 672 and 23 × 27 = 621; 672 − 621 = 51. Verified.
In simple words: Distributing twice gives four products; with both numbers raised by 1, the product grows by a + b + 1.

Teacher's Note:
1. The distributive property is used once for each term of the second bracket.
2. Always verify a general result with numbers.

Question 2. The figure shows a square of side 74 units cut into four parts by splitting each side as 70 + 4.
(i) Write the area of each of the four parts I, II, III and IV.
(ii) Add the four areas to find 74².
(iii) Write the general identity for (a + b)² that this picture proves, and use it to find 104².
Answer:
(i) Part I: 70 × 70 = 4900 sq. units; Part II: 4 × 70 = 280 sq. units;
Part III: 70 × 4 = 280 sq. units; Part IV: 4 × 4 = 16 sq. units.
(ii) 74² = 4900 + 280 + 280 + 16 = 5476 sq. units.
The two rectangles are equal, so together they give 2 × (70 × 4) = 560.
(iii) Splitting a square of side a + b the same way gives a square a², a square b² and two rectangles of area ab each. So (a + b)² = a² + 2ab + b².
For 104², take a = 100, b = 4: 104² = 10000 + 800 + 16 = 10816.
In simple words: The big square is made of two squares and two equal rectangles, which is exactly a² + 2ab + b².

Teacher's Note:
1. The picture is a proof of Identity 1A.
2. The two rectangles explain the 2ab term.

Question 3. Three students simplified expressions on the blackboard as shown. For each one, say whether there is a mistake, explain what went wrong, and write the correct expression.
(i) −2p(−4p + 3q) = −2p + 4p − 3q = 2p − 3q
(ii) 6w² + 5w = 11w²
(iii) (−t + 3)² = t² − 6t + 9
Answer:
(i) Mistake. The multiplier −2p was dropped and the signs were only changed. Every term must be multiplied by −2p:
−2p(−4p + 3q) = (−2p)(−4p) + (−2p)(3q) = 8p² − 6pq.
(ii) Mistake. 6w² and 5w are unlike terms (w × w against w), so they cannot be joined. The expression stays as 6w² + 5w.
Testing with w = 2: 24 + 10 = 34, while 11w² would give 44.
(iii) No mistake. (−t + 3)² = (3 − t)² = 9 − 6t + t² = t² − 6t + 9.
Check with t = 2: (−2 + 3)² = 1 and 4 − 12 + 9 = 1.
In simple words: Multiply every term by the full multiplier, add only like terms, and remember (−t + 3) = (3 − t).

Teacher's Note:
1. A negative times a negative gives a positive.
2. Substituting a number is a quick way to test working.

Question 4. The identity (a + b)(a − b) = a² − b².
(i) Expand (a + b)(a − b) using the distributive property and show how the middle terms cancel.
(ii) Use the identity to evaluate 96 × 104 and 57 × 63.
(iii) Sridharacharya wrote the same identity as a² = (a + b)(a − b) + b². Use this form to find 62².
Answer:
(i) (a + b)(a − b) = a(a − b) + b(a − b) = a² − ab + ba − b².
Since ba = ab, the terms −ab and +ab add to zero, leaving a² − b².
(ii) 96 × 104 = (100 − 4)(100 + 4) = 10000 − 16 = 9984.
57 × 63 = (60 − 3)(60 + 3) = 3600 − 9 = 3591.
In each case a is the number midway between the two factors.
(iii) Adding b² to both sides of (a + b)(a − b) = a² − b² gives a² = (a + b)(a − b) + b².
For 62² choose b = 2: 62² = 64 × 60 + 4 = 3840 + 4 = 3844.
In simple words: The middle terms cancel, so a product around a round number is a square minus a small square.

Teacher's Note:
1. Pick b so that a factor ends in 0.
2. Sridharacharya's rule is the same identity rearranged.

