Class 9 · MathematicsSet 3 · Answer key

Answer Key – Set 3 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 3, for teachers and parents to value the paper. Accept any correct answer in the student’s own words; give method marks even when the final answer is wrong.

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ANSWER KEY & MARKING SCHEME · SET 3
Standard IX · Mathematics · Annual Examination
Total score: 60 · Time: 2 hours
Section AAnswer all questions · 10 × 1 = 10 marks
1.
The sum of two numbers is 36 and their difference is 10. The smaller number is
(a) 13(b) 23(c) 18(d) 26
[1]

Step-by-step solution

  1. Sum (തുക) − difference (വ്യത്യാസം) = twice the smaller number: 36 − 10 = 26.
  2. Smaller number = 26 ÷ 2 = 13. (Larger = (36 + 10) ÷ 2 = 23. Check: 23 + 13 = 36, 23 − 13 = 10 ✓.)
✅ Answer: (a) 13

Marking: 1 mark for (a).

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

2.
Using √2 ≈ 1.414 and √3 ≈ 1.732, the value of √2 + √3 correct to two decimal places is
(a) 2.24(b) 3.14(c) 3.15(d) 3.46
[1]

Step-by-step solution

  1. Add the approximations: 1.414 + 1.732 = 3.146.
  2. Correct to two decimal places, 3.146 → 3.15 (the third decimal 6 is 5 or more, so the second decimal goes up).
  3. Note: √2 + √3 is not √5 (≈ 2.24); square roots are not added by adding the numbers under the root.
✅ Answer: (c) 3.15

Marking: 1 mark for (c).

📖 Revise: Ch 2 · Adding square roots · Chapter 2 – New Numbers

3.
The orthocentre (the point where the altitudes meet) of a right triangle is
(a) the midpoint of the hypotenuse(b) the vertex of the right angle(c) the centroid(d) a point outside the triangle
[1]

Step-by-step solution

  1. In a right triangle each of the two perpendicular sides is perpendicular to the other side through the opposite vertex, so each of them is an altitude.
  2. These two altitudes meet at the right-angle vertex, so the orthocentre (ലംബകേന്ദ്രം) is the vertex of the right angle. (The midpoint of the hypotenuse is the circumcentre, not the orthocentre.)
✅ Answer: (b) the vertex of the right angle

Marking: 1 mark for (b).

📖 Revise: Ch 3 · Altitudes and orthocentre · Chapter 3 – Parallel Lines

4.
Fill in the blanks:   99 × 98 = (100 − 1)(100 − 2) = 10000 − 200 − 100 + =
[1]

Step-by-step solution

  1. (x − y)(u − v) = xu − xv − yu + yv with x = u = 100, y = 1, v = 2.
  2. The last term is yv = 1 × 2 = 2.
  3. 10000 − 200 − 100 + 2 = 10000 − 300 + 2 = 9702.
✅ Answer: 2; 9702

Marking: ½ mark for 2, ½ mark for 9702.

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

5.
√6 × √15 =
(a) √21(b) 3√10(c) 9√10(d) 90
[1]

Step-by-step solution

  1. √x × √y = √(xy), so √6 × √15 = √90.
  2. 90 = 9 × 10, so √90 = √9 × √10 = 3√10.
✅ Answer: (b) 3√10

Marking: 1 mark for (b).

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

6.
The area of an equilateral triangle of side 4 cm is
(a) 4√3 cm²(b) 8√3 cm²(c) 16√3 cm²(d) 2√3 cm²
[1]

Step-by-step solution

  1. Area of an equilateral triangle = √3 × (half the side)².
  2. Half the side = 2 cm, so area = √3 × 2² = 4√3 cm².
✅ Answer: (a) 4√3 cm²

Marking: 1 mark for (a).

📖 Revise: Ch 5 · Area of an equilateral triangle · Chapter 5 – Irrational Numbers

7.
A point moves along a number line towards the left at 6 m/s and passes the point O (zero) at time 0. Using s = vt, its position 4 seconds before it reached O was
(a) −24(b) 24(c) −10(d) 10
[1]

Step-by-step solution

  1. Moving left, so v = −6. Four seconds before, so t = −4.
  2. s = vt = (−6) × (−4) = 24: the point was 24 m to the right of O (it was coming from the right).
✅ Answer: (b) 24

Marking: 1 mark for (b).

