Model Test Paper – Solved ഉത്തരങ്ങൾ
Every question of the model paper, with a full solution, an animated figure wherever a figure helps, and a marking hint so you can score your own paper honestly.
📖 How marks are usually given
1-mark questions: the correct answer only (exact forms like 6√2 are expected unless a decimal is asked for). 2- and 3-mark questions: about one mark for writing the right rule or formula, the rest for the working and the answer with units. 5-mark questions: marks for each part as shown. A correct method with an arithmetic slip usually loses only one mark.
SECTION A8 × 1 = 8 marks
1.
Written in its simplest form, √72 = ______.
(1)Answer: 6√2
72 = 36 × 2, so √72 = √36 × √2 = 6√2 (≈ 8.49).2.
√3 × √27 = (a) √30 (b) 9 (c) 3√3 (d) 81
(1)Answer: (b) 9
√3 × √27 = √(3 × 27) = √81 = 9. (Or: √27 = 3√3, so √3 × 3√3 = 3 × 3 = 9.)3.
The height of an equilateral triangle of side 6 cm is ______ cm.
(1)Answer: 3√3 cm
Height = √3 × (half the side) = √3 × 3 = 3√3 cm ≈ 5.20 cm.Review: Page 9 · Equilateral triangles
4.
Which of these is a natural number? (a) √5 × √10 (b) √2 × √8 (c) √3 × √5 (d) √6 ÷ √2
(1)Answer: (b) √2 × √8
√2 × √8 = √16 = 4. The others are √50 = 5√2, √15 and √3 – all irrational.Review: Page 6 · Exercise 1, Q6
5.
Written without a square root in the denominator, 1/√5 = ______.
(1)Answer: √5/5
1/√x = √x/x, so 1/√5 = √5/5 (≈ 2.236 ÷ 5 = 0.447).Review: Page 7 · 1/√x = √x/x
6.
(√5 + 2)(√5 − 2) = (a) 1 (b) 3 (c) 5 − 2√5 (d) 9
(1)Answer: (a) 1
(x + y)(x − y) = x² − y² works for irrational numbers too: (√5)² − 2² = 5 − 4 = 1.Review: Page 7 · The identity (x+y)(x−y)
7.
√50 ÷ √2 = ______.
(1)Answer: 5
√50 ÷ √2 = √(50 ÷ 2) = √25 = 5. (Or 5√2 ÷ √2 = 5.)Review: Page 7 · Division
8.
In Heron's formula for the area of a triangle with sides a, b, c, the number s stands for (a) a + b + c (b) ½(a + b + c) (c) ⅓(a + b + c) (d) abc
(1)Answer: (b) ½(a + b + c)
s is half the perimeter; area = √(s(s−a)(s−b)(s−c)).Review: Page 9 · Heron's formula
SECTION B5 × 2 = 10 marks
9.
Using √2 ≈ 1.414 and √3 ≈ 1.732, find √6 correct to three decimal places. Explain which rule you used.
(2)Answer: √6 ≈ 2.449
Rule: √x × √y = √(xy), so √6 = √2 × √3 ≈ 1.414 × 1.732 = 2.449048… ≈ 2.449. (Check: 2.449² = 5.998 ≈ 6.)Marks: 1 for √6 = √2 × √3, 1 for the multiplication and rounding to 2.449.
Review: Page 5 · √2 × √3 = √6
10.
Find the perimeter and the area of an equilateral triangle of side 6 cm.
(2)Answer: perimeter 18 cm; area 9√3 sq cm ≈ 15.59 sq cm
Perimeter = 3 × 6 = 18 cm. Area = √3 × (half the side)² = √3 × 3² = 9√3 sq cm. (Or ½ × 6 × 3√3 = 9√3.)Marks: 1 for the perimeter, 1 for the area in the form 9√3 (or 15.59).
Review: Page 9 · Equilateral triangles
11.
Two squares of areas 48 sq cm and 27 sq cm are placed side by side on a line as in the figure. Find the exact length of the bottom line.