Question 5. A dot pattern grows as shown. At each step the dots form a square array from which a 2 × 2 block of dots has been removed at one corner.
(i) Count the dots in Steps 1, 2 and 3, and find the number of dots in Step 4.
(ii) Three students describe Step k differently: Ila writes (k + 2)² − 4, Balu writes k(k + 2) + 2k, and Chetan writes k(k + 4). Explain how each student could have seen the figure.
(iii) Show by expanding that all three expressions are equal, and use the simplified form to find the number of dots in Step 10.
Answer:
(i) Step 1 has 5 dots, Step 2 has 12 and Step 3 has 21.
Step 4 is a 6 × 6 array with the 2 × 2 corner block missing: 36 − 4 = 32 dots.
(ii) Ila sees the full square of (k + 2) rows and (k + 2) columns and takes away the 4 missing dots: (k + 2)² − 4.
Balu cuts the figure into a full lower block of k rows with (k + 2) dots each, and a top block of 2 rows with k dots each: k(k + 2) + 2k.
Chetan uses the difference of two squares: (k + 2)² − 2² = (k + 2 + 2)(k + 2 − 2) = k(k + 4).
(iii) (k + 2)² − 4 = k² + 4k + 4 − 4 = k² + 4k.
k(k + 2) + 2k = k² + 2k + 2k = k² + 4k.
k(k + 4) = k² + 4k.
All three simplify to k² + 4k, so they describe the same pattern.
Step 10: 10² + 4 × 10 = 100 + 40 = 140 dots.
In simple words: All three ways of counting give k² + 4k, which is 140 at Step 10.

Teacher's Note:
1. Different ways of seeing one figure lead to equal expressions.
2. Check Step 2: 4 + 8 = 12.

Question 6. In the square mosaic shown, four identical rectangular tiles of sides m and n (with n > m) are laid around a small shaded square.
(i) Tadang finds the shaded area by subtracting the four tiles from the big square. Write his expression.
(ii) Yusuf says the shaded part is itself a square of side n − m. Write his expression.
(iii) Show by expanding that the two expressions are equal, and find the shaded area when m = 3 units and n = 8 units.
Answer:
(i) The outer figure is a square of side m + n, so its area is (m + n)². The four tiles have area mn each, 4mn in all.
Tadang: shaded area = (m + n)² − 4mn.
(ii) Along a line across the middle piece, the length left over is n − m, both across and down. So it is a square of side n − m.
Yusuf: shaded area = (n − m)².
(iii) (m + n)² − 4mn = m² + 2mn + n² − 4mn = m² − 2mn + n² = (n − m)².
Both students are right; they looked at the same region in two ways.
For m = 3, n = 8: Tadang gets 11² − 4 × 3 × 8 = 121 − 96 = 25; Yusuf gets (8 − 3)² = 5² = 25.
The shaded area is 25 square units.
In simple words: Big square minus four tiles equals the small square of side n − m.

Teacher's Note:
1. Two correct methods must give the same answer.
2. The figure is a picture proof of (m + n)² − 4mn = (n − m)².

Question 7. Fast multiplication using the distributive property.
(i) Taking a three-digit number cba, show why cba × 11 can be written in one line by adding neighbouring digits. Use cba × 11 = cba × 10 + cba.
(ii) Use the one-line rule to find 6253 × 11.
(iii) Find 4786 × 1001 and 7625 × 999 by splitting the multiplier suitably. Name the ancient Indian mathematician who first stated this “multiplication by parts” rule.
Answer:
(i) cba × 11 = cba × (10 + 1) = cba × 10 + cba. Write the shifted number cba0 above cba and add column by column:
units digit = a, tens digit = a + b, hundreds digit = b + c, thousands digit = c.
So the answer keeps the end digits of the number and puts the sums of neighbouring digits between them, carrying over when a sum is more than 9.
(ii) End digits 6 and 3; neighbour sums 6 + 2 = 8, 2 + 5 = 7, 5 + 3 = 8. So 6253 × 11 = 68783.
Check: 62530 + 6253 = 68783.
(iii) 4786 × 1001 = 4786 × (1000 + 1) = 4786000 + 4786 = 4790786.
7625 × 999 = 7625 × (1000 − 1) = 7625000 − 7625 = 7617375.
Each shortcut is the distributive property at work. Brahmagupta first stated it explicitly, calling multiplication by parts khaṇḍa-guṇanam; he described such fast methods as iṣṭa-guṇana.
In simple words: Splitting 11, 1001 or 999 into easy parts turns each product into a quick addition or subtraction.

Teacher's Note:
1. Watch for carries when a neighbour sum is 10 or more.
2. 999 = 1000 − 1 uses distributivity over subtraction.

Teacher's Note:
1. Name the dates with one letter to prove a general fact.
2. Examples suggest a pattern; algebra proves it.