📖 Revise: Ch 7 · Speed, time and multiplication · Chapter 7 – Negative Numbers

8.
True or false? Give the reason:   (−1)20 + (−1)21 = 0
[1]

Step-by-step solution

  1. An even power of −1 is 1, so (−1)20 = 1.
  2. An odd power of −1 is −1, so (−1)21 = −1.
  3. 1 + (−1) = 0, so the statement is true.
✅ Answer: True

Marking: ½ mark for “true”, ½ mark for the reason.

📖 Revise: Ch 7 · Powers of −1 · Chapter 7 – Negative Numbers

9.
How many integers x satisfy |x − 2| < 4?
(a) 4(b) 7(c) 8(d) 9
[1]

Step-by-step solution

  1. |x − 2| < 4 means the distance between x and 2 is less than 4, so −2 < x < 6.
  2. The integers are −1, 0, 1, 2, 3, 4, 5: 7 integers (−2 and 6 are not included).
✅ Answer: (b) 7

Marking: 1 mark for (b).

📖 Revise: Ch 9 · Distances and |x − a| &lt; b · Chapter 9 – Real Numbers

10.
If p(x) = x³ − 2x + 5, then p(−1) =
[1]

Step-by-step solution

  1. p(−1) = (−1)³ − 2 × (−1) + 5 = −1 + 2 + 5 = 6.
✅ Answer: 6

Marking: 1 mark.

📖 Revise: Ch 10 · Evaluating polynomials · Chapter 10 – Polynomials

Section BAnswer all questions · 8 × 2 = 16 marks
11.
If 1 is added to the numerator of a fraction, it becomes ½. If 3 is added to its denominator, it becomes ⅓. Find the fraction.
[2]

Step-by-step solution

  1. Let the numerator (അംശം) be x and the denominator (ഛേദം) be y.
  2. x + 1y = ½ means y is twice x + 1:   y = 2x + 2.
  3. xy + 3 = ⅓ means y + 3 is three times x:   y + 3 = 3x.
  4. Put y = 2x + 2 in the second equation: 2x + 2 + 3 = 3x ⇒ 2x + 5 = 3x ⇒ x = 5.
  5. y = 2 × 5 + 2 = 12. The fraction is 512.
  6. Check: 612 = ½ ✓ and 515 = ⅓ ✓.
✅ Answer: 512

Marking: Two equations: 1 mark (½ each). Numerator ½, denominator ½.

📖 Revise: Ch 1 · Fraction problems · Ch 1 · Solving two equations · Chapter 1 – Pairs of Equations

12.
State whether each statement is true or false, with the reason.   (a) √8 + √2 is bigger than √10.   (b) (√5 + 2) − (√5 − 1) is not a fraction (rational number).
[2]

Step-by-step solution

  1. (a) Draw a right triangle with perpendicular sides √8 and √2. Its hypotenuse² = 8 + 2 = 10, so the hypotenuse is √10.
  2. In any triangle two sides together are longer than the third side, so √8 + √2 > √10: true. (Decimals: 2.828 + 1.414 = 4.242 > 3.162 ✓.)
  3. (b) (√5 + 2) − (√5 − 1) = √5 + 2 − √5 + 1 = 3.
  4. The √5 parts cancel and the result 3 is a natural number (so also a fraction): the statement is false.
✅ Answer: (a) True   (b) False – it equals 3

Marking: (a) “true” ½, reason ½. (b) “false” ½, working showing 3 ½.

