Class 10 Maths · 3 hours · 80 marks

CBSE Class 10 Maths Sample Paper 1: marking scheme and solutions

Every answer with its marking scheme, the way an examiner awards the marks. Chapters in this paper: 1 Real Numbers; 2 Polynomials; 3 Pair of Linear Equations in Two Variables; 4 Quadratic Equations; 5 Arithmetic Progressions; 6 Triangles; 7 Coordinate Geometry; 8 Introduction to Trigonometry; 9 Some Applications of Trigonometry; 10 Circles; 11 Areas Related to Circles; 12 Surface Areas and Volumes; 13 Statistics; 14 Probability.

Section A (20 × 1 mark)

Question 1 · 1 mark
The line drawn from the eye of an observer to the point of the object being viewed is called the

Answer: (c) By definition this line is the line of sight; the angles of the chapter are measured between it and the horizontal.

Question 2 · 1 mark
Assertion (A): 5x2 − 3 = 0 is a quadratic equation. Reason (R): ax2 + bx + c = 0 is a quadratic equation whenever a ≠ 0, even if b = 0.

Answer: (a) Here a = 5, b = 0, c = −3. Only a = 0 would destroy the x2 term, so R is exactly why A holds.

Question 3 · 1 mark
To draw the line 3x − 2y = 0, a suitable pair of points is

Answer: (d) c = 0, so x = 0 and y = 0 give the same point (0, 0) twice. Another value is needed: x = 2 gives 6 − 2y = 0, y = 3, so (2, 3). (3, 2) gives 9 − 4 = 5 ≠ 0.

Question 4 · 1 mark
Two angles of a triangle are 48° and 67°. A triangle similar to it has angles

Answer: (d) The third angle is 180° − 48° − 67° = 65°. A similar triangle is equiangular with it: 48°, 67°, 65° in any order.

Question 5 · 1 mark
In a right triangle one acute angle is 45° and the hypotenuse is 10√2 cm. The length of each of the other two sides is

Answer: (d) The other acute angle is also 45°, so the legs are equal. Each leg = hypotenuse × sin 45° = 10√2 × 1/√2 = 10 cm.

Question 6 · 1 mark
Which term of the AP 8, 14, 20, . . . is 176?

Answer: (c) 176 = 8 + (n − 1) × 6 gives n − 1 = 28, so n = 29. (28 forgets to add 1.)

Question 7 · 1 mark
For a grouped distribution, Σfᵢ = 40 and Σfᵢxᵢ = 1260. The mean is

Answer: (a) x̄ = Σfᵢxᵢ ÷ Σfᵢ = 1260 ÷ 40 = 31.5.

Question 8 · 1 mark
From a point on level ground 18 m from the foot of a water tower, the angle of elevation of its top is 45°. The height of the water tower is

Answer: (a) tan 45° = height / 18 and tan 45° = 1, so the height is 18 m. (18√3 m comes from wrongly using tan 60°, 6√3 m from tan 30°.)

Question 9 · 1 mark
The probability that the morning bus from Bhopal reaches Indore late is 0.37. The probability that it is not late is

Answer: (b) P(not late) = 1 − P(late) = 1 − 0.37 = 0.63. 1.37 adds instead of subtracting; 0.73 reverses the digits.

Question 10 · 1 mark
If 2a × 3b = 432, where a and b are natural numbers, then (a, b) is

Answer: (b) 432 = 2 × 216 = 2 × 23 × 27 = 24 × 33. The factorisation is unique, so a = 4 and b = 3.

Question 11 · 1 mark
ΔABC and ΔDEF are equiangular, with A, B, C matching D, E, F. AB = 4 cm, BC = 6 cm and DE = 10 cm. Then EF is

Answer: (c) In equiangular triangles corresponding sides are in one ratio: EF/BC = DE/AB = 10/4, so EF = 6 × 2.5 = 15 cm. (12 cm adds 6 instead of multiplying.)

Question 12 · 1 mark
In the triangle with vertices A(2, 3), B(6, 6) and C(−1, 7), the right angle is at

Answer: (a) AB² = 16 + 9 = 25, AC² = 9 + 16 = 25, BC² = 49 + 1 = 50. AB² + AC² = BC², so by the converse of Pythagoras the angle opposite BC, that is ∠A, is 90°.

Question 13 · 1 mark
A circular mirror with centre O and radius 28 cm has a strip cut off along a straight line AB, where ∠AOB = 60°. The area of the strip cut off is. Take π = 22/7.