Question 9. Prove the following two claims using algebra, and verify each with one example.
(i) The square of every even number is a multiple of 4.
(ii) The square of every odd number is 1 more than a multiple of 8.
(iii) Using (ii), explain why no odd number has a square that leaves remainder 5 on division by 8.
Answer:
(i) Every even number is 2n for some whole number n. Then (2n)² = 4n², which is 4 times a whole number, so it is a multiple of 4.
Example: 14 = 2 × 7 and 14² = 196 = 4 × 49.
(ii) Every odd number is 2n + 1. Then (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.
n and n + 1 are consecutive numbers, so one of them is even and n(n + 1) = 2k for some whole number k.
So (2n + 1)² = 4 × 2k + 1 = 8k + 1, which is 1 more than a multiple of 8.
Examples: 9² = 81 = 8 × 10 + 1 and 7² = 49 = 8 × 6 + 1.
(iii) By (ii), the square of an odd number always leaves remainder 1 on division by 8. A number cannot leave two different remainders on division by 8, so no odd square can leave remainder 5.
(An even square is a multiple of 4, so it leaves remainder 0 or 4, and never 5 either.)
In simple words: Even squares are 4n², and odd squares are 8k + 1, so remainder 5 is impossible.

Teacher's Note:
1. Write even numbers as 2n and odd numbers as 2n + 1.
2. The product of two consecutive numbers is always even.

Question 10. Riya claims: “If the sum of two numbers is fixed, their product is largest when the two numbers are equal.”
(i) Test the claim by writing three pairs of whole numbers whose sum is 20 and finding their products.
(ii) Let the fixed sum be S, and write the two numbers as \( \frac{S}{2} + d \) and \( \frac{S}{2} - d \). Using the identity (a + b)(a − b) = a² − b², write their product in terms of S and d, and explain why the claim is true.
(iii) Hence decide, without multiplying fully, which is larger: 49 × 51 or 48 × 52, and by how much.
Answer:
(i) 10 + 10: 10 × 10 = 100; 9 + 11: 9 × 11 = 99; 6 + 14: 6 × 14 = 84.
The product falls as the numbers move apart, and it is largest for the equal pair.
(ii) Any two numbers with sum S can be written as \( \frac{S}{2} + d \) and \( \frac{S}{2} - d \), where d is half the gap between them. Using the identity with \( a = \frac{S}{2} \) and b = d:
\( \left(\frac{S}{2} + d\right)\left(\frac{S}{2} - d\right) = \frac{S^2}{4} - d^2 \)
\( \frac{S^2}{4} \) is fixed and d² is never negative. So the product is largest exactly when d = 0, that is, when the two numbers are equal. Riya’s claim is correct.
(iii) Both pairs add up to 100, so \( \frac{S}{2} = 50 \).
49 × 51 = 50² − 1² = 2500 − 1 = 2499 (here d = 1).
48 × 52 = 50² − 2² = 2500 − 4 = 2496 (here d = 2).
So 49 × 51 is larger, by 3.
In simple words: The product is (half the sum)² minus (half the gap)², so it is biggest when the gap is zero.

Teacher's Note:
1. d² can never be negative, so subtracting it can only reduce the product.
2. Closer factors with the same sum give a bigger product.

Case Study Based Questions

Question 11. Mental arithmetic at the stationery counter. At a small stationery shop in Dhauli, the cashier never uses a calculator. When a customer buys 48 pens costing ₹11 each, he says “₹480 and ₹48, that is ₹528” almost at once. He explains his trick to the children waiting in the queue: “Eleven is just ten and one, so I multiply by ten, then add the number once more.” For a price of ₹99 he does the opposite — he multiplies by 100 and then takes the number away once. The children realise that the shopkeeper is using a property of multiplication they have just learnt in school.
(i) Write 48 × 11 as 48 × (10 + 1) and find the total.
(ii) Using the same idea, find the cost of 236 erasers at ₹11 each.
(iii) Find 125 × 99 by writing 99 as 100 − 1.
(iv) Name the property of multiplication that makes both shortcuts correct.
Answer:
(i) 48 × 11 = 48 × 10 + 48 × 1 = 480 + 48 = 528. The bill is ₹528.
(ii) 236 × 11 = 2360 + 236 = 2596. The cost is ₹2596.
(iii) 125 × 99 = 125 × (100 − 1) = 12500 − 125 = 12375.
(iv) The distributive property of multiplication over addition (and over subtraction), a(b + c) = ab + ac. The multiplier is broken into parts and the results are added or subtracted.
In simple words: Split 11 as 10 + 1 and 99 as 100 − 1, then multiply each part and combine.

