CBSE Class 10 Mathematics Triangles, Circles, Statistics, Quadratic Equations, AP and Probability Assignment Set 13

Answer: (d) 4. By the basic proportionality theorem \( \dfrac{x-2}{3x} = \dfrac{x+2}{9x} \), so 3(x − 2) = x + 2, giving x = 4.

Teacher's Note:
1. The line parallel to one side divides the other two sides in the same ratio.
2. Cross-multiply after cancelling x.

Answer: (a) 6.5 m. Length = \( \sqrt{2.5^2 + 6^2} = \sqrt{42.25} = 6.5 \) m.

Teacher's Note:
1. The wall, ground and ladder form a right triangle.
2. Apply the Pythagoras theorem.

Answer: (b) median. The median class is found from the cumulative frequencies.

Teacher's Note:
1. The median class is the class in which N/2 lies in the cumulative column.
2. Mean and mode do not need it.

Answer: (b) 7. Mean = (5x + 20)/5 = x + 4 = 11, so x = 7.

Teacher's Note:
1. Add the five terms and divide by 5.
2. The deviations 0, 2, 4, 6, 8 add to 20.

Answer: (d) mean = median = mode. In a symmetrical distribution all three measures coincide.

Teacher's Note:
1. Symmetry places the centre at the same point for all three averages.
2. The empirical relation is for skewed data.

Answer: (d) √119 cm. The radius OP is perpendicular to the tangent, so \( PQ = \sqrt{12^2 - 5^2} = \sqrt{119} \).

Teacher's Note:
1. The tangent is perpendicular to the radius at the point of contact.
2. Use the Pythagoras theorem in triangle OPQ.

Answer: (a) none. Every line through an interior point cuts the circle at two points.

Teacher's Note:
1. A tangent touches the circle at only one point.
2. No such line passes through an interior point.

Answer: (b) AB² + QB². The tangent TB is perpendicular to the diameter AB at B, so in the right triangle ABQ, QA² = AB² + QB².

Teacher's Note:
1. The angle between a tangent and the radius or diameter through the point of contact is 90°.
2. Then apply Pythagoras.

Answer: Let ABCD be the parallelogram with sides touching the circle. Tangent lengths from a vertex are equal, so AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD. Adjacent sides are equal, so ABCD is a rhombus.

Teacher's Note:
1. Add the four pairs of equal tangents to get AB + CD = AD + BC.
2. Then use the property of opposite sides.

Answer: Let ABC be right-angled at B, and draw BD perpendicular to AC. Triangles ADB and ABC are similar, so AB² = AD × AC. Triangles BDC and ABC are similar, so BC² = DC × AC. Adding, AB² + BC² = AC(AD + DC) = AC².

Teacher's Note:
1. The altitude to the hypotenuse creates two triangles similar to the whole triangle.
2. Write the proportional sides carefully.

Answer: Total N = 40, so N/2 = 20. Cumulative frequencies are 5, 15, 30, 36, 40, so the median class is 20-30 with l = 20, cf = 15, f = 15, h = 10. Median = 20 + (20 − 15)/15 × 10 = 23.33 (approximately).

Teacher's Note:
1. Locate the class whose cumulative frequency first exceeds N/2.
2. This value is also where the two ogives meet.

Answer: Let PA and PB be tangents from P to a circle with centre O. OA ⊥ PA and OB ⊥ PB. In right triangles OAP and OBP, OA = OB (radii), OP is common and the angles at A and B are 90°. So the triangles are congruent by RHS and PA = PB.

Teacher's Note:
1. Join the centre to the point of contact to get right triangles.
2. RHS congruence finishes the proof.

Answer: In right triangle ABL, BL² = AB² + AL² = AB² + AC²/4. In right triangle ACM, CM² = AC² + AM² = AC² + AB²/4. Adding, BL² + CM² = (5/4)(AB² + AC²) = (5/4)BC². So 4(BL² + CM²) = 5BC².

Teacher's Note:
1. L and M are mid-points, so AL = AC/2 and AM = AB/2.
2. Pythagoras in each small triangle does the rest.

Answer: Total: f₁ + 5 + 9 + 12 + f₂ + 3 + 2 = 40, so f₁ + f₂ = 9. The median 32.5 lies in the class 30-40, with cumulative frequency before it = f₁ + 14. Median = 30 + (20 − f₁ − 14)/12 × 10 = 32.5, so (6 − f₁)/12 = 1/4 and f₁ = 3. Then f₂ = 6.