📖 Revise: Ch 2 · Adding and subtracting; sum and square root · Chapter 2 – New Numbers

13.
In triangle ABC, AD is the median to BC and G is the centroid (figure). Prove that the area of triangle BGD is ⅙ of the area of triangle ABC.ABCDG
[2]

Step-by-step solution

  1. D is the midpoint of BC, so BD = DC. Triangles ABD and ACD have equal bases on BC and the same height from A, so area ABD = ½ × area ABC.
  2. The centroid (മധ്യമകേന്ദ്രം) divides each median in the ratio 2 : 1, so AG = 2 GD, i.e. GD = ⅓ AD.
  3. Triangles BGD and BAD have their bases GD and AD on the same line AD and the same vertex B, so they have the same height; their areas are in the ratio of the bases: area BGD = ⅓ × area BAD.
  4. So area BGD = ⅓ × ½ × area ABC = ⅙ × area ABC. ∎
✅ Answer: area BGD = ⅓ × area ABD = ⅓ × ½ × area ABC = ⅙ area ABC

Marking: Area ABD = ½ ABC: ½. AG : GD = 2 : 1: ½. Same height ⇒ ratio of bases: ½. Conclusion: ½.

📖 Revise: Ch 3 · Medians and centroid · Ch 3 · Medians and areas · Chapter 3 – Parallel Lines

14.
Numbers that are 1 less than a multiple of 5 (4, 9, 14, 19, …) can be written as 5n − 1. Prove that the product of two such numbers is always 1 more than a multiple of 5. Check with 9 × 14.
[2]

Step-by-step solution

  1. Take two such numbers 5m − 1 and 5n − 1 (m, n natural numbers).
  2. (x − y)(u − v) = xu − xv − yu + yv, so (5m − 1)(5n − 1) = 25mn − 5m − 5n + 1.
  3. = 5(5mn − m − n) + 1, which is 1 more than a multiple of 5. ∎
  4. Check: 9 × 14 = 126 = 125 + 1 = 5 × 25 + 1 ✓ (here m = 2, n = 3 and 5mn − m − n = 30 − 2 − 3 = 25).
✅ Answer: (5m − 1)(5n − 1) = 5(5mn − m − n) + 1

Marking: Expansion 1, writing as 5(…) + 1 ½, check ½.

📖 Revise: Ch 4 · (x − y)(u − v) · Ch 4 · Proving number facts · Chapter 4 – Multiplication Identities

15.
In the roof truss ABC (figure), the rafters meet at a right angle at C and the post CD is perpendicular to the beam AB. AD = 2 m and DB = 8 m. (a) Why do triangles ADC and CDB have the same angles? (b) Find the length of the post CD and the area of triangle ABC.ABCD2 m8 m
[2]

Step-by-step solution

  1. (a) Let ∠A = x. In triangle ADC, ∠ADC = 90°, so ∠ACD = 90° − x.
  2. ∠ACB = 90°, so ∠BCD = 90° − (90° − x) = x; then in triangle CDB, ∠B = 90° − x.
  3. So both triangles have the angles x, 90° − x and 90°: the same angles.
  4. (b) In triangles with the same angles (സദൃശത്രികോണങ്ങൾ), sides opposite equal angles are in the same ratio. Opposite x: CD (in ADC) and DB (in CDB). Opposite 90° − x: AD (in ADC) and CD (in CDB).
  5. So CDDB = ADCD, i.e. CD8 = 2CD ⇒ CD² = 16 ⇒ CD = 4 m.
  6. AB = 2 + 8 = 10 m. Area = ½ × AB × CD = ½ × 10 × 4 = 20 m².
✅ Answer: (a) both have angles x, 90° − x, 90°   (b) CD = 4 m; area = 20 m²

Marking: (a) ½. (b) ratio equation ½, CD ½, area ½.

📖 Revise: Ch 6 · Same angles, sides in the same ratio · Ch 6 · Perpendicular to the hypotenuse · Chapter 6 – Similar Triangles

16.
The surface of a circular temple pond has an area of 616 m². Taking π = 22/7, find (a) the radius of the pond and (b) the length of its boundary (circumference).
[2]

Step-by-step solution

  1. (a) Area = πr², so 22/7 × r² = 616.
  2. r² = 616 × 7/22 = 28 × 7 = 196, so r = 14 m.
  3. (b) Circumference = 2πr = 2 × 22/7 × 14 = 2 × 22 × 2 = 88 m.
✅ Answer: (a) 14 m   (b) 88 m

Marking: (a) equation ½, radius ½. (b) formula ½, answer ½.