Teacher's Note:
1. Multiplying by 10 or 100 is just adding zeros.
2. Subtracting works because a(b − c) = ab − ac.

Question 12. A square plot made bigger. A village panchayat owns a square plot of land of side 40 m. It buys a strip of land 5 m wide along the eastern edge and another strip 5 m wide along the northern edge, together with the small square corner piece where the two strips meet. After the purchase the plot is again a square, now of side 45 m. The secretary draws the new plot and marks the four pieces: the old square, the two strips and the corner square.
(i) Write the area of the old square plot.
(ii) Write the area of each 40 m × 5 m strip and of the corner square.
(iii) Add the four areas to find the area of the new plot.
(iv) By how many square metres has the area increased?
Answer:
(i) Old plot = 40 × 40 = 1600 m².
(ii) Each strip = 40 × 5 = 200 m²; corner square = 5 × 5 = 25 m².
(iii) New plot = 1600 + 200 + 200 + 25 = 2025 m², which is indeed 45².
(iv) Increase = 2025 − 1600 = 425 m². This equals 2 × (40 × 5) + 5² = 400 + 25, exactly the extra terms in (a + b)² = a² + 2ab + b².
In simple words: The new square is the old square plus two strips and a small corner square.

Teacher's Note:
1. This is Identity 1A drawn on land.
2. The increase is not just 5 × 45 or 2 × 5 × 40; the corner counts too.

Question 13. A tiled patio. A hotel is laying square tiles of side 1 m on a patio. The patio is a square of side 30 m. In one corner of the patio a square sandpit of side 20 m is left untiled for the children’s play area. The contractor wants to order the correct number of tiles. His assistant says, “You need not subtract the big numbers. Just multiply the sum of the two sides by their difference.”
(i) How many tiles would cover the whole patio?
(ii) How many tiles are saved because of the sandpit?
(iii) Find the number of tiles actually needed by using the assistant’s method, and show it gives the same answer.
(iv) Which identity of the chapter is the assistant using?
Answer:
(i) Whole patio: 30 × 30 = 900 tiles.
(ii) Sandpit: 20 × 20 = 400 tiles saved.
(iii) Assistant’s method: (30 + 20)(30 − 20) = 50 × 10 = 500 tiles. By subtraction, 900 − 400 = 500. The two agree.
(iv) He is using (a + b)(a − b) = a² − b² (Identity 1C) with a = 30 and b = 20.
In simple words: Tiles needed = 30² − 20² = (30 + 20)(30 − 20) = 500.

Teacher's Note:
1. Each 1 m tile covers 1 m², so tiles = area.
2. Difference of squares avoids large squares.

Question 14. Mending mistakes on the blackboard. Before a class test, a teacher writes four answers given by different students on the blackboard and asks the class to check each one. She reminds them, “A wrong step is not a disaster; finding it and mending it is real mathematics.” The four answers on the board are:
(i) 4(x + 3) = 4x + 3
(ii) (a + 5)² = a² + 25
(iii) 3y² + 2y = 5y³
(iv) (x + 2)(x + 6) = x² + 8x + 12
For each answer, state whether it is right or wrong, explain briefly what went wrong where it is wrong, and write the correct expression.
Answer:
(i) Wrong. The 4 was multiplied only with x, not with 3. Correct: 4(x + 3) = 4x + 12.
(ii) Wrong. Each term was squared and the middle term was left out. Correct: (a + 5)² = a² + 10a + 25.
(iii) Wrong. 3y² and 2y are unlike terms, so they cannot be combined, and powers are never added like this. Correct: the expression stays as 3y² + 2y.
(iv) Right. (x + 2)(x + 6) = x² + 6x + 2x + 12 = x² + 8x + 12.
In simple words: Multiply every term, keep the middle term of a square, and add only like terms.

Teacher's Note:
1. Testing with a number such as x = 1 quickly shows a wrong answer.
2. These are the three most common slips of the chapter.

