CBSE Class 10 Mathematics Basic Maths Short and Long Answer Questions Assignment Set 17

Answer: Let the point be (x, 0). Then (x − 2)² + 25 = (x + 2)² + 81, so −4x + 29 = 4x + 85, giving x = −7. The point is (−7, 0).

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Equate the squares of the distances to avoid square roots.

Answer: = 1/(1/√2) + 1/2 = √2 + 1/2 = \( \dfrac{2\sqrt{2}+1}{2} \).

Teacher's Note:
1. Use sin 90° = 1, cos 45° = 1/√2 and cosec 30° = 2.
2. Combine into a single fraction at the end.

Answer: Let the ratio be k : 1. Then (6k + 2)/(k + 1) = 4, so k = 1. Using the y-coordinate, m = (−3 × 1 + 3 × 1)/2 = 0. The ratio is 1 : 1 (P is the mid-point) and m = 0.

Teacher's Note:
1. Use the x-coordinate to find the ratio and the y-coordinate to find m.
2. A ratio of 1 : 1 means P is the mid-point.

Answer: The diagonal of the square equals the diameter = 16 cm. Area = (diagonal)²/2 = 256/2 = 128 cm².

Teacher's Note:
1. The diagonal of an inscribed square is the diameter.
2. Area of a square = d²/2.

Answer: Minor sector angle = 45°, so the major sector angle = 315°. Area = (315/360) × (22/7) × 49 = (7/8) × 154 = 134.75 cm².

Teacher's Note:
1. Subtract the minor angle from 360° to get the major angle.
2. Use area = (θ/360°) × πr².

Answer: Without x, the values 15, 20, 25 and 18 each occur 3 times. For 20 to be the unique mode it must occur more often, so x = 20 (making its frequency 4).

Teacher's Note:
1. Count the frequency of each value first.
2. The mode is the value with the highest frequency.

Answer: 92690 = 7378 × 12 + 4154; 7378 = 4154 × 1 + 3224; 4154 = 3224 × 1 + 930; 3224 = 930 × 3 + 434; 930 = 434 × 2 + 62; 434 = 62 × 7 + 0. So HCF(92690, 7378) = 62. Now 7161 = 62 × 115 + 31 and 62 = 31 × 2 + 0, so the required HCF is 31.

Teacher's Note:
1. Find the HCF of two numbers first, then of that result and the third number.
2. The algorithm stops when the remainder is 0.

Answer: Let the zeroes be α and 7α. Sum: 8α = 8/3, so α = 1/3. Product: 7α² = (2k + 1)/3, so 7/9 = (2k + 1)/3, giving 2k + 1 = 7/3 and k = 2/3.

Teacher's Note:
1. Use the sum of zeroes to find α first.
2. Then use the product of zeroes to find k.

Answer: From the second equation, y = 6 − 4x. Substituting: 3x + 12 − 8x = 7, so −5x = −5 and x = 1. Then y = 6 − 4 = 2. Solution: x = 1, y = 2.

Teacher's Note:
1. Express one variable in terms of the other from the simpler equation.
2. Verify by putting the values in the first equation.

Answer: The diagonals AC and BD bisect each other. Mid-point of AC = (1, (1 + b)/2); mid-point of BD = ((a + 1)/2, 1). So a = 1 and b = 1. Then AB = √(9 + 1) = √10, BC = √(9 + 1) = √10, CD = √(9 + 1) = √10 and DA = √(9 + 1) = √10. All sides are √10 units.

Teacher's Note:
1. Equate the mid-points of the two diagonals.
2. All four sides come out equal, so it is also a rhombus.

Answer: The numbers form the AP 105, 112, …, 994. Here a = 105, d = 7 and l = 994. 994 = 105 + (n − 1)7 gives n − 1 = 127, so n = 128.

Teacher's Note:
1. Find the first and last three-digit multiples of 7.
2. Use the nth term formula to count them.

Answer: Let triangles ABC and PQR be similar, so area(ABC)/area(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². The areas are equal, so the ratio is 1, hence AB = PQ, BC = QR and CA = RP. By SSS, the triangles are congruent.

Teacher's Note:
1. The ratio of the areas of similar triangles equals the square of the ratio of corresponding sides.
2. Equal sides give congruence by SSS.

Answer: By the section formula: −1 = (3x + 4 × 2)/7, so 3x + 8 = −7 and x = −5. Also 2 = (3y + 4 × 5)/7, so 3y + 20 = 14 and y = −2. Hence B is (−5, −2).

Teacher's Note:
1. Treat x and y as unknown coordinates of B and form two equations.
2. In the section formula, the first point's coordinates are multiplied by n.

Answer: Let the height be h and the angles be θ and (90° − θ). Then tan θ = h/4 and tan(90° − θ) = cot θ = h/9. Since tan θ × cot θ = 1, (h/4)(h/9) = 1, so h² = 36 and h = 6 m.

Teacher's Note:
1. Use tan θ for the nearer point and cot θ for the farther point.
2. The product tan θ cot θ = 1 removes the angle.