📖 Revise: Ch 8 · Area of a circle · Ch 8 · Circumference 2πr · Chapter 8 – Measures of Circles

17.
Solve:   (a) |x − 7| = |x + 2|   (b) |x + 2| = 5
[2]

Step-by-step solution

  1. (a) |x − 7| is the distance between x and 7; |x + 2| = |x − (−2)| is the distance between x and −2.
  2. Equal distances from 7 and −2 means x is their midpoint (മധ്യബിന്ദു): x = ½(7 + (−2)) = ½ × 5 = 2½.
  3. (b) |x + 2| = |x − (−2)| = 5: x is at distance 5 from −2.
  4. To the right: −2 + 5 = 3; to the left: −2 − 5 = −7.
✅ Answer: (a) x = 2½   (b) x = 3 or x = −7

Marking: (a) reasoning ½, answer ½. (b) each value ½.

📖 Revise: Ch 9 · Midpoint and |x − a| = |x − b| · Ch 9 · Distances · Chapter 9 – Real Numbers

18.
A rectangle has breadth x cm and length 2x cm. Write as expressions in x: (a) the length of its diagonal, (b) its perimeter divided by its area. Is each a polynomial? Give the reason, and the degree where it is one.
[2]

Step-by-step solution

  1. (a) By Pythagoras, diagonal² = x² + (2x)² = x² + 4x² = 5x², so diagonal = √5 × x = √5 x cm.
  2. Here x is only multiplied by the fixed number √5, so it is a polynomial (ബഹുപദം) of degree 1. (√5 is a coefficient; it does not change with x.)
  3. (b) Perimeter = 2(2x + x) = 6x; area = 2x × x = 2x². Perimeter ÷ area = 6x2x² = 3x.
  4. This takes the reciprocal of the changing number x, so it is not a polynomial.
✅ Answer: (a) √5 x – polynomial of degree 1   (b) 3/x – not a polynomial

Marking: (a) expression ½, “polynomial, degree 1” with reason ½. (b) expression ½, “not a polynomial” with reason ½.

📖 Revise: Ch 10 · What is a polynomial · Ch 10 · Algebra of measures · Chapter 10 – Polynomials

Section CAnswer all questions · 6 × 3 = 18 marks
19.
Two positive numbers have sum s and difference d. (a) Show that the larger number is ½(s + d) and the smaller is ½(s − d). (b) A 7.5 m rope is cut into two pieces whose lengths differ by 2.5 m. Use (a) to find the pieces. (c) Someone says: “A rectangle has perimeter 18 cm and its length is 10 cm more than its breadth.” Show that this is impossible.
[3]

Step-by-step solution

  1. (a) Let the numbers be x (larger) and y: x + y = s and x − y = d.
  2. Adding: 2x = s + d ⇒ x = ½(s + d). Subtracting: 2y = s − d ⇒ y = ½(s − d).
  3. (b) s = 7.5, d = 2.5: longer piece = ½(7.5 + 2.5) = ½ × 10 = 5 m; shorter piece = ½(7.5 − 2.5) = ½ × 5 = 2.5 m. (Check 5 + 2.5 = 7.5, 5 − 2.5 = 2.5 ✓.)
  4. (c) Length + breadth = half the perimeter = 9 cm, and length − breadth = 10 cm.
  5. Breadth = ½(9 − 10) = −½ cm. A length cannot be negative: the difference of two positive numbers can never be more than their sum. So such a rectangle is impossible.
✅ Answer: (a) add / subtract the equations   (b) 5 m and 2.5 m   (c) breadth would be −½ cm

Marking: (a) 1. (b) both pieces 1 (½ each). (c) equations ½, conclusion with reason ½.