Answer: (b) Sector = (60/360) × (22/7) × 28² = 1232/3 cm²; triangle (equilateral, side 28) = (√3/4) × 784 = 196√3 cm²; segment = sector − triangle.

Question 14 · 1 mark
Which of the following is a zero of x2 + 2x − 15?

Answer: (a) p(3) = 9 + 6 − 15 = 0, so 3 is a zero. p(5) = 20, p(−3) = −12, p(15) = 240. (x2 + 2x − 15 = (x + 5)(x − 3).)

Question 15 · 1 mark
After simplifying, the equation (x + 3)2 = x(x − 2) + 5 is

Answer: (d) x2 + 6x + 9 = x2 − 2x + 5. The x2 terms cancel, leaving 8x + 4 = 0, which is linear.

Question 16 · 1 mark
In practice at a court in Kochi, a basketball player attempts 80 free throws and scores on 52 of them. The empirical probability that she scores on a free throw is

Answer: (b) Empirical probability = 52/80 = 13/20. 7/20 is the misses; 0.52 treats 52 as a percentage; 20/13 exceeds 1 and is upside down.

Question 17 · 1 mark
To factorise 6x2 + 7x − 3 by splitting the middle term, 7x is written as

Answer: (a) ac = 6 × (−3) = −18 and b = 7. Since ac < 0 the parts have opposite signs: 9 × (−2) = −18 and 9 + (−2) = 7.

Question 18 · 1 mark
A solid cone has base radius 3 cm and height 7 cm. Take π = 22/7. Its volume is

Answer: (b) V = (1/3)πr²h = (1/3) × (22/7) × 9 × 7 = 66 cm³. 198 cm³ forgets the 1/3.

Question 19 · 1 mark
A line l is at a distance of 7 cm from the centre of a circle of radius 6.5 cm. Then l is

Answer: (c) The distance 7 cm is greater than the radius 6.5 cm, so every point of l lies outside the circle: l and the circle have no common point.

Question 20 · 1 mark
Assertion (A): A line drawn parallel to one side of a triangle, meeting the other two sides at distinct points, divides them in the same ratio. Reason (R): Any line meeting two sides of a triangle divides them in the same ratio.

Answer: (c) A is the Basic Proportionality Theorem. R is false: only a line parallel to the third side does this; a slanting line generally cuts the sides in different ratios.

Section B (5 × 2 marks)

Question 21 · 2 marks
A circle touches all four sides of a quadrilateral PQRS in which PQ = (2x + 3) cm, QR = 3x cm, RS = 9 cm and SP = 10 cm. Find x and the side QR.

For a quadrilateral whose sides touch a circle, PQ + RS = QR + SP: 2x + 3 + 9 = 3x + 10, so x = 2 and QR = 6 cm.

  • PQ + RS = QR + SP written (1)
  • x = 2, QR = 6 cm (1)
Question 22 · 2 marks
If tan 2θ = √3, where 0° < 2θ < 90°, find θ. Hence show that tan 2θ is not equal to 2 tan θ.

tan 2θ = √3 = tan 60°. As 2θ is acute, where tan takes each value only once, 2θ = 60° and θ = 30°. Then 2 tan θ = 2 tan 30° = 2/√3, while tan 2θ = tan 60° = √3 = 3/√3. Since 2/√3 ≠ 3/√3, tan 2θ ≠ 2 tan θ.

  • 2θ = 60°, so θ = 30° (1)
  • 2 tan 30° = 2/√3 against tan 60° = √3, so they differ (1)
Question 23 · 2 marks
A storage tank is a cylinder 10 m long with a hemisphere of radius 7 m at each end. Take π = 22/7. Find the outer surface area of the tank.
Or A boiler is a cylinder 6 m long with a hemisphere of radius 3.5 m at each end. Take π = 22/7. Find the outer surface area of the boiler.

Surface = CSA of cylinder + CSA of two hemispheres = 2πrh + 4πr² = 440 + 616 = 1056 m².

  • uses 2πrh + 2 × 2πr², with no flat faces (1)
  • gets 1056 m² (1)
Or:

Surface = 2πrh + 4πr² = 2(22/7)(3.5)(6) + 4(22/7)(12.25) = 132 + 154 = 286 m².

  • uses 2πrh + 2 × 2πr², with no flat faces (1)
  • gets 286 m² (1)
Question 24 · 2 marks
An AP has first term −12, common difference 5 and last term 108. Find the number of terms and its middle term.