Question 15. A picture in a frame. A photograph measuring 18 cm by 12 cm is put into a wooden frame. The frame is of the same width all around the photograph. For a frame of width 3 cm, the shopkeeper must know how much wood surface is visible, because he charges for the area of the frame itself and not for the photograph.
(i) Write the outer length and the outer width of the framed picture when the frame is 3 cm wide.
(ii) Find the outer area and hence the area of the wooden frame.
(iii) If the frame has width w cm, show that the area of the frame is 4w² + 60w square centimetres.
(iv) Use this expression to find the area of the frame when w = 2, and check your answer by subtracting areas.
Answer:
(i) The frame adds 3 cm on both sides, so the outer length is 18 + 6 = 24 cm and the outer width is 12 + 6 = 18 cm.
(ii) Outer area = 24 × 18 = 432 cm². Photograph = 18 × 12 = 216 cm². Frame = 432 − 216 = 216 cm².
(iii) Outer area = (18 + 2w)(12 + 2w) = 216 + 36w + 24w + 4w². Subtracting the photograph (216 cm²), area of frame = 4w² + 60w cm².
(iv) For w = 2: 4(4) + 60(2) = 16 + 120 = 136 cm². Check: 22 × 16 = 352 and 352 − 216 = 136 cm².
In simple words: Frame area = outer rectangle − photograph = 4w² + 60w.

Teacher's Note:
1. The width is added on both sides, so add 2w.
2. Expand with Identity 1, then subtract.

Question 16. A garden path border. A rectangular lawn in a school garden measures 15 m by 10 m. A path of uniform width is to be laid all around the lawn, just outside it, and the path is to be covered with gravel. The gardener wants the area of the path for two different widths before he orders the gravel. One helper plans to subtract the lawn from the big rectangle; another plans to cut the path into four rectangular strips and add them.
(i) For a path 2 m wide, write the outer length and the outer breadth.
(ii) Find the area of the path by the first helper’s method.
(iii) Find the same area by the second helper’s method, showing the four strips.
(iv) Show that for a path of width w m the area is 4w² + 50w square metres, and use it when w = 3.
Answer:
(i) The path adds 2 m on each side, so the outer rectangle is 19 m by 14 m.
(ii) Path = 19 × 14 − 15 × 10 = 266 − 150 = 116 m².
(iii) Top and bottom strips: 19 × 2 = 38 m² each. Two side strips (beside the lawn only): 10 × 2 = 20 m² each.
Total = 38 + 38 + 20 + 20 = 116 m², the same answer.
(iv) For width w: (15 + 2w)(10 + 2w) − 150 = 150 + 30w + 20w + 4w² − 150 = 4w² + 50w.
For w = 3: 36 + 150 = 186 m².
In simple words: Both methods give the same path area; in general it is 4w² + 50w.

Teacher's Note:
1. In the strip method the corners must be counted only once.
2. Check (iv) with w = 2: 16 + 100 = 116.

Question 17. A multiplication grid on the wall. A multiplication table from 1 to 10 is painted on a classroom wall. Each number inside the grid is the product of its row number and its column number. A teacher places a cardboard window over the grid so that exactly three numbers of one row are visible. If the middle visible number is in row p and column q, it stands for pq.
(i) Write expressions for the numbers to the left and to the right of the middle number.
(ii) Write the product of these two neighbours as a single expression and simplify it using a suitable identity.
(iii) Show that this product is always p² less than (pq)².
(iv) Check your result for p = 4 and q = 6.
Answer:
(i) Left: p(q − 1); right: p(q + 1).
(ii) Product = p(q − 1) × p(q + 1) = p²(q − 1)(q + 1) = p²(q² − 1), using (a + b)(a − b) = a² − b².
(iii) p²(q² − 1) = p²q² − p² = (pq)² − p². So the product of the two neighbours is always p² less than the square of the middle number.
(iv) For p = 4, q = 6 the numbers are 20, 24, 28. 20 × 28 = 560, and 24² − 4² = 576 − 16 = 560. Verified.
In simple words: The neighbours multiply to (middle number)² − p².

Teacher's Note:
1. Numbers in one row go up by p each time.
2. Taking out p² first makes the identity easy to use.