📖 Revise: Ch 1 · Sum and difference; impossible data · Chapter 1 – Pairs of Equations

20.
Two right-triangular flower beds are made in the corners of a courtyard. Bed P has perpendicular sides 1 m and 3 m; bed Q has perpendicular sides 2 m and 2 m. (a) Find the length of the third side (hypotenuse) of each bed. (b) Without adding up either perimeter, show that the perimeter of P is longer than that of Q by √10 − √8 m. (c) Using √10 ≈ 3.162 and √8 ≈ 2.828, find this difference in centimetres.
[3]

Step-by-step solution

  1. (a) By Pythagoras: bed P, hypotenuse² = 1² + 3² = 10 ⇒ hypotenuse √10 m. Bed Q, hypotenuse² = 2² + 2² = 8 ⇒ hypotenuse √8 m.
  2. (b) Perimeter of P = 1 + 3 + √10 = 4 + √10; perimeter of Q = 2 + 2 + √8 = 4 + √8.
  3. Difference = (4 + √10) − (4 + √8) = 4 + √10 − 4 − √8 = √10 − √8 m: the two perpendicular sides add to 4 m in both beds, so only the hypotenuses differ.
  4. (c) √10 − √8 ≈ 3.162 − 2.828 = 0.334 m = 33.4 cm.
✅ Answer: (a) √10 m and √8 m   (b) (4 + √10) − (4 + √8) = √10 − √8   (c) 33.4 cm

Marking: (a) ½ + ½. (b) both perimeters and the subtraction 1. (c) 1.

📖 Revise: Ch 2 · Adding and subtracting perimeters · Ch 2 · Square roots as lengths · Chapter 2 – New Numbers

21.
Draw triangle ABC with BC = 7 cm, CA = 5 cm and AB = 6 cm. Find the midpoints D of BC and E of CA by drawing their perpendicular bisectors. Draw the medians AD and BE and mark their meeting point G. Measure AD and AG, and check that AG = 2 GD.
[3]

Step-by-step solution

  1. Draw BC = 7 cm. With centre B and radius 6 cm draw an arc; with centre C and radius 5 cm draw an arc cutting it at A. Join AB and AC.
  2. Perpendicular bisector (ലംബസമഭാജി) of BC: with centres B and C and the same radius (more than half of BC, say 4.5 cm) draw arcs on both sides of BC; join the two crossing points. It cuts BC at its midpoint D.
  3. In the same way draw the perpendicular bisector of CA (radius say 3.5 cm); it cuts CA at its midpoint E.
  4. Join AD and BE (നടുവരകൾ, medians). Mark their meeting point G – the centroid (മധ്യമകേന്ദ്രം).
  5. Measure: AD ≈ 4.3 cm, AG ≈ 2.8 cm and GD ≈ 1.4 cm. So AG ≈ 2 × GD: G divides the median AD in the ratio 2 : 1 (and BE as well). The third median from C also passes through G.
  6. Finished construction: ABCDEG7 cm5 cm6 cmMeasured: AD ≈ 4.3 cmAG ≈ 2.8 cm, GD ≈ 1.4 cm
✅ Answer: G is the centroid; AD ≈ 4.3 cm, AG ≈ 2.8 cm, GD ≈ 1.4 cm, so AG = 2 GD

Marking: Triangle 1. Perpendicular bisectors giving D and E 1. Medians and G ½. Measurements and the check AG = 2 GD ½ (allow ±0.1 cm).

📖 Revise: Ch 3 · Perpendicular bisector and SSS triangle · Ch 3 · Centroid · Chapter 3 – Parallel Lines

22.
A rectangular plot has length x m and breadth y m (x > 3). It is reshaped: the length is decreased by 3 m and the breadth is increased by 3 m (figure). (a) Use an identity to show that the new area is xy + 3(x − y) − 9. (b) Show that the new area is more than the old area when x − y > 3, the same when x − y = 3 and less when x − y < 3. (c) For a plot 20 m long and 12 m broad, find the new area using (a), and check by direct multiplication.xy33old plotnew plot (dashed)
[3]

Step-by-step solution

  1. (a) New sides: x − 3 and y + 3. Using (x + y)(u − v) = xu − xv + yu − yv:
  2. (y + 3)(x − 3) = yx − 3y + 3x − 9 = xy + 3(x − y) − 9.
  3. (b) New area − old area = 3(x − y) − 9 = 3(x − y − 3).
  4. If x − y > 3 this is positive (new area bigger); if x − y = 3 it is 0 (same area); if x − y < 3 it is negative (new area smaller).
  5. (c) xy = 20 × 12 = 240 and x − y = 8: new area = 240 + 3 × 8 − 9 = 240 + 24 − 9 = 255 m².
  6. Check: the new plot is 17 m × 15 m = 255 m² ✓ (bigger than 240, since x − y = 8 > 3).
✅ Answer: (a) (y + 3)(x − 3) = xy + 3(x − y) − 9   (b) compare 3(x − y) with 9   (c) 255 m²

Marking: (a) 1. (b) 1. (c) 1 (½ using the identity, ½ check).