108 = −12 + (n − 1) × 5 gives n − 1 = 24, n = 25. The middle term is the 13th: a₁₃ = −12 + 12 × 5 = 48.

  • n = 25 (1)
  • middle (13th) term = 48 (1)
Question 25 · 2 marks
In a grouped distribution with class size 5, the modal class is 25-30 with frequency 18. The class before it has frequency 12 and the class after it has frequency 9. Find the mode.

l = 25, f₁ = 18, f₀ = 12, f₂ = 9, h = 5. Mode = 25 + ((18 − 12) ÷ (36 − 12 − 9)) × 5 = 25 + (6 ÷ 15) × 5 = 27.

  • substitutes l, f₁, f₀, f₂, h correctly (1)
  • mode = 27 (1)

Section C (6 × 3 marks)

Question 26 · 3 marks
A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is (i) a black card that is not a face card, (ii) a number card (2 to 10) of hearts, (iii) neither a club nor a king.

(i) 26 black cards less 6 black face cards = 20: P = 20/52 = 5/13. (ii) 9 cards: P = 9/52. (iii) Clubs 13 and kings of the other three suits 3 give 16, so P(club or king) = 16/52 and P(neither) = 1 − 16/52 = 36/52 = 9/13.

  • P(black non-face) = 5/13 (1)
  • P(number card of hearts) = 9/52 (1)
  • P(neither) = 9/13 (1)
Question 27 · 3 marks
α and β are the zeroes of ax2 + bx + c, a ≠ 0. Starting from ax2 + bx + c = k(x − α)(x − β), show that α + β = −ba and αβ = ca.

Since α and β are zeroes, x − α and x − β are factors, so ax2 + bx + c = k(x − α)(x − β) = kx2 − k(α + β)x + kαβ. Comparing coefficients: a = k, b = −k(α + β), c = kαβ. Hence α + β = −bk = −ba and αβ = ck = ca.

  • expands k(x − α)(x − β) = kx2 − k(α + β)x + kαβ (1)
  • compares coefficients: a = k, b = −k(α + β), c = kαβ (1)
  • concludes α + β = −ba, αβ = ca (1)
Question 28 · 3 marks
A circular window with centre O and radius 30 cm has a straight wooden bar AB fixed across it, where ∠AOB = 120°. Find the length of the bar and the area of the smaller glass pane. (Take π = 3.14 and √3 = 1.73.)

Sector = (120/360) × 3.14 × 900 = 942 cm². Draw OM ⊥ AB; ∠AOM = 60°, so OM = 30 cos 60° = 15 cm and AM = 30 sin 60° = 15√3 cm, so the bar AB = 30√3 = 51.9 cm. Triangle = ½ × 30√3 × 15 = 225√3 = 389.25 cm². Smaller pane = minor segment = 942 − 389.25 = 552.75 cm².

  • OM = 15 cm and bar AB = 30√3 = 51.9 cm (1)
  • sector 942 cm² and triangle 389.25 cm² (1)
  • pane 552.75 cm² (1)
Question 29 · 3 marks
A mobile mast CM stands on the flat roof of a building BC. From a point P on the ground 12 m from the foot B of the building, the angles of elevation of the foot C and the top M of the mast are 30° and 60°. Find the height of the building and the height of the mast. (Use √3 ≈ 1.73.)
30° 60° P B C M 12 m

In right ΔPBC, BC = 12 tan 30° = 12/√3 = 4√3 m ≈ 6.92 m. In right ΔPBM, BM = 12 tan 60° = 12√3 m. Mast CM = BM − BC = 12√3 − 4√3 = 8√3 m ≈ 13.84 m.

  • building BC = 12 tan 30° = 4√3 ≈ 6.92 m (1)
  • BM = 12 tan 60° = 12√3 m (1)
  • mast = 8√3 ≈ 13.84 m (1)
Question 30 · 3 marks
In ΔABC, D is a point on AB. The line through D parallel to BC meets AC at E, and the line through E parallel to AB meets BC at F, as in the figure. Prove that ADDB = BFFC.
A B C D E F

In ΔABC, DE ∥ BC, so by BPT AD/DB = AE/EC ...(1). In ΔCAB, EF ∥ AB with E on CA and F on CB, so by BPT CE/EA = CF/FB, that is AE/EC = BF/FC ...(2). From (1) and (2), AD/DB = BF/FC.