Question 18. A growing border of tiles. A designer makes square patios that grow step by step. At Step n the patio is a square of side (n + 2) tiles. The inner n × n square is laid with plain cream tiles and the single ring of tiles around it is laid with blue tiles. The designer needs to order blue tiles separately for each patio she builds.
(i) How many tiles in all are used at Step 3, and how many of them are blue?
(ii) Write an expression for the number of blue tiles at Step n using the two squares.
(iii) Simplify the expression and check it against your answer in (i).
(iv) For which step will the designer need exactly 100 blue tiles?
Answer:
(i) At Step 3 the side is 3 + 2 = 5 tiles, so 25 tiles in all. The inner square uses 3 × 3 = 9 cream tiles, so 25 − 9 = 16 tiles are blue.
(ii) Blue tiles at Step n = (n + 2)² − n².
(iii) (n + 2)² − n² = n² + 4n + 4 − n² = 4n + 4. For n = 3: 12 + 4 = 16, matching (i).
(Also (n + 2)² − n² = (n + 2 + n)(n + 2 − n) = 2(2n + 2), the same thing.)
(iv) 4n + 4 = 100 gives 4n = 96, so n = 24. She needs exactly 100 blue tiles at Step 24.
In simple words: Blue tiles = big square − inner square = 4n + 4.

Teacher's Note:
1. The ring is the difference of two squares.
2. 4n + 4 also counts four sides of n plus four corners.

Question 19. The cashier’s squaring trick. An old cashier at a cloth shop computes squares in his head using a rule from Sridharacharya: to find a², pick a small number b that makes a + b or a − b a round number, then use a² = (a + b)(a − b) + b². He also claims that among two multiplications whose two factors add up to the same total, the one whose factors are closer together always gives the bigger product.
(i) Use the cashier’s rule to find 49², taking b = 1.
(ii) Use it to find 203², choosing a suitable b.
(iii) Explain, using the identity (a + b)(a − b) = a² − b², why the cashier’s rule is correct.
(iv) Both 37 × 43 and 38 × 42 have factors adding to 80. Find both products using the identity and say which is larger.
Answer:
(i) 49² = (49 + 1)(49 − 1) + 1² = 50 × 48 + 1 = 2400 + 1 = 2401.
(ii) Take b = 3, so that 203 − 3 = 200: 203² = 206 × 200 + 9 = 41200 + 9 = 41209.
(iii) Adding b² to both sides of (a + b)(a − b) = a² − b² gives a² = (a + b)(a − b) + b². So the rule is just the difference-of-squares identity written the other way round.
(iv) 37 × 43 = (40 − 3)(40 + 3) = 1600 − 9 = 1591 and 38 × 42 = (40 − 2)(40 + 2) = 1600 − 4 = 1596.
So 38 × 42 is larger, by 5. This supports the cashier’s claim: the closer the factors, the smaller the square taken away from 1600.
In simple words: Squares are found by moving to a round number, and closer factors with the same sum give a bigger product.

Teacher's Note:
1. Choose b to make one factor end in zeros.
2. Both products are 40² minus a small square.

Question 20. Reshaping a square field. A farmer owns a square field of side x metres. He agrees to give up a 5 m strip along one side to widen a village road, and in exchange he is given a 5 m strip along the side at right angles to it. His field is now a rectangle. The farmer believes that since he lost 5 m on one side and gained 5 m on the other, the area of his field cannot have changed.
(i) Write an expression for the area of the new rectangular field in terms of x.
(ii) Using an identity, find by how much the area has changed.
(iii) Does the change depend on the size of the original field? Justify your answer with algebra.
(iv) The farmer now says that if instead one side were increased by 5 m and the other by 4 m, the area would certainly grow. Is he right for every positive x? Justify.
Answer:
(i) The new field measures (x + 5) m by (x − 5) m, so its area is (x + 5)(x − 5) m².
(ii) By (a + b)(a − b) = a² − b², the new area is x² − 25. The old area was x², so the area has decreased by 25 m². The farmer is wrong.
(iii) No. The change is −25, which contains no x, because +5x and −5x cancel in the expansion. A field of side 60 m loses exactly 25 m², just like a field of side 200 m: 65 × 55 = 3575 while 60² = 3600.
(iv) Yes, this time he is right. (x + 5)(x + 4) = x² + 9x + 20, which is 9x + 20 more than x². For every positive x, 9x + 20 is positive, so the area always grows.
In simple words: Gaining and losing 5 m loses 25 m² every time, but adding to both sides always increases the area.

Teacher's Note:
1. Here x must be more than 5 for the field to exist.
2. Equal gain and loss on the sides does not mean equal area.