Chapter 10 · Exercise 1Textbook pages 180–181

Exercise 1 – Solutions അളവുകളുടെ ബീജഗണിതം

Three questions: rectangles with sides that differ by 1 cm, an open box folded from a card, and rectangles made with a 1 m rope. In each one you turn a measurement into a rule in x and then use it. Try each part in your notebook first, then step through the animation.

📖 What you need (pages 5 and 6 of this guide)

Question 1 · Sides that differ by 1 cm

Q1. ഒരു വശത്തിന്റെ നീളം മറ്റേ വശത്തിന്റെ നീളത്തേക്കാൾ 1 സെന്റിമീറ്റർ കുറവായ ചതുരങ്ങൾ – take rectangles in which one side is 1 cm less than the other. Take the shorter side as x cm.

  1. Take their perimeters as p(x) cm and write the relation between x and p(x) as an equation.
  2. Take their areas as a(x) sq cm and write the relation between x and a(x) as an equation.
  3. Find p(1), p(2), p(3), p(4), p(5). Do you see any pattern?
  4. Find a(1), a(2), a(3), a(4), a(5). Do you see any pattern?

If the shorter side is x, the longer side is x + 1. For the patterns, subtract each value from the next one.

🎬 Solution animation – (i) and (ii)
🎬 Solution animation – (iii) and (iv): the tables and their patterns
🧪 Try other rectangles
Answer: (i) p(x) = 2(x + x + 1) = 4x + 2. (ii) a(x) = x(x + 1) = x² + x.
(iii) p(1) = 6, p(2) = 10, p(3) = 14, p(4) = 18, p(5) = 22: each is 4 more than the one before (an arithmetic sequence; each is 2 more than a multiple of 4) – making x 1 cm bigger adds 1 cm to each of the four sides.
(iv) a(1) = 2, a(2) = 6, a(3) = 12, a(4) = 20, a(5) = 30: they are products of two consecutive numbers (1 × 2, 2 × 3, 3 × 4, …); the increases are 4, 6, 8, 10 – each 2 more than the last; all are even (twice 1, 3, 6, 10, 15).
💡 Why this matters

The first-degree rule 4x + 2 grows by the same step every time; the second-degree rule x² + x grows by steps that themselves grow. Looking at differences tells you which kind of rule you are dealing with – a trick used on data from experiments.

Question 2 · An open box from a card

Q2. ഒരു ചതുരത്തിന്റെ നാലു മൂലകളിൽ നിന്നും ഒരേ വലുപ്പമുള്ള ചെറു സമചതുരങ്ങൾ വെട്ടിമാറ്റി, മേലോട്ട് മടക്കി, ഒരു പെട്ടി ഉണ്ടാക്കുന്നു. Equal small squares are cut from the four corners of a rectangle 7 cm long and 5 cm wide, and the sides are folded up to make a box (as in the figure).

  1. Take the side of the square cut out as x cm and write the three measures of the box.
  2. Take the volume of the box as v(x) cubic cm and write the relation between x and v(x) as an equation.
  3. Find v(½), v(1), v(1½).

A square of side x is cut from both ends of each side, so each side loses 2x. The height of the box is the side of the square, x.

🎬 Solution animation – cut, fold, measure
🧪 Change the cut – watch the box and its volume

A tiny cut gives a flat tray; a big cut gives a tall, thin box. Somewhere in between the volume is biggest – try to find it.

Answer: (i) length 7 − 2x cm, width 5 − 2x cm, height x cm (x must be between 0 and 2½).
(ii) v(x) = x(7 − 2x)(5 − 2x) = 4x³ − 24x² + 35x.
(iii) v(½) = ½ × 6 × 4 = 12; v(1) = 1 × 5 × 3 = 15; v(1½) = 1½ × 4 × 2 = 12 cubic cm. (Check with the polynomial: v(1) = 4 − 24 + 35 = 15 ✓.)
💡 Why this matters

Packaging factories make boxes exactly this way. The rule v(x) lets them choose the cut that wastes the least card for the most space – here a cut of about 0.96 cm gives the biggest box, just over 15 cubic cm.

Question 3 · Rectangles from a 1 m rope

Q3. ഒരു മീറ്റർ നീളമുള്ള കയറുകൊണ്ട് ഉണ്ടാക്കാവുന്ന ചതുരങ്ങൾ – rectangles are made with a rope 1 metre long. Take the length of one side as x cm and the area inside as a(x) sq cm.

  1. Write the relation between x and a(x) as an equation.
  2. Why are a(10) and a(40) the same number?
  3. When two different numbers are taken as x and give the same a(x), what must be the relation between the numbers?

1 m = 100 cm is the perimeter, so two neighbouring sides add up to 50 cm. If one side is 10, what is the other?

🎬 Solution animation
🧪 Pull the rope into different rectangles
Answer: (i) The other side is 50 − x, so a(x) = x(50 − x) = 50x − x².
(ii) a(10) = 10 × 40 = 400 and a(40) = 40 × 10 = 400: when one side is 10 cm the other is 40 cm, and when one side is 40 cm the other is 10 cm – it is the same rectangle, turned around.
(iii) The two numbers must add up to 50 (x + y = 50): each is the other side of the same rectangle. Algebra: 50x − x² = 50y − y² gives 50(x − y) = (x − y)(x + y), and since x ≠ y, x + y = 50.
💡 Why this matters

A farmer with a fixed length of fencing wants the biggest field. The table of a(x) is symmetric about x = 25, so the biggest area, 625 sq cm, comes from the square. Fixed perimeter, biggest area: the square wins.

Check yourself

Next: Polynomials and degree ▶