📖 Revise: Ch 4 · (x + y)(u − v) · Ch 4 · Rectangles with changed sides · Chapter 4 – Multiplication Identities

23.
A straight road is taken as a number line; the bus stand is at 0 and distances are in km, east positive (figure). Bus stop A is at −7 and bus stop B is at 5. (a) Find the distance between A and B. (b) A new stop M is to be equally far from A and B. Write this condition with absolute values and find M. (c) A school bus must always stay less than 4 km from M. Write this with absolute value, then as an inequality, and show it on the number line.−8−7−6−5−4−3−2−10123456ABwesteast
[3]

Step-by-step solution

  1. (a) Distance = bigger − smaller = 5 − (−7) = 5 + 7 = 12 km (= |−7 − 5|).
  2. (b) Distance of x from A is |x − (−7)| = |x + 7|; from B it is |x − 5|. Condition: |x + 7| = |x − 5|.
  3. Equal distances ⇒ x is the midpoint: M = ½(−7 + 5) = ½ × (−2) = −1.
  4. (c) Distance from M less than 4: |x + 1| < 4 (since |x − (−1)| = |x + 1|).
  5. So x lies between −1 − 4 = −5 and −1 + 4 = 3: −5 < x < 3.
  6. On the number line (open circles: −5 and 3 themselves are not allowed): −8−7−6−5−4−3−2−10123456ABwesteastM
✅ Answer: (a) 12 km   (b) |x + 7| = |x − 5|, M = −1   (c) |x + 1| < 4, i.e. −5 < x < 3

Marking: (a) ½. (b) equation ½, M = −1 ½. (c) absolute value form ½, inequality ½, number line ½.

📖 Revise: Ch 9 · Distances; |x − a| &lt; b · Ch 9 · Midpoint · Chapter 9 – Real Numbers

24.
A square photo of side x cm is put in a frame 2 cm wide all around it. (a) Write the area f(x) of the frame alone as a polynomial. (b) What is its degree? Find f(10). (c) Write the length of the diagonal of the outer square as an expression in x. Is it a polynomial? Give the reason.
[3]

Step-by-step solution

  1. (a) Outer square side = x + 2 + 2 = x + 4. Frame area = outer square − photo = (x + 4)² − x².
  2. (x + 4)² = x² + 8x + 16, so f(x) = x² + 8x + 16 − x² = 8x + 16.
  3. (b) The highest power of x is 1: degree 1 (ഒന്നാംകൃതി ബഹുപദം). f(10) = 8 × 10 + 16 = 96 cm² (check: 14² − 10² = 196 − 100 = 96 ✓).
  4. (c) The diagonal of a square of side a is √2 a, so the outer diagonal = √2 (x + 4) = √2 x + 4√2 cm.
  5. x is only multiplied by the fixed number √2 and the fixed number 4√2 is added, so it is a polynomial (degree 1, coefficients √2 and 4√2 – ഗുണകങ്ങൾ).
✅ Answer: (a) f(x) = 8x + 16   (b) degree 1; f(10) = 96 cm²   (c) √2 x + 4√2 – a polynomial

Marking: (a) 1. (b) degree ½, f(10) ½. (c) expression ½, “polynomial” with reason ½.

📖 Revise: Ch 10 · Algebra of measures · Ch 10 · Polynomials and degree · Chapter 10 – Polynomials

Section DAnswer all questions · 4 × 4 = 16 marks
25.
Meera makes a paper kite ABDC for Onam (figure). Triangle ABC is equilateral with side 12 cm. Triangle BCD on the other side of BC has BD = 9 cm and CD = 15 cm. (a) Find the height of triangle ABC from A. (b) Find the area of triangle ABC. (c) Find the area of triangle BCD using Heron’s formula. (d) Find the total area of the kite correct to two decimal places, taking √3 ≈ 1.732.ABCD12 cm9 cm15 cm
[4]

