Answer Key – Set 1 ഉത്തരസൂചിക

Step-by-step solution

1. The height cuts the triangle into two right triangles with hypotenuse 2 cm and base 1 cm (half the side).
2. By Pythagoras, height² = 2² − 1² = 4 − 1 = 3, so the height is √3 cm.

Marking: 1 mark for (c).

📖 Revise: Ch 2 · Lengths and numbers · Ch 2 · Pythagoras · Chapter 2 – New Numbers

Step-by-step solution

1. Mark 5 equal steps A₁, …, A₅ on the slanted line and join A₅ to B.
2. Lines through A₁, A₂, A₃, A₄ parallel to A₅B cut AB into 5 equal parts, because parallel lines that cut one line into equal parts cut every line into equal parts.

Marking: 1 mark for (b).

📖 Revise: Ch 3 · Equal division of a line · Chapter 3 – Parallel Lines

Step-by-step solution

1. (x − 1)(y − 1) = xy − (x + y) + 1.
2. = 300 − 40 + 1 = 261. (The numbers are 10 and 30; 9 × 29 = 261.)

Marking: 1 mark for (b).

📖 Revise: Ch 4 · Product of the previous numbers · Chapter 4 – Multiplication Identities

Step-by-step solution

1. (x + ½)(y + ½) = xy + ½(x + y) + ¼.
2. 10½ × 4½ = (10 × 4) + ½(10 + 4) + ¼ = 40 + 7 + ¼ = 47¼, not 40¼. So the statement is false. (40¼ leaves out the two products 10 × ½ and ½ × 4.)

Marking: ½ mark for "false", ½ mark for the correct product 47¼ with the reason.

📖 Revise: Ch 4 · Multiplying sums · Chapter 4 – Multiplication Identities

Step-by-step solution

1. √x × √y = √(xy), so √5 × √20 = √100 = 10.

Marking: 1 mark for (b).

📖 Revise: Ch 5 · Multiplying square roots · Chapter 5 – Irrational Numbers

Step-by-step solution

1. A move to the left is a negative change: −9.
2. Final position = 4 + (−9) = 4 − 9 = −(9 − 4) = −5.

Marking: 1 mark for (b).

📖 Revise: Ch 7 · Position and number · Chapter 7 – Negative Numbers

Step-by-step solution

1. Archimedes of Greece (3rd century BC) used regular polygons with 96 sides and found that π lies between 31071 and 317.

Marking: 1 mark for (c).

📖 Revise: Ch 8 · π – a new number · Chapter 8 – Measures of Circles

Step-by-step solution

1. Circumference = 2πr = 2 × 227 × 3.5 = 2 × 22 × 0.5 = 22 cm.

Marking: 1 mark for 22 cm.

📖 Revise: Ch 8 · Circumference c = 2πr · Chapter 8 – Measures of Circles

Step-by-step solution

1. √9 = 3, −4 = −4/1, 0.25 = ¼ and ¾ can all be written as fractions: they are rational numbers.
2. No fraction squares to 6, so √6 cannot be written as a fraction: it is an irrational number.

Marking: ½ mark for √6, ½ mark for the reason.

📖 Revise: Ch 9 · Rational and irrational numbers · Chapter 9 – Real Numbers

Step-by-step solution

1. Area = (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9 sq cm.

Marking: 1 mark for (c).

📖 Revise: Ch 10 · Area as a polynomial · Chapter 10 – Polynomials

Step-by-step solution

1. Let the daughter’s present age be d and the mother’s m years.
2. Five years ago: m − 5 = 7(d − 5), so m = 7d − 30.
3. Five years from now: m + 5 = 3(d + 5), so m = 3d + 10.
4. So 7d − 30 = 3d + 10 ⇒ 4d = 40 ⇒ d = 10; m = 3 × 10 + 10 = 40.
5. Check: 5 years ago 35 = 7 × 5 ✓; in 5 years 45 = 3 × 15 ✓.

Marking: Two equations 1 mark; daughter’s age ½, mother’s age ½.

📖 Revise: Ch 1 · Two equations · Ch 1 · Age problems · Chapter 1 – Pairs of Equations

Step-by-step solution

1. (a) By Pythagoras, hypotenuse² = 1² + 3² = 1 + 9 = 10, so the hypotenuse is √10 m. (10 is not a perfect square, so this length is not a fraction.)
2. (b) Perimeter = 1 + 3 + √10 = 4 + √10 m ≈ 4 + 3.162 = 7.162 m.
3. Correct to a centimetre (two decimal places): 7.16 m.

Marking: (a) 1 mark; (b) 4 + √10 ½ mark, 7.16 m ½ mark.

