Model Test Paper – Solved ഉത്തരങ്ങൾ
Every question of the model paper, with a full solution, an animated figure wherever a picture or a chain of steps helps, and a marking hint so you can score your own paper honestly.
1-mark questions: the correct answer only. 2- and 3-mark questions: about one mark for the right idea (which fact, which picture), the rest for the working and the final answer with the correct rounding. 5-mark questions: marks for each part as shown. In a proof, every step must give its reason ("odd × odd is odd", "lowest terms means no common factor"); a correct chain of reasons earns marks even if the last line is missing.
Area 2: draw a right triangle with perpendicular sides 1 cm and 1 cm. By Pythagoras the square on its hypotenuse has area 1 + 1 = 2 sq cm; draw that square. Its side is the hypotenuse, √2 cm.
Area 3: draw a right triangle with perpendicular sides 1 cm and √2 cm (use the hypotenuse just drawn as one side). The square on its hypotenuse has area 1 + 2 = 3 sq cm; its side is √3 cm. (Or: a right triangle with hypotenuse 2 cm and one side 1 cm – the square on the other side has area 4 − 1 = 3.)
(1⅓)² = (4/3)² = 16/9 = 1 7/9, which is less than 2 (by 2/9). (1½)² = (3/2)² = 9/4 = 2¼, which is more than 2 (by ¼). Picture: (1⅓)² = 1 + ⅓ + ⅓ + 1/9 and (1½)² = 1 + ½ + ½ + ¼.
√2 is the number whose square is 2. Since squares grow with the number, √2 lies between 1⅓ and 1½ – so √2 is between 1.33… and 1.5, and neither of these fractions is √2.
By Pythagoras, 3² = 2² + (third side)², so (third side)² = 9 − 4 = 5: the third side is √5 cm. 2.2² = 4.84 and 2.3² = 5.29, and 2.23² = 4.9729, 2.24² = 5.0176, so √5 ≈ 2.236 ≈ 2.24.
Perimeter = 3 + 2 + √5 = 5 + √5 = 5 + 2.236… = 7.236… ≈ 7.24 cm.
- 1 + √2 + √3
- (2 + √3) − (1 + √2)
(i) 1 + 1.414 + 1.732 = 4.146 → 4.15 (the third decimal is 6, so round up).
(ii) (2 + √3) − (1 + √2) = 2 − 1 + √3 − √2 = 1 + √3 − √2 = 1 + 1.732 − 1.414 = 1.318 → 1.32. (Keeping the roots until the end avoids rounding twice.)
⅓ = 0.333… (the digit 3 repeats for ever); ⅙ = 0.1666… (6 repeats). √2 = 1.41421356… – the digits go on without end, but no block of digits ever repeats. That is the difference between a fraction and a new number like √2: every fraction's decimal terminates or repeats.
- Between which tenths does √3 lie?
- Between which hundredths?
- Write √3 correct to 2 decimal places.
(i) 2.89 < 3 < 3.24, i.e. 1.7² < 3 < 1.8², so 1.7 < √3 < 1.8. (ii) 2.9929 < 3 < 3.0276, i.e. 1.73² < 3 < 1.74², so 1.73 < √3 < 1.74. (iii) √3 is between 1.73 and 1.74; since 1.735² = 3.010225 > 3, √3 is below 1.735, so correct to 2 decimal places √3 ≈ 1.73. (In fact √3 = 1.73205…)
- Find its hypotenuse.
- Find its perimeter correct to a centimetre.
- A second right triangle has perpendicular sides √5 m and 2 m. Find its hypotenuse, and how much longer its perimeter is than that of the first triangle.
(i) hypotenuse² = 1² + 2² = 5, so the hypotenuse is √5 m (≈ 2.236 m).
(ii) perimeter = 1 + 2 + √5 = 3 + √5 = 3 + 2.236… = 5.236… → 5.24 m correct to a centimetre.
(iii) hypotenuse² = (√5)² + 2² = 5 + 4 = 9, so the hypotenuse is √9 = 3 m – a natural number. Its perimeter is √5 + 2 + 3 = 5 + √5. Difference: (5 + √5) − (3 + √5) = 2 m exactly – no decimals needed, because √5 cancels.
- Find the height.
- Find the perimeter of one half, correct to 2 decimal places.
- How much less is this than the perimeter of the whole triangle?
(i) The height meets the base at its midpoint, so each half is a right triangle with hypotenuse 4 m and base 2 m. height² = 4² − 2² = 16 − 4 = 12, height = √12 m. 3.4² = 11.56, 3.5² = 12.25; 3.46² = 11.9716, 3.47² = 12.0409, so √12 ≈ 3.464 ≈ 3.46 m.