  • AD/DB = AE/EC from DE ∥ BC (1)
  • CE/EA = CF/FB from EF ∥ AB (1)
  • combines to AD/DB = BF/FC (1)
Question 31 · 3 marks
Solve graphically: x + y = 5 and 2x − y = 4.

x + y = 5: (0, 5), (5, 0), (2, 3). 2x − y = 4: (0, −4), (2, 0), (3, 2). Plot and join each set. The lines cross at (3, 2), so x = 3, y = 2. Check: 3 + 2 = 5, 6 − 2 = 4.

  • table of at least two points for each line (1)
  • both lines drawn correctly (1)
  • reads the crossing point x = 3, y = 2 (1)

Section D (4 × 5 marks)

Question 32 · 5 marks
The daily earnings of 50 vegetable vendors at a market in Mumbai are given below. (a) Find the mean by the step deviation method. (b) Find the mode. (c) Compare the two.
Earnings (₹)Number of vendors
200-3004
300-4008
400-50014
500-60012
600-7008
700-8004
Or The lengths of 40 fish in a catch landed at Kochi harbour are given below. (a) Find the mean by the step deviation method. (b) Find the mode. (c) Compare the two.

(a) a = 450, h = 100; uᵢ = −2, −1, 0, 1, 2, 3; fᵢuᵢ = −8, −8, 0, 12, 16, 12; Σfᵢuᵢ = 24, Σfᵢ = 50. x̄ = 450 + (24 ÷ 50) × 100 = ₹498. (b) Modal class 400-500: l = 400, f₁ = 14, f₀ = 8, f₂ = 12. Mode = 400 + ((14 − 8) ÷ (28 − 8 − 12)) × 100 = ₹475. (c) The mean is greater than the mode; the two are different measures and the mode need not equal the mean.

  • uᵢ and fᵢuᵢ columns, Σfᵢuᵢ = 24 (1)
  • mean = 498 (1)
  • modal class 400-500 with l, f₁, f₀, f₂ (1)
  • mode = 475 (1)
  • compares: the mean and the mode differ, either may be larger (1)
Or:

(a) a = 150, h = 20; uᵢ = −2, −1, 0, 1, 2, 3; fᵢuᵢ = −6, −6, 0, 10, 10, 12; Σfᵢuᵢ = 20, Σfᵢ = 40. x̄ = 150 + (20 ÷ 40) × 20 = 160 mm. (b) Modal class 140-160: l = 140, f₁ = 12, f₀ = 6, f₂ = 10. Mode = 140 + ((12 − 6) ÷ (24 − 6 − 10)) × 20 = 155 mm. (c) The mean is greater than the mode; the two are different measures and the mode need not equal the mean.

  • uᵢ and fᵢuᵢ columns, Σfᵢuᵢ = 20 (1)
  • mean = 160 mm (1)
  • modal class 140-160 with l, f₁, f₀, f₂ (1)
  • mode = 155 mm (1)
  • compares: the mean and the mode differ, either may be larger (1)
Question 33 · 5 marks
(a) Prove that the lengths of the two tangents drawn from an external point to a circle are equal. (b) Show also that the line joining the external point to the centre bisects the angle between the tangents. (c) For a circle of radius 33 cm whose centre O is 65 cm from the external point P, find the length of each tangent and the distance from P to the nearest point of the circle.
Or (a) PA and PB are tangents from P to a circle with centre O. Prove that the centre lies on the bisector of ∠APB. (b) Deduce that PA = PB. (c) If the radius is 11 cm and PA = 60 cm, find OP and the distance from P to the nearest point of the circle.

(a) Let PA, PB be the tangents and O the centre. Join OA, OB, OP. ∠OAP = ∠OBP = 90°, OA = OB and OP is common, so ΔOAP ≅ ΔOBP (RHS) and PA = PB (CPCT). (b) From the same congruence, ∠APO = ∠BPO. (c) PA = PB = √(65² − 33²) = √3136 = 56 cm. The nearest point C lies on PO: PC = 65 − 33 = 32 cm.

  • joins OA, OB, OP; ∠OAP = ∠OBP = 90° (1)
  • ΔOAP ≅ ΔOBP (RHS), PA = PB (1)
  • ∠APO = ∠BPO (CPCT) (1)
  • PA = 56 cm (1)
  • PC = 32 cm (1)
Or:

(a) Join OA, OB, OP. ∠OAP = ∠OBP = 90°, OA = OB, OP common, so ΔOAP ≅ ΔOBP (RHS). Then ∠APO = ∠BPO, so O lies on the bisector of ∠APB. (b) From the same congruence, PA = PB (CPCT). (c) OP = √(60² + 11²) = √3721 = 61 cm. The nearest point lies on PO: 61 − 11 = 50 cm.