📖 Revise: Ch 2 · Measures and numbers · Ch 2 · Perimeters like 2 + √2 · Chapter 2 – New Numbers

Step-by-step solution

1. (a) The parallel lines l, m, n cut the line AC into equal parts (AB = BC).
2. Parallel lines that cut one line into equal parts cut every line into equal parts. So PQ = QR, that is QR = 3.2 cm (and PR = 6.4 cm).
3. (b) Yes. Parallel lines that cut some line into equal parts are equally spaced: the distance between l and m equals the distance between m and n.

Marking: (a) QR = 3.2 cm ½, reason ½; (b) yes ½, reason ½.

📖 Revise: Ch 3 · Equally spaced parallel lines · Chapter 3 – Parallel Lines

Step-by-step solution

1. (a) Area = √8 × √18 = √(8 × 18) = √144 = 12 sq cm.
2. (b) √8 = √(4 × 2) = √4 × √2 = 2√2 and √18 = √(9 × 2) = 3√2.
3. Perimeter = 2(2√2 + 3√2) = 2 × 5√2 = 10√2 cm (≈ 14.14 cm).

Marking: (a) 1 mark; (b) simplifying the sides ½, 10√2 ½.

📖 Revise: Ch 5 · Multiplication and simplifying · Chapter 5 – Irrational Numbers

Step-by-step solution

1. Two pairs of angles are equal, so the third angles are equal too: ∠C = ∠R. The triangles have the same angles.
2. So sides opposite equal angles are in the same ratio. AB (opposite ∠C) and PQ (opposite ∠R): 6 ÷ 4 = 1½. Every side of PQR is 1½ times the matching side of ABC.
3. QR is opposite ∠P, matching BC (opposite ∠A): QR = 1½ × 6 = 9 cm.
4. PR is opposite ∠Q, matching CA (opposite ∠B): PR = 1½ × 5 = 7.5 cm.

Marking: Same angles and scale factor 1½: 1 mark; QR ½, PR ½.

📖 Revise: Ch 6 · Same angles ⇒ sides in the same ratio · Chapter 6 – Similar Triangles

Step-by-step solution

1. (a) y − z = 5 − (−4) = 5 + 4 = 9 (since x − (−y) = x + y). So x − (y − z) = −3 − 9 = −(3 + 9) = −12.
2. x − y = −3 − 5 = −8, and (−8) + (−4) = −(8 + 4) = −12. Both sides are equal.
3. (b) x − y − z = −3 − 5 − (−4) = −8 + 4 = −4, which is not −12. So no: subtracting (y − z) is the same as subtracting y and then adding z.

Marking: (a) each side ½; (b) value −4 ½, conclusion ½.

📖 Revise: Ch 7 · Subtracting negative numbers · Ch 7 · Practice set 2 · Chapter 7 – Negative Numbers

Step-by-step solution

1. (a) Distance = bigger − smaller = 3½ − (−2¾) = 3½ + 2¾ = 6¼.
2. (b) Midpoint = ½(sum) = ½(−2¾ + 3½) = ½ × ¾ = ⅜.
3. Check: ⅜ + 3⅛ = 3½ and ⅜ − 3⅛ = −2¾ – the midpoint is 3⅛ (half of 6¼) from each end.

Marking: (a) 1 mark; (b) 1 mark.

📖 Revise: Ch 9 · Distances · Ch 9 · Midpoint · Chapter 9 – Real Numbers

Step-by-step solution

1. The new edge is x + 2, so v(x) = (x + 2)(x + 2)(x + 2).
2. (x + 2)(x + 2) = x² + 2x + 2x + 4 = x² + 4x + 4.
3. (x + 2)(x² + 4x + 4) = (x³ + 4x² + 4x) + (2x² + 8x + 8) = x³ + 6x² + 12x + 8.
4. The highest power of x is 3, so the degree is 3. (Check: v(1) = 1 + 6 + 12 + 8 = 27 = 3³ ✓.)

Marking: (x + 2)² ½, the polynomial 1 mark, degree ½.

📖 Revise: Ch 10 · Volume as a polynomial · Ch 10 · Degree · Chapter 10 – Polynomials

Step-by-step solution

1. (a) (3½)² = 12¼ and (2½)² = 6¼, so (3½)² − (2½)² = 12¼ − 6¼ = 6.
2. (b) Draw AB = 2.5 cm. Draw the perpendicular to AB at B.
3. With A as centre and radius 3.5 cm, draw an arc cutting the perpendicular at C. Join AC.
4. ABC is a right triangle with hypotenuse AC = 3.5 cm, so BC² = 3.5² − 2.5² = 6 (part (a)): BC = √6 cm.
5. On BC draw the square BCDE (perpendiculars at B and C, arcs of radius BC). Its area is (√6)² = 6 sq cm.
6. (c) 2.4² = 5.76 < 6 < 6.25 = 2.5², so 2.4 < √6 < 2.5.