Answer: Let PQ = x, so PR = x + 1. Then (x + 1)² = x² + 81, so 2x + 1 = 81 and x = 40; PR = 41. sin R = 40/41 and cos R = 9/41, so sin R + cos R = 49/41.

Teacher's Note:
1. Use the Pythagoras theorem to find the unknown sides.
2. sin R = PQ/PR and cos R = QR/PR.

Answer: LHS = (cos θ/sin θ − 1/sin θ)² = (cos θ − 1)²/sin²θ = (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)] = (1 − cos θ)/(1 + cos θ) = RHS.

Teacher's Note:
1. Convert everything into sin and cos.
2. Write sin²θ as (1 − cos θ)(1 + cos θ).

Answer: r = 12 m, h = 3.5 m. Volume = (1/3)π r² h = (1/3) × (22/7) × 144 × 3.5 = 528 m³. Slant height l = √(144 + 12.25) = √156.25 = 12.5 m. Canvas = curved surface area = π r l = (22/7) × 12 × 12.5 = 471.43 m² (approximately).

Teacher's Note:
1. The canvas covers only the curved surface of the cone.
2. Find the slant height using l² = r² + h².

Answer: The largest hemisphere has diameter = side of the cube = 10 cm, so r = 5 cm. TSA = 6 × 100 − πr² + 2πr² = 600 + πr² = 600 + 3.14 × 25 = 678.5 cm². Cost = 678.5/100 × 5 = ₹33.93 (approximately).

Teacher's Note:
1. The hemisphere covers a circle of the top face, which must be subtracted and replaced by its curved surface.
2. Net change is +πr².

Answer: Let the first prize be a. Sum of 7 terms: (7/2)[2a + 6(−20)] = 700, so 2a − 120 = 200 and a = 160. The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

Teacher's Note:
1. The prizes form an AP with d = −20.
2. Check that the seven values add up to 700.

Answer: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides the two sides in the same ratio. Proof: In triangle ABC let DE || BC with D on AB and E on AC. Join BE and CD and draw DM ⟂ AC and EN ⟂ AB. Area(ADE)/Area(BDE) = (½ AD × EN)/(½ DB × EN) = AD/DB. Also Area(ADE)/Area(CED) = (½ AE × DM)/(½ EC × DM) = AE/EC. Triangles BDE and CED are on the same base DE and between the same parallels, so their areas are equal. Hence AD/DB = AE/EC.

Teacher's Note:
1. Compare areas of triangles with a common height.
2. Triangles on the same base and between the same parallels are equal in area.

Answer: Let triangle ABC be right angled at B. Draw BD ⟂ AC. Triangles ADB and ABC are similar (AA), so AD/AB = AB/AC, giving AB² = AD × AC. Triangles BDC and ABC are similar, so DC/BC = BC/AC, giving BC² = DC × AC. Adding: AB² + BC² = AC(AD + DC) = AC × AC = AC².

Teacher's Note:
1. Draw the perpendicular from the right-angle vertex to the hypotenuse.
2. Use the similarity of the two smaller triangles with the whole triangle.

Answer: Let the circle touch the legs at P and Q, and the hypotenuse at R, with the right angle at the vertex B. OPBQ is a square, so BP = BQ = r. Tangent lengths from a point are equal, so the other two tangent lengths are (a − r) and (b − r), and these sum to the hypotenuse: (a − r) + (b − r) = c. Hence 2r = a + b − c, so r = (a + b − c)/2.

Teacher's Note:
1. Use the fact that tangents from an external point are equal in length.
2. The centre and the right-angle vertex form a square with the points of contact.

Answer: Let the speed of the stream be x km/h. 24/(18 − x) − 24/(18 + x) = 1, so 24(2x) = 324 − x², i.e. x² + 48x − 324 = 0. Then (x + 54)(x − 6) = 0, so x = 6 (rejecting −54). The speed of the stream is 6 km/h.

Teacher's Note:
1. Upstream speed is boat speed minus stream speed, and downstream is plus.
2. Reject the negative root.

Answer: Class marks: 6, 8, 10, 12, 14, 16, 18. Σfx = 420 + 960 + 320 + 1200 + 630 + 448 + 90 = 4068, so mean = 4068/400 = 10.17 years. N/2 = 200. Cumulative frequencies: 70, 190, 222, … so the median class is 9–11 with l = 9, cf = 190, f = 32, h = 2. Median = 9 + (200 − 190)/32 × 2 = 9.625 years.

Teacher's Note:
1. The median class is the first class whose cumulative frequency is at least N/2.
2. Check that the frequencies add to 400 before starting.

Answer: Total outcomes = 36. (i) Doublets: (1,1) to (6,6) = 6, so P = 6/36 = 1/6. (ii) 5 on at least one die: 11 outcomes, so P = 11/36. (iii) Sum 9: (3,6), (4,5), (5,4), (6,3) = 4 outcomes, so P = 4/36 = 1/9.

Teacher's Note:
1. List the favourable outcomes systematically.
2. For part (ii), count 6 + 6 and subtract the common outcome (5,5).