(2)🎬 Solution
Answer: 7√3 cm (≈ 12.12 cm)
Side of the first square = √48 = √(16 × 3) = 4√3 cm; side of the second = √27 = √(9 × 3) = 3√3 cm. Bottom line = 4√3 + 3√3 = 7√3 cm.Marks: 1 for simplifying both square roots, 1 for adding the like terms to 7√3.
Review: Page 6 · Exercise 1, Q3
12.
Write 1/(√5 − 2) in a form that has no square root in the denominator. (Use question 6.)
(2)Answer: √5 + 2 (≈ 4.236)
Since (√5 + 2)(√5 − 2) = 1, dividing both sides by (√5 − 2) gives 1/(√5 − 2) = √5 + 2. (Same thing: multiply top and bottom by √5 + 2; the bottom becomes 1.)Marks: 1 for multiplying numerator and denominator by √5 + 2 (or quoting Q6), 1 for the answer √5 + 2.
Review: Page 8 · Exercise 2, Q1
13.
For each of the following, find whether the result is a natural number, a fraction, or an irrational number, and write its value in simplest form: (i) √8 × √18 (ii) √12 ÷ √3 (iii) √7 × √14 (iv) √(2/3) × √(3/8)
(2)Answers
- √8 × √18 = √144 = 12 – natural number.
- √12 ÷ √3 = √(12/3) = √4 = 2 – natural number.
- √7 × √14 = √98 = √(49 × 2) = 7√2 – irrational (≈ 9.90).
- √(2/3) × √(3/8) = √(6/24) = √(1/4) = ½ – a fraction.
Marks: ½ for each part (value and type).
SECTION C4 × 3 = 12 marks
14.
The height of an equilateral triangle is 6√3 cm. (i) Find the length of its side. (ii) Find its perimeter. (iii) Find its area.
(3)🎬 Solution
Answer: side 12 cm; perimeter 36 cm; area 36√3 sq cm ≈ 62.35 sq cm
(i) Height = √3 × (half the side) ⇒ half the side = 6√3 ÷ √3 = 6 ⇒ side = 12 cm. (ii) Perimeter = 3 × 12 = 36 cm. (iii) Area = ½ × 12 × 6√3 = 36√3 sq cm (or √3 × 6² = 36√3).Marks: 1 each for the side, the perimeter and the area.
Review: Page 10 · Exercise 3, Q3
15.
The sides of a triangle are 10 cm, 10 cm and 12 cm. (i) Find the height of the triangle, measured to the 12 cm side. (ii) Find its area. (iii) Check your area using Heron's formula.
(3)🎬 Solution
Answer: height 8 cm; area 48 sq cm
(i) The triangle is isosceles, so the height from the top vertex bisects the 12 cm base into 6 cm and 6 cm. h² = 10² − 6² = 100 − 36 = 64 ⇒ h = 8 cm. (ii) Area = ½ × 12 × 8 = 48 sq cm. (iii) Heron: s = ½(10 + 10 + 12) = 16; area = √(16 × 6 × 6 × 4) = √2304 = 48 sq cm ✓.Marks: 1 for the height (with the bisected base), 1 for the area, 1 for the Heron check.
16.
A regular hexagon has side 4 cm. Find its area, exactly and correct to one decimal place.
(3)🎬 Solution
Answer: 24√3 sq cm ≈ 41.6 sq cm
Joining the centre to the six vertices splits the hexagon into 6 equilateral triangles of side 4 cm. Each has area √3 × 2² = 4√3 sq cm. Total = 6 × 4√3 = 24√3 = 24 × 1.732 ≈ 41.6 sq cm.Marks: 1 for splitting into 6 equilateral triangles, 1 for one triangle's area 4√3, 1 for 24√3 ≈ 41.6.
Review: Page 10 · Exercise 3, Q2
17.
Use Heron's formula to find the area of a triangle with sides 7 cm, 8 cm and 9 cm. Write the answer in simplest form and also as a decimal correct to one place.
(3)🎬 Solution
Answer: 12√5 sq cm ≈ 26.8 sq cm
s = ½(7 + 8 + 9) = 12; s − a = 5, s − b = 4, s − c = 3. Area = √(12 × 5 × 4 × 3) = √720 = √(144 × 5) = 12√5 = 12 × 2.236 ≈ 26.8 sq cm.Marks: 1 for s = 12 and the three differences, 1 for √720, 1 for simplifying to 12√5 ≈ 26.8.