Marking: (a) 1 mark; (b) right triangle with the correct sides ½, BC = √6 with reason ½, square ½; (c) ½ mark.

📖 Revise: Ch 2 · Drawing and square roots · Ch 2 · √x as a side · Chapter 2 – New Numbers

Step-by-step solution

1. Side of the square = 13 ÷ 4 = 3¼ cm – so we divide a 13 cm line into 4 equal parts.
2. Draw AB = 13 cm. From A draw a slanted ray AX. With the compass mark 4 equal steps of 3 cm on it: A₁, A₂, A₃, A₄.
3. Join A₄B. Through A₁, A₂, A₃ draw lines parallel to A₄B (slide a set square). They meet AB at P, Q, R. Then AP = PQ = QR = RB = 3¼ cm.
4. Draw perpendiculars at the two ends of AP, cut them with arcs of radius AP, and join: this is the square of side 3¼ cm, perimeter 4 × 3¼ = 13 cm.
5. Why: the parallel lines through A₁, A₂, A₃, A₄ (and one through A) cut AX into equal parts. Parallel lines that cut one line into equal parts cut every line into equal parts, so they cut AB into 4 equal parts.

Marking: Side 3¼ cm ½; dividing the line (ray, equal steps, parallels) 1½; square drawn ½; reason ½.

📖 Revise: Ch 3 · Dividing a line into equal parts · Ch 3 · Exercise 1 constructions · Chapter 3 – Parallel Lines

Step-by-step solution

1. (a) √45 = √(9 × 5) = √9 × √5 = 3√5; √80 = √(16 × 5) = √16 × √5 = 4√5.
2. (b) Hypotenuse² = 45 + 80 = 125, so hypotenuse = √125 = √(25 × 5) = 5√5 cm.
3. (c) Perimeter = 3√5 + 4√5 + 5√5 = 12√5 cm (≈ 26.83 cm).
4. Area = ½ × 3√5 × 4√5 = ½ × 12 × (√5 × √5) = ½ × 12 × 5 = 30 sq cm.

Marking: (a) ½ + ½; (b) 1 mark; (c) perimeter ½, area ½.

📖 Revise: Ch 5 · √x × √y = √(xy) and simplifying · Ch 5 · Exercise 1 · Chapter 5 – Irrational Numbers

Step-by-step solution

1. (a) Start: −4. First move: −4 + 7 = 7 − 4 = 3.
2. Second move (10 km west = −10): 3 + (−10) = 3 − 10 = −(10 − 3) = −7.
3. (b) Third move: −7 + 2 = −(7 − 2) = −5. The bus ends 5 km west of the bus stand.
4. (c) Change = 7 + (−10) + 2 = −1: one move of 1 km west. Check: −4 + (−1) = −(4 + 1) = −5 ✓.

Marking: (a) additions ½, −7 ½; (b) −5 ½, meaning ½; (c) 1 mark.

📖 Revise: Ch 7 · Positions and addition · Chapter 7 – Negative Numbers

Step-by-step solution

1. (a) −√16 = −4. Integers: −3, 0, −√16. Rational (can be written as fractions): −3, 2/5, 0, −√16, 1.6 (1.6 = 16/10). Irrational: √7, π.
2. (b) √7 is between 2 and 3 (2² = 4 < 7 < 9 = 3²; in fact 2.6² = 6.76 < 7), π ≈ 3.14. Order: −√16 < −3 < 0 < 2/5 < 1.6 < √7 < π.
3. (c) |x + 1| = |x − (−1)| is the distance between x and −1. It is less than 3 when x is between −1 − 3 = −4 and −1 + 3 = 2: −4 < x < 2.
4. Of the given numbers, −3, 0, 2/5 and 1.6 lie in this part. (−√16 = −4 is at distance exactly 3, not less; √7 and π are bigger than 2.)

Marking: (a) 1 mark (integers ½, rational/irrational ½); (b) 1 mark; (c) interval ½, the four numbers ½.

📖 Revise: Ch 9 · Real numbers on the number line · Ch 9 · |x − a| < b as “between” · Chapter 9 – Real Numbers

Step-by-step solution

1. (a) The frame adds x on each side: outer length 2x + 15, outer breadth 2x + 10 (cm).
2. p(x) = 2((2x + 15) + (2x + 10)) = 2(4x + 25) = 8x + 50.
3. (b) Outer area = (2x + 15)(2x + 10) = 4x² + 20x + 30x + 150 = 4x² + 50x + 150.
4. Frame area a(x) = (4x² + 50x + 150) − 150 = 4x² + 50x. Degree 2.
5. (c) a(2) = 4 × 2² + 50 × 2 = 16 + 100 = 116 sq cm.
6. Check: outer measures 19 cm × 14 cm, area 266; 266 − 150 = 116 ✓.

