Chapter 2 · Practice set 1Textbook pages 25–26

Practice Set 1 – Animated Solutions ചിന്തിക്കാം

Four questions from the "think" box on pages 25–26, each turning an area into a side: squares of area 7, 10 and 11, the fan of right triangles, and fractions trapped between √2 and √3. Read each question, try it in your notebook (use the hint if you are stuck), then press Next ▶ and watch the construction and the reasoning grow. The page ends with the textbook's side box on drawing √x for any x.

📖 Tools you will use

Pythagoras (page 3): square on the hypotenuse = squares on the two perpendicular sides; so (third side)² = (hypotenuse)² − (side)².
Odd numbers are differences of consecutive squares (page 4): 2n + 1 = (n + 1)² − n², e.g. 7 = 16 − 9, 11 = 36 − 25.
Squeezing (page 6): find the decimals whose squares are just below and just above the area.
Drawing: a perpendicular at a point and a compass arc give a right triangle with a chosen hypotenuse and side.

Question 1 · Squares of area 7 and 11

Q1In Class 8 you saw that every odd number is the difference of two perfect squares. Using this, draw squares of area 7 sq cm and 11 sq cm. What are the lengths of their sides?

(ഏത് ഒറ്റസംഖ്യയെയും രണ്ടു പൂർണവർഗങ്ങളുടെ വ്യത്യാസമായി എഴുതാമെന്ന് എട്ടാം ക്ലാസിൽ കണ്ടല്ലോ. ഇതുപയോഗിച്ച്, 7 ചതുരശ്രസെന്റിമീറ്റർ, 11 ചതുരശ്രസെന്റിമീറ്റർ എന്നീ പരപ്പളവുകളുള്ള സമചതുരങ്ങൾ വരയ്ക്കുക. അവയുടെ വശങ്ങളുടെ നീളം എന്തൊക്കെയാണ്?)

7 = 16 − 9 = 4² − 3². A right triangle with hypotenuse 4 cm and one side 3 cm has its third side squared equal to 16 − 9 = 7. Draw that triangle, then the square on the third side. For 11 use 36 − 25.

🎬 Solution animation

Squeezing the two sides

below	square		above	square	so
2.6	6.76	< 7 <	2.7	7.29	2.6 < √7 < 2.7
2.64	6.9696	< 7 <	2.65	7.0225	√7 ≈ 2.65
3.3	10.89	< 11 <	3.4	11.56	3.3 < √11 < 3.4
3.31	10.9561	< 11 <	3.32	11.0224	√11 ≈ 3.32

Answer: 7 = 4² − 3²: a right triangle with hypotenuse 4 cm and one side 3 cm has third side √7 cm; the square on it has area 7 sq cm, side √7 ≈ 2.65 cm. 11 = 6² − 5²: hypotenuse 6 cm, side 5 cm, third side √11 cm; square of area 11 sq cm, side √11 ≈ 3.32 cm.

🌍 Why this matters

You cannot measure √7 cm with a ruler, but you can draw it exactly with ruler and compass. Draughtsmen and tile-cutters use exactly this: build the right triangle, and the awkward length appears by itself.

Question 2 · The square in the figure

Q2What is the area of the square in the figure? And the length of its side?

The square stands on one perpendicular side of a right triangle whose hypotenuse is 3.5 cm and whose other side is 1.5 cm. Pythagoras gives the area at once; squeeze for the side.

🎬 Solution animation

Answer: area = 3.5² − 1.5² = 12.25 − 2.25 = 10 sq cm (or (3.5 + 1.5)(3.5 − 1.5) = 5 × 2 = 10). Side = √10 cm; 3.1² = 9.61 < 10 < 10.24 = 3.2² and 3.16² = 9.9856 < 10 < 10.0489 = 3.17², so the side is √10 ≈ 3.16 cm.

Question 3 · The fan of triangles

Q3Find the area of the square in each of the figures below, and the length of its side.

(The three textbook figures: one, two and three right triangles, each with a 1 m side, with a square on the last hypotenuse.)

Each new triangle uses the previous hypotenuse as one side and 1 m as the other. So each hypotenuse² is the previous hypotenuse² plus 1.

🎬 Solution animation

🎚️ Keep adding triangles

number of triangles

With n triangles the last hypotenuse² is n + 1, so the square has area n + 1 sq m and side √(n + 1) m. Natural numbers appear at n = 3 (√4 = 2) and n = 8 (√9 = 3); everything in between is a new number.

Answer: (a) one triangle: hypotenuse² = 1 + 1 = 2 ⇒ square of area 2 sq m, side √2 ≈ 1.41 m. (b) two triangles: 2 + 1 = 3 ⇒ area 3 sq m, side √3 ≈ 1.73 m. (c) three triangles: 3 + 1 = 4 ⇒ area 4 sq m, side √4 = 2 m exactly.

Question 4 · Between √2 and √3

Q4Find three fractions larger than √2 and smaller than √3.

A number s is between √2 and √3 exactly when its square is between 2 and 3. So pick decimals and square them.

🎚️ Is s between √2 and √3? Square it!

try:

Answer (one of many): 1.5, 1.6 and 1.7, because 1.5² = 2.25, 1.6² = 2.56 and 1.7² = 2.89 all lie between 2 and 3. As mixed numbers: 1½, 1⅗, 1 7/10. (Also 1.42, 1.45, 1.73, …)

Side box · Drawing √x for any x വരയും വർഗമൂലവും

Question 1 worked because 7 and 11 are odd. The textbook's side box (page 26) shows that every positive number is a difference of two squares, so √x can always be drawn.

🎬 Animation: x = (½(x + 1))² − (½(x − 1))², shown for x = 5

🎚️ Draw √x for your own x

Hypotenuse ½(x + 1), one side ½(x − 1): the third side is always √x, because (½(x + 1))² − (½(x − 1))² = x. For x between 0 and 1 the side would be negative – use ½(1 − x) instead, which is the same length.

⭐ The identity behind it (Class 8)

(x + 1)² − (x − 1)² = 4x so x = (½(x + 1))² − (½(x − 1))²

Check with x = 5: 3² − 2² = 9 − 4 = 5 ✔. With x = 10: 5.5² − 4.5² = 30.25 − 20.25 = 10 ✔. With x = 2: 1.5² − 0.5² = 2.25 − 0.25 = 2 ✔ – the triangle of Practice set 2, Question 1.

Check yourself

🧭 What these questions taught

To draw a square of area A, find a right triangle whose hypotenuse² − side² = A; the third side is √A.
Odd A: use consecutive squares (A = (n + 1)² − n²). Any A: use hypotenuse ½(A + 1) and side ½(A − 1).
In the fan of unit triangles the areas go 2, 3, 4, 5, … and the sides √2, √3, 2, √5, …
s lies between √2 and √3 exactly when s² lies between 2 and 3.

Next: Section 3 – adding and subtracting lengths like 2 + √2 ▶