SECTION D2 × 5 = 10 marks
18.
In the figure, squares are drawn on the two perpendicular sides of a right triangle. Their areas are 32 sq cm and 18 sq cm. (i) Write the lengths of the two perpendicular sides in simplest form. (ii) Find the length of the hypotenuse in simplest form. (iii) Find the perimeter of the triangle, exactly and correct to one decimal place. (iv) Find the area of the triangle.
(5)🎬 Solution
Answer: legs 4√2 cm and 3√2 cm; hypotenuse 5√2 cm; perimeter 12√2 ≈ 17.0 cm; area 12 sq cm
(i) √32 = √(16 × 2) = 4√2 cm and √18 = √(9 × 2) = 3√2 cm.(ii) Hypotenuse² = 32 + 18 = 50 (the areas of the two squares add up, by Pythagoras) ⇒ hypotenuse = √50 = 5√2 cm.
(iii) Perimeter = 4√2 + 3√2 + 5√2 = 12√2 = 12 × 1.414 ≈ 17.0 cm.
(iv) Area = ½ × 4√2 × 3√2 = ½ × 12 × (√2 × √2) = ½ × 12 × 2 = 12 sq cm.
Marks: 1 for the two legs, 1 for 50 and 5√2, 1 for adding like terms to 12√2, 1 for the decimal 17.0, 1 for the area (using √2 × √2 = 2).
Review: Page 6 · Exercise 1, Q5
19.
A plot of land has the shape of the quadrilateral ABCD in the figure, with AB = 5 m, BC = 12 m, CD = 15 m, DA = 14 m and the diagonal AC = 13 m. (i) Show that the angle at B is a right angle, and find the area of triangle ABC. (ii) Find the area of triangle ACD using Heron's formula. (iii) Find the total area of the plot.
(5)🎬 Solution
Answer: 30 sq m + 84 sq m = 114 sq m
(i) AB² + BC² = 25 + 144 = 169 = 13² = AC², so by the converse of Pythagoras the angle at B is 90°. Area of ΔABC = ½ × 5 × 12 = 30 sq m.(ii) ΔACD has sides 13, 14, 15: s = ½(13 + 14 + 15) = 21; s − 13 = 8, s − 14 = 7, s − 15 = 6. Area = √(21 × 8 × 7 × 6) = √7056 = 84 sq m.
(iii) Total area = 30 + 84 = 114 sq m.
Marks: 1 for 5² + 12² = 13², 1 for 30 sq m, 1 for s = 21 with the differences, 1 for √7056 = 84, 1 for the total 114 sq m.
Review: Page 9 · Land area box · Page 11 · Surveying
Marks summary
| Section | Questions | Marks | What it tests |
|---|---|---|---|
| A | 1 – 8 | 8 | Simplifying roots, the product and quotient rules, rationalising, the identity (x+y)(x−y), Heron's s |
| B | 9 – 13 | 10 | Decimal of √6, equilateral perimeter and area, adding like roots, 1/(√5−2), classifying products |
| C | 14 – 17 | 12 | Height → side, isosceles height and area, hexagon area, Heron's formula with simplification |
| D | 18 – 19 | 10 | Squares on a right triangle (all four skills together), land area by two methods |
| Total | 40 |
⏱️ Time management (1½ hours = 90 minutes)
- Section A: about 10 minutes – write the exact form (6√2, √5/5); do not waste time on decimals unless asked.
- Section B: about 20 minutes (4 minutes a question). Always write the rule you are using before the arithmetic.
- Section C: about 30 minutes. Draw the figure first – a correct figure with the height marked earns a mark by itself.
- Section D: about 25 minutes. Label every part (i), (ii), (iii)… so the examiner can give part marks.
- Keep the last 5 minutes to check units (cm, sq cm, m, sq m) and that every square root is simplified.
🎉 Finished the chapter?
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 three sections again with the animations – then retake the test in a week.