Teacher's Note:
1. Use the total to get one equation and the median formula for the other.
2. Check that 3 + 14 = 17 is below 20.

Answer: The statement is false. For example x² + 1 = 0 has discriminant b² − 4ac = −4 < 0, so it has no real root.

Teacher's Note:
1. A quadratic has real roots only when the discriminant is non-negative.
2. One counter-example is enough to disprove the statement.

Answer: Yes. Take \( \sqrt{2}(x-1)(x-2) = \sqrt{2}x^2 - 3\sqrt{2}x + 2\sqrt{2} = 0 \). Its coefficients \( \sqrt{2}, -3\sqrt{2}, 2\sqrt{2} \) are distinct irrational numbers, while the roots 1 and 2 are rational.

Teacher's Note:
1. Multiply a quadratic with rational roots by an irrational constant.
2. The roots stay the same.

Answer: Yes. The equation x² − 2 = 0 has rational coefficients, but its roots are \( \pm\sqrt{2} \), which are irrational.

Teacher's Note:
1. The discriminant here is 8, which is positive but not a perfect square.
2. Roots are therefore real and irrational.

Answer: a + 50d = 0 and a + 10d = 8. Subtracting, 40d = −8, so \( d = -\dfrac{1}{5} \) and a = 10. For the last term, \( 10 + (n-1)\left(-\dfrac{1}{5}\right) = -\dfrac{1}{5} \), so n − 1 = 51 and n = 52.

Teacher's Note:
1. Solve the two term equations together for a and d.
2. Then use the last-term formula for n.

Answer: Let the numbers be a − d, a, a + d. The sum 3a = 3 gives a = 1. Product: (1 − d)(1)(1 + d) = 1 − d² = −35, so d² = 36 and d = ±6. The numbers are −5, 1, 7 (or 7, 1, −5).

Teacher's Note:
1. Writing the terms as a − d, a, a + d removes the middle variable from the sum.
2. Both signs of d give the same set.

Answer: (a + 14d) − (a + 10d) = 4d = 16, so d = 4. Then a + 7d = 131 gives a = 131 − 28 = 103. The AP is 103, 107, 111, 115, ...

Teacher's Note:
1. The difference of two terms depends only on d.
2. Find d first and then a.

Answer: For consecutive terms, 2(k + 5) = (2k − 7) + (3k + 2). So 2k + 10 = 5k − 5 and k = 5. The terms are 3, 10, 17 with common difference 7.

Teacher's Note:
1. The middle term is the average of its neighbours.
2. Verify by checking equal differences.

Answer: (i) There are 2 red kings, so \( P = \dfrac{2}{52} = \dfrac{1}{26} \).
(ii) There are 12 face cards, so \( P = \dfrac{12}{52} = \dfrac{3}{13} \).
(iii) There is 1 queen of diamonds, so \( P = \dfrac{1}{52} \).

Teacher's Note:
1. Count favourable cards and divide by 52.
2. Face cards are kings, queens and jacks.

Answer: Let there be b blue balls, so the total is 5 + b. \( P(\text{blue}) = \dfrac{b}{5+b} \) and \( P(\text{red}) = \dfrac{5}{5+b} \). Given b/(5 + b) = 2 × 5/(5 + b), so b = 10. There are 10 blue balls.

Teacher's Note:
1. Both probabilities have the same denominator.
2. Equate the numerators.

Answer: Let the sides be x and y with x > y. 4x − 4y = 24 gives x = y + 6. Then (y + 6)² + y² = 468, so y² + 6y − 216 = 0, i.e. (y + 18)(y − 12) = 0 and y = 12. The sides are 12 m and 18 m.

Teacher's Note:
1. Use the perimeter condition to substitute for x.
2. Reject the negative root of the quadratic.

Answer: Let the integers be x and x + 2. x² + (x + 2)² = 290 gives x² + 2x − 143 = 0, so (x + 13)(x − 11) = 0. Since x is positive, x = 11. The integers are 11 and 13.

Teacher's Note:
1. Consecutive odd numbers differ by 2.
2. Check: 121 + 169 = 290.

Answer: Let the speed be x km/h. 360/x − 360/(x + 5) = 1, so 360 × 5 = x(x + 5) and x² + 5x − 1800 = 0. Then (x + 45)(x − 40) = 0, and the speed is 40 km/h (the negative root is rejected).

Teacher's Note:
1. Time equals distance divided by speed.
2. Form the equation from the difference of the two times.