(ii) perimeter of one half = 2 + 4 + √12 = 6 + √12 = 6 + 3.464 = 9.464 → 9.46 m.
(iii) whole triangle: 3 × 4 = 12 m. Less by 12 − (6 + √12) = 6 − √12 = 6 − 3.464 = 2.536 → ≈ 2.54 m.
- Write the sides of the 5th triangle.
- How much longer is the perimeter of the 5th triangle than that of the 4th?
- After the 3rd triangle, which is the next triangle whose hypotenuse is a natural number?
The hypotenuses are √2, √3, √4, √5, …: the nth triangle has sides 1, √n and √(n + 1). (i) 5th triangle: 1 m, √5 m, √6 m.
(ii) 4th triangle: 1, √4 = 2, √5. Perimeters: 5th − 4th = (1 + √5 + √6) − (1 + 2 + √5) = √6 − 2 = 2.449 − 2 = 0.449 m ≈ 0.45 m.
(iii) The hypotenuse of the nth triangle is √(n + 1); it is a natural number when n + 1 is a perfect square: n + 1 = 4 (3rd triangle), then n + 1 = 9, i.e. the 8th triangle, with hypotenuse 3 m (then the 15th, with hypotenuse 4 m).
Suppose some fraction has square 2, and write it in lowest terms as x/y, so that x and y have no common factor. Then (x/y)² = 2 gives x²/y² = 2, so x² = 2y². Thus x² is even. The square of an odd number is odd (odd × odd = odd), so x cannot be odd: x is even.
Since x and y have no common factor and x is even, y cannot be even too (2 would be a common factor): y is odd.
Now x is even, so x = 2z for some natural number z. Substituting: (2z)² = 2y², i.e. 4z² = 2y², so y² = 2z². Thus y² is even, and as before y is even.
So y is odd and y is even – impossible. Our supposition must be wrong: no fraction has square 2. ∎
- Explain why the square drawn on the third side of this triangle has area x + y.
- Write the lengths of the three sides of the triangle.
- Why is √x + √y > √(x + y)?
- Check this with x = 9 and y = 16.
- For x = 3 and y = 6, compute √x + √y − √(x + y) correct to 3 decimal places.
(i) The two squares meet at a right angle, so the triangle between them is a right triangle whose perpendicular sides are the sides of the squares. By Pythagoras, the square on the hypotenuse equals the two squares on the perpendicular sides together: area x + y.
(ii) The sides of the squares are √x and √y (a square of area x has side √x), and the hypotenuse is the side of the square of area x + y: √x, √y, √(x + y).
(iii) In any triangle, two sides together are longer than the third side. Here that gives √x + √y > √(x + y).
(iv) x = 9, y = 16: √9 + √16 = 3 + 4 = 7 and √25 = 5; indeed 7 > 5 (a 3-4-5 triangle).
(v) x = 3, y = 6: √3 + √6 − √9 = 1.732 + 2.449 − 3 = 1.181.
Marks summary
| Section | Questions | Marks | What it tests |
|---|---|---|---|
| A | 1 – 8 | 8 | The key facts by heart: the square on the diagonal, √ of perfect-square fractions, squeezing between whole numbers and decimals, which numbers are fractions, simple hypotenuses and perimeters, √x + √y > √(x + y) |
| B | 9 – 13 | 10 | Drawing squares of area 2 and 3, comparing squares of fractions with 2, Pythagoras with a new-number side, adding and subtracting with rounding, decimal forms |
| C | 14 – 17 | 12 | The squeeze for √3, chains of right triangles and their perimeters, the half equilateral triangle, the fan of triangles 1, √n, √(n + 1) |
| D | 18 – 19 | 10 | The appendix proof, and the sum-and-root picture with its inequality |
| Total | 40 |
- Section A: about 10 minutes – one-line recalls; if one is unclear, move on and come back.
- Section B: about 20 minutes (4 minutes a question). Write the Pythagoras line first, then the root, then the decimal – marks go to each step.
- Section C: about 30 minutes. Keep √ in the working and turn it into a decimal only at the end; round to the places asked (cm = 2 places of a metre, mm = 3).
- Section D: about 25 minutes. For the proof, write the five steps in order with a reason for each; for the picture question, draw the figure first.
- Keep the last 5 minutes to check every rounding and every ≈ sign.
Scored 32+? Excellent. Between 24 and 31? Revisit the pages named under the questions you lost marks on. Below 24? Go back to page 5 and work through the section pages again with the animations – then retake the test in a week.