  • joins OA, OB, OP; right angles at A and B (1)
  • ΔOAP ≅ ΔOBP, ∠APO = ∠BPO (1)
  • PA = PB (1)
  • OP = 61 cm (1)
  • nearest distance 50 cm (1)
Question 34 · 5 marks
A science museum in Kolkata sells adult and student tickets. On Saturday, 40 adult and 60 student tickets brought in ₹5,200. On Sunday, 50 adult and 90 student tickets brought in ₹7,100. (a) Form the pair of equations. (b) Find the price of each ticket. (c) Find what a group of 12 adults and 30 students pays.
Or At a plant nursery in Pune, 5 rose plants and 8 tulsi plants cost ₹490, while 3 rose plants and 10 tulsi plants cost ₹450. (a) Form the pair of equations. (b) Find the price of each plant. (c) Find the cost of 4 rose and 6 tulsi plants.

(a) 40x + 60y = 5200 and 50x + 90y = 7100, i.e. 2x + 3y = 260 and 5x + 9y = 710. (b) Multiply the first by 3: 6x + 9y = 780. Subtract the second: x = 70, then 3y = 260 − 140, y = 40. Adult ₹70, student ₹40. Check: 2800 + 2400 = 5200, 3500 + 3600 = 7100. (c) 12 × 70 + 30 × 40 = 840 + 1200 = ₹2,040.

  • both equations formed (1)
  • simplified and coefficients matched (1)
  • adult ticket ₹70 (1)
  • student ticket ₹40, checked against both days (1)
  • group pays ₹2,040 (1)
Or:

(a) 5x + 8y = 490 and 3x + 10y = 450. (b) Multiply by 3 and 5: 15x + 24y = 1470, 15x + 50y = 2250. Subtract: 26y = 780, y = 30; 5x = 490 − 240, x = 50. Rose ₹50, tulsi ₹30. Check: 150 + 300 = 450. (c) 4 × 50 + 6 × 30 = 200 + 180 = ₹380.

  • both equations formed (1)
  • x-coefficients matched and subtracted (1)
  • tulsi ₹30 (1)
  • rose ₹50, checked against the second purchase (1)
  • cost ₹380 (1)
Question 35 · 5 marks
In ΔPQR, right-angled at Q, PQ = 12 cm and sin R = 35 (see the figure). (a) Find PR and QR. (b) Find cos P and tan P. (c) Verify that 1 + tan²R = sec²R.
Q R P 12 cm
Or In ΔABC, right-angled at C, AC = 20 cm and cos A = 45 (see the figure). (a) Find AB and BC. (b) Find sin B and tan B. (c) Verify that cot²A + 1 = cosec²A.
C A B 20 cm

(a) sin R = PQ/PR = 3/5, so PR = 12 × 5/3 = 20 cm. QR = √(20² − 12²) = √256 = 16 cm. (b) For angle P the adjacent side is PQ and the opposite side is QR: cos P = 12/20 = 3/5, tan P = 16/12 = 4/3. (c) tan R = PQ/QR = 12/16 = 3/4 and sec R = PR/QR = 20/16 = 5/4. 1 + tan²R = 1 + 9/16 = 25/16 = (5/4)² = sec²R.

  • PR = 20 cm (1)
  • QR = 16 cm (1)
  • cos P = 3/5 and tan P = 4/3 (1)
  • tan R = 3/4 and sec R = 5/4 (1)
  • both sides equal 25/16 (1)
Or:

(a) cos A = AC/AB = 4/5, so AB = 20 × 5/4 = 25 cm. BC = √(25² − 20²) = √225 = 15 cm. (b) For angle B the opposite side is AC and the adjacent side is BC: sin B = 20/25 = 4/5, tan B = 20/15 = 4/3. (c) cot A = AC/BC = 20/15 = 4/3 and cosec A = AB/BC = 25/15 = 5/3. cot²A + 1 = 16/9 + 1 = 25/9 = (5/3)² = cosec²A.