Step-by-step solution

  1. (a) The height of an equilateral triangle bisects the base: (half side)² + h² = side², so h² = 12² − 6² = 144 − 36 = 108.
  2. h = √108 = √(36 × 3) = 6√3 cm (= √3 × half the side).
  3. (b) Area = ½ × 12 × 6√3 = 36√3 cm² (= √3 × 6²).
  4. (c) Sides 12, 9, 15: s = ½(12 + 9 + 15) = 18; s − 12 = 6, s − 9 = 9, s − 15 = 3.
  5. Area = √(18 × 6 × 9 × 3) = √2916 = 54 cm². (Check: 9² + 12² = 81 + 144 = 225 = 15², so BCD is right-angled at B and its area is ½ × 9 × 12 = 54 ✓.)
  6. (d) Total = 36√3 + 54 ≈ 36 × 1.732 + 54 = 62.352 + 54 = 116.352 ≈ 116.35 cm².
✅ Answer: (a) 6√3 cm   (b) 36√3 cm²   (c) 54 cm²   (d) ≈ 116.35 cm²

Marking: (a) 1. (b) 1. (c) s ½, area ½. (d) 1.

📖 Revise: Ch 5 · Equilateral triangles and Heron’s formula · Ch 5 · Simplifying √108 · Chapter 5 – Irrational Numbers

26.
Children are decorating the temple ground for the festival. (a) At 4 p.m. a vertical stick 1.2 m tall casts a shadow 0.8 m long, and at the same time the flagstaff (കൊടിമരം) casts a shadow 6 m long. Draw a rough figure, explain why the two triangles have the same angles, and find the height of the flagstaff. (b) For the pandal, two bamboo poles start from a point O (figure): OA = 2 m, AC = 1 m, OB = 2.4 m, BD = 1.2 m, and the cross piece AB = 1.8 m. (i) Show that the sides of triangle OCD around ∠O are the same multiple of those of triangle OAB. (ii) Find the length of the cross piece CD. (iii) Is CD parallel to AB? Why?OACBD2 m1 m2.4 m1.2 m1.8 m
[4]

Step-by-step solution

  1. (a) The stick and the flagstaff are vertical, so each makes a right angle with the ground.
  2. At the same moment the sun’s rays are parallel, so they make equal angles with the level ground. The two triangles (object, shadow, ray) therefore have the same angles.
  3. Sides opposite equal angles are in the same ratio: heightshadow is the same: h6 = 1.20.8 = 1.5 ⇒ h = 6 × 1.5 = 9 m.
  4. (b)(i) OC = 2 + 1 = 3 m and OD = 2.4 + 1.2 = 3.6 m. OC/OA = 3/2 = 1½ and OD/OB = 3.6/2.4 = 1½. ∠COD = ∠AOB (the same angle at O).
  5. (ii) Two sides change by the same scale 1½ and the angles between them are equal, so by the third way (മൂന്നാംവഴി) the third side also changes by 1½: CD = 1½ × 1.8 = 2.7 m.
  6. (iii) The triangles also have the other angles equal: ∠OCD = ∠OAB. These are corresponding angles made by the line OC with AB and CD, so CD ∥ AB.
✅ Answer: (a) 9 m   (b) (i) 1½ times (ii) CD = 2.7 m (iii) yes

Marking: (a) reason for equal angles 1, height 1. (b)(i) ½ (ii) rule named ½, CD ½ (iii) ½.

📖 Revise: Ch 6 · The third way · Ch 6 · Heights by shadows · Chapter 6 – Similar Triangles

27.
In a Vallam Kali race the straight canal is taken as a number line in metres: the finishing post is O (zero) and east is positive. Two boats, Karichal and Chundan, pass O at the same moment; this moment is time 0, times after it are positive and times before it are negative. Position s = vt. (a) Karichal moves west at 8 m/s. Write its v and find its position 3 s after and 3 s before passing O. (b) With the distributive law (x + y)z = xz + yz, taking x = 10, y = −8, z = −3, explain why (−8) × (−3) must be 24. (c) 4 s before passing O, Chundan was at 36. Find its v. Which boat is faster? (d) Find both positions 5 s after passing O. Which boat is ahead, and by how much?
[4]