Marking: (a) outer sides ½, p(x) ½; (b) a(x) ½, degree ½; (c) a(2) ½, check ½.

📖 Revise: Ch 10 · Algebra of measures · Ch 10 · Area rules · Chapter 10 – Polynomials

Step-by-step solution

1. (a) 3x + 2y = 1300 … (1)   2x + 3y = 1100 … (2)
2. (b) Add (1) and (2): 5x + 5y = 2400, so x + y = 480.
3. Subtract (2) from (1): x − y = 200.
4. Sum + difference = twice the larger: 2x = 680 ⇒ x = 340; y = 480 − 340 = 140. Adult ticket ₹340, child ticket ₹140. (Check: 1020 + 280 = 1300 ✓, 680 + 420 = 1100 ✓.)
5. (c) Let the group buy a adult and c child tickets: a + c = 50 and 340a + 140c = 10000.
6. Multiply the first by 140: 140a + 140c = 7000. Subtract it from the second: 200a = 3000 ⇒ a = 15, c = 50 − 15 = 35.
7. Check: 15 × 340 + 35 × 140 = 5100 + 4900 = 10000 ✓.

Marking: (a) 1 mark; (b) method ½, each price ½; (c) equations ½, the two numbers ½.

📖 Revise: Ch 1 · Two equations – elimination · Ch 1 · Sum and difference · Chapter 1 – Pairs of Equations

Step-by-step solution

1. Perimeter 46 m ⇒ x + y = 23; area xy = 120.
2. (a) (x + 1)(y + 1) = xy + (x + y) + 1 = 120 + 23 + 1 = 144 sq m.
3. (b) (x + ½)(y + ½) = xy + ½(x + y) + ¼ = 120 + 11½ + ¼ = 131¾ sq m.
4. (c) (x − y)² = 23² − 4 × 120 = 529 − 480 = 49, so x − y = 7.
5. x = ½(23 + 7) = 15 m, y = ½(23 − 7) = 8 m.
6. (d) 16 × 9 = 144 ✓ and 15½ × 8½ = 120 + ½(23) + ¼ = 131¾ ✓ (or 15½ × 8½ = 131.75).

Marking: (a) 1 mark; (b) 1 mark; (c) x − y = 7 ½, the sides ½; (d) 1 mark.

📖 Revise: Ch 4 · Multiplying sums · Ch 4 · Sum and product ⇒ the numbers · Chapter 4 – Multiplication Identities

Step-by-step solution

1. (a) ∠TAB = ∠DCB = 90°. ∠ABT = ∠CBD, since they are vertically opposite angles (TD and AC are straight lines).
2. Two angles are equal, so the third angles ∠ATB and ∠CDB are equal too (angles of a triangle add up to 180°). So the triangles have the same angles.
3. (b) Sides opposite equal angles are in the same ratio: opposite ∠B: TA and DC; opposite ∠T and ∠D: AB and CB; opposite the right angles: TB and DB. So TA/DC = AB/CB = TB/DB.
4. (c) AB/CB = 50/10 = 5, so TA = 5 × DC = 5 × 12 = 60 m.
5. (d) The river is still 60 m wide and now AB/CB = 50/5 = 10, so CD = TA ÷ 10 = 60 ÷ 10 = 6 m.

Marking: (a) right angles and vertically opposite angles ½, third angle ½; (b) 1 mark; (c) 1 mark; (d) 1 mark.

📖 Revise: Ch 6 · Same angles ⇒ sides in the same ratio · Ch 6 · Exercise 1 Q3 (vertically opposite angles) · Chapter 6 – Similar Triangles

Step-by-step solution

1. (a) Circumference of a circle of diameter 28 m = π × 28 = 227 × 28 = 88 m. A semicircle is half: 44 m.
2. (b) One round = 66 + 66 + 44 + 44 = 220 m.
3. (c) One turn of the wheel covers its circumference: π × 70 = 227 × 70 = 220 cm = 2.2 m.
4. Number of turns = 220 ÷ 2.2 = 100.
5. (d) The two semicircles of the outer track make a full circle of diameter 30 m: π × 30 = 6607 = 9427 m. One round = 132 + 9427 = 22627 m.
6. Extra length = 22627 − 220 = 627 m. (In general the difference is π × 30 − π × 28 = π × 2 = 2π, because the diameter grows by 2 m; 2π ≈ 6.28 m.)

Marking: (a) 1 mark; (b) 1 mark; (c) circumference of the wheel ½, 100 turns ½; (d) outer round ½, difference ½.

📖 Revise: Ch 8 · c = πd · Ch 8 · Practice set 2 (circles one inside another) · Chapter 8 – Measures of Circles