  • AB = 25 cm (1)
  • BC = 15 cm (1)
  • sin B = 4/5 and tan B = 4/3 (1)
  • cot A = 4/3 and cosec A = 5/3 (1)
  • both sides equal 25/9 (1)

Section E (3 × 4 marks)

Question 36 · 4 marks
A coastguard drone D hovers 90 m above the sea, directly above the point F on the water. Its camera picks up a fishing boat A on one side at an angle of depression of 60°, and a second boat B on the other side, in the same straight line with F, at an angle of depression of 45°, as in the figure. The crew on the boats use these readings to fix their positions. (Use √3 ≈ 1.73.)
60° 45° D F A B 90 m
(a) The angle of elevation of the drone, as seen from boat A, is [1]

Answer: (c) Angle FDA = 90° − 60° = 30°, so angle DAF = 180° − 90° − 30° = 60°; it equals the depression (alternate angles).

(b) The distance FA is [1]

Answer: (b) tan 60° = 90/FA, so FA = 90/√3 = 30√3 m.

(c) Find the distance between the boats A and B. [2]
Or A third boat C, on the same side of F as A and in line with it, is seen at an angle of depression of 30°. Find the distance AC.

FB = 90 cot 45° = 90 m. F lies between the boats, so AB = FA + FB = 30√3 + 90 m ≈ 51.9 + 90 = 141.9 m.

  • FB = 90 m (1)
  • adds, since F is between: AB = 90 + 30√3 ≈ 141.9 m (1)
Or:

FC = 90 cot 30° = 90√3 m. A and C are on the same side, so AC = FC − FA = 90√3 − 30√3 = 60√3 m ≈ 103.8 m.

  • FC = 90√3 m (1)
  • subtracts: AC = 60√3 ≈ 103.8 m (1)
Question 37 · 4 marks
A gram panchayat near Hassan is fencing a rectangular plot for a community vegetable garden. The surveyor notes that the length of the plot is 6 m more than its breadth and that its area is 160 m². The panchayat needs the dimensions to order the fencing wire and to plan the beds. Let the breadth of the plot be x metres.
(a) The quadratic equation satisfied by x is [1]

Answer: (a) Length x + 6, area x(x + 6) = 160, so x2 + 6x − 160 = 0.

(b) The discriminant of this equation is [1]

Answer: (a) D = 62 − 4 × 1 × (−160) = 36 + 640 = 676.

(c) Find the length and breadth of the plot. [2]
Or Find the length of wire needed to fence the plot once around.

x = (−6 ± √676)/2 = (−6 ± 26)/2, so x = 10 or x = −16. A breadth cannot be negative, so the breadth is 10 m and the length is 16 m.

  • solves to x = 10 or −16 (1)
  • rejects −16; breadth 10 m, length 16 m (1)
Or:

x2 + 6x − 160 = 0 gives (x + 16)(x − 10) = 0; breadth 10 m (−16 rejected), length 16 m. Perimeter = 2(16 + 10) = 52 m.

  • finds breadth 10 m, length 16 m, rejecting −16 (1)
  • perimeter 52 m (1)
Question 38 · 4 marks
A printing unit in Surat makes a triangular logo ABC for a cooperative dairy and a large banner PQR of exactly the same shape. The banner's longest side RP had not been measured when the order sheet below was written. A third piece, a sticker XYZ, was cut by a new worker from his own measurements.
ItemSides
Logo ABCAB = 4 cm, BC = 6 cm, CA = 7 cm
Banner PQRQR = 12 cm, PQ = 18 cm, RP = ?
Sticker XYZXY = 8 cm, YZ = 12 cm, ZX = 13 cm
(a) The scale factor of the banner to the logo is [1]

Answer: (a) Matching by size, QR ↔ AB and PQ ↔ BC: 12/4 = 18/6 = 3.

(b) The length of RP is [1]

Answer: (a) RP matches the longest side CA: RP = 3 × 7 = 21 cm.

(c) Write the similarity between the logo and the banner with the vertices in the correct order, and name the criterion. [2]
Or Is the sticker XYZ similar to the logo ABC? Justify.

AB/QR = BC/PQ = CA/RP = 1/3, so the triangles are similar by SSS. B lies on AB and BC, matching Q on QR and PQ; A matches R and C matches P. So ΔABC ~ ΔRQP.

  • three equal ratios, SSS (1)
  • ΔABC ~ ΔRQP (1)
Or:

Match by size: XY/AB = 8/4 = 2 and YZ/BC = 12/6 = 2, but ZX/CA = 13/7 ≠ 2. The sides are not in one ratio, so the sticker is not similar to the logo.

  • computes the three ratios (1)
  • 13/7 ≠ 2, so not similar (1)
Take the free Class 10 Maths testSee the Class 10 flashcards