Step-by-step solution

  1. (a) West is the negative direction, so v = −8. After 3 s: s = (−8) × 3 = −24 (24 m west of O).
  2. 3 s before: t = −3, s = (−8) × (−3) = 24 (24 m east of O – the boat was coming from the east).
  3. (b) Left side: (10 + (−8)) × (−3) = 2 × (−3) = −6.
  4. Right side: 10 × (−3) + (−8) × (−3) = −30 + (−8) × (−3). For both sides to be equal, −30 + (−8) × (−3) = −6, and −30 + 24 = −6, so (−8) × (−3) = 24.
  5. (c) s = vt gives v = s ÷ t = 36 ÷ (−4) = −9 (ന്യൂനഹരണം: positive ÷ negative is negative). Chundan also moves west, at 9 m/s.
  6. Chundan’s speed 9 m/s is more than Karichal’s 8 m/s, so Chundan is faster.
  7. (d) t = 5: Karichal s = (−8) × 5 = −40; Chundan s = (−9) × 5 = −45.
  8. Both go west and −45 is further west, so Chundan is ahead by −40 − (−45) = −40 + 45 = 5 m.
✅ Answer: (a) v = −8; −24 and 24   (b) −30 + ? = −6 ⇒ 24   (c) v = −9; Chundan   (d) −40 and −45; Chundan by 5 m

Marking: (a) v ½, two positions ½. (b) 1. (c) v ½, faster boat ½. (d) positions ½, lead ½.

📖 Revise: Ch 7 · Time, speed, multiplication and division · Ch 7 · Change of position · Chapter 7 – Negative Numbers

28.
A kolam is painted on a square tile of side 20 cm (figure). At each corner there is a quarter circle of radius 5 cm (shaded grey), painted red. At the centre there is a ring between circles of radii 7 cm and 5 cm (shaded blue), painted yellow; the inner circle and the rest of the tile are left unpainted. (a) Find the total red area, as a multiple of π. (b) Find the yellow area, as a multiple of π. (c) Taking π ≈ 3.14, find the unpainted area. (d) A circle is to be painted with the same area as the yellow ring. Find its radius in simplest form.5 cm7 cm5 cm20 cm20 cm
[4]

Step-by-step solution

  1. (a) Four quarter circles of radius 5 make one full circle: red area = π × 5² = 25π cm².
  2. (b) Ring (വൃത്തവലയം) = big circle − small circle = π × 7² − π × 5² = 49π − 25π = 24π cm².
  3. (c) Square = 20 × 20 = 400 cm². Painted = 25π + 24π = 49π ≈ 49 × 3.14 = 153.86 cm².
  4. Unpainted = 400 − 153.86 = 246.14 cm².
  5. (d) πr² = 24π ⇒ r² = 24 ⇒ r = √24 = √(4 × 6) = 2√6 cm (≈ 4.9 cm).
  6. This is the third side of a right triangle with hypotenuse 7 and one side 5: 7² − 5² = 24 – the textbook’s “ring = circle” idea.
✅ Answer: (a) 25π cm²   (b) 24π cm²   (c) 246.14 cm²   (d) 2√6 cm

Marking: (a) 1. (b) 1. (c) painted area ½, answer ½. (d) r² = 24 ½, 2√6 ½.

📖 Revise: Ch 8 · Area of a circle and rings · Chapter 8 – Measures of Circles

Question numbers by chapter

ChapterA (1)B (2)C (3)D (4)Total
1. Pairs of Equations11119–6
2. New Numbers21220–6
3. Parallel Lines31321–6
4. Multiplication Identities41422–6
5. Irrational Numbers5, 6––256
6. Similar Triangles–15–266
7. Negative Numbers7, 8––276
8. Measures of Circles–16–286
9. Real Numbers91723–6
10. Polynomials101824–6
Total1016181660

Score to grade

Kerala grading scale: A+ 54–60 (90 %+), A 48–53, B+ 42–47, B 36–41, C+ 30–35, C 24–29, D+ 18–23, D 12–17, E below 12. For every question the student lost marks on, the linked guide page explains the idea again.