CBSE Class 10 Mathematics Real Numbers, Polynomials, Triangles and Trigonometry Assignment Set 09

Answer: 225 = 135 × 1 + 90; 135 = 90 × 1 + 45; 90 = 45 × 2 + 0.
So HCF = 45.

Teacher's Note:
1. Euclid's division algorithm stops when the remainder becomes zero.
2. The last divisor is the HCF.

Answer: Polynomial = x² − (sum)x + product = x² − √5 x + \( \dfrac{1}{3} \).
Multiplying by 3 gives 3x² − 3√5 x + 1 (any non-zero multiple is correct).

Teacher's Note:
1. Use x² − (sum of zeroes)x + (product of zeroes).
2. Multiply by 3 to remove the fraction if you wish.

Answer: Here \( \dfrac{a_1}{a_2}=\dfrac{1}{2} \), \( \dfrac{b_1}{b_2}=\dfrac{-2}{-4}=\dfrac{1}{2} \) and \( \dfrac{c_1}{c_2}=\dfrac{-5}{-1}=5 \).
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and there is no solution.

Teacher's Note:
1. Compare the three ratios a₁/a₂, b₁/b₂ and c₁/c₂.
2. Equal first two ratios with a different third ratio mean no solution.

Answer: For parallel lines \( \dfrac{4}{2}=\dfrac{-6}{k}\neq\dfrac{1}{7} \).
So 2 = −6/k, giving k = −3.

Teacher's Note:
1. Parallel lines need equal x and y coefficient ratios.
2. Check that the third ratio is different.

Answer: A polynomial of degree n has at most n zeroes. The degree here is 7, so it has at most 7 zeroes.

Teacher's Note:
1. The number of real zeroes cannot exceed the degree.
2. Here the graph can cut the x-axis at most 7 times.

Answer: cosec (A − 20°) = sec [90° − (A − 20°)] = sec (110° − A).
So 4A = 110° − A, i.e. 5A = 110° and A = 22°.

Teacher's Note:
1. Convert cosec to sec using the complementary angle relation.
2. Then equate the two angles.

Answer: The longest side is 10. 8² + 6² = 64 + 36 = 100 = 10².
By the converse of Pythagoras theorem the triangle is right-angled.

Teacher's Note:
1. Always test the square of the longest side.
2. The converse of Pythagoras theorem is used here.

Answer: sin 55° = sin (90° − 35°) = cos 35°.
So the expression = sin² 35° + cos² 35° = 1.

Teacher's Note:
1. Write one angle as the complement of the other.
2. Then use sin²θ + cos²θ = 1.

Answer: Distance = \( \sqrt{3^2+5^2}=\sqrt{9+25}=\) \( \sqrt{34} \) units.

Teacher's Note:
1. Distance from the origin is √(x² + y²).
2. Leave the answer in surd form unless asked for a decimal.

Answer: \( \dfrac{1375}{8000}=\dfrac{11}{64} \). Since 64 = 2⁶, the denominator has only the factor 2, so the decimal terminates.
\( \dfrac{11}{64}=\) 0.171875.

Teacher's Note:
1. A fraction terminates if its denominator has only 2 and 5 as prime factors.
2. Here 11/64 gives six decimal places because 2 occurs six times.

Answer: Combine: \( \dfrac{(x-2)-x}{x(x-2)}=3 \Rightarrow -2=3x^2-6x \Rightarrow 3x^2-6x+2=0 \).
D = 36 − 24 = 12, so \( x=\dfrac{6\pm 2\sqrt3}{6}=\) \( 1\pm\dfrac{\sqrt3}{3} \).

Teacher's Note:
1. Clear the denominators first and bring the equation to ax² + bx + c = 0.
2. Note that x cannot be 0 or 2.

Answer: Let QR = x, so PR = 25 − x. By Pythagoras, (25 − x)² = 5² + x², so 625 − 50x = 25 and x = 12.
QR = 12 cm and PR = 13 cm. Hence sin P = 12/13, cos P = 5/13, tan P = 12/5.

Teacher's Note:
1. Find the missing sides with Pythagoras theorem.
2. Then apply the ratios from the angle at P.

Answer: Let the pole break at height h, so the broken part has length 9 − h. Then sin 30° = h/(9 − h), i.e. 1/2 = h/(9 − h).
So 9 − h = 2h and h = 3 m.

Teacher's Note:
1. The broken part acts as the hypotenuse of the right triangle.
2. Use sin 30° = 1/2.

Answer: The tangent is perpendicular to the radius at the point of contact, so OA² = r² + 4².
25 = r² + 16, so r² = 9 and the radius is 3 cm.

Teacher's Note:
1. Draw the right triangle formed by the centre, the point of contact and A.
2. OA is the hypotenuse.

Answer: Let A(−4, 0), B(4, 0), C(0, 3). AB = 8, BC = √(16 + 9) = 5, AC = √(16 + 9) = 5.
Since BC = AC, the triangle is isosceles.

Teacher's Note:
1. Find all three sides with the distance formula.
2. Two equal sides make the triangle isosceles.

Answer: Take any positive integer a and b = 2. By Euclid's division lemma, a = 2q + r with r = 0 or r = 1.
If r = 0, a = 2q, which is even. If r = 1, a = 2q + 1, which is odd. Hence every even integer is 2m and every odd integer is 2m + 1.

Teacher's Note:
1. Apply the division lemma with divisor 2.
2. Only two remainders are possible.

Answer: Multiply the first by 2 and the second by 3: 4x + 6y = 14 and 9x − 6y = 15. Adding: 13x = 29, so x = 29/13.
Then 3y = 7 − 58/13 = 33/13, so y = 11/13. Thus x = 29/13, y = 11/13.

Teacher's Note:
1. Use elimination to remove y.
2. Check by substituting in 3x − 2y.

Answer: Let the fraction be \( \dfrac{x-3}{x} \). Then \( \dfrac{x+1}{x+4}-\dfrac{x-3}{x}=\dfrac{1}{8} \), which simplifies to \( \dfrac{12}{x(x+4)}=\dfrac{1}{8} \).
So x² + 4x − 96 = 0, i.e. (x + 12)(x − 8) = 0. As x = 8 (x is positive), the fraction is 5/8.

Teacher's Note:
1. Form the quadratic equation from the stated increase.
2. Reject the negative root.

Answer: Let the stream speed be x km/h. Then \( \dfrac{24}{18-x}-\dfrac{24}{18+x}=1 \), so 24 × 2x = 324 − x².
x² + 48x − 324 = 0, so (x + 54)(x − 6) = 0. The speed of the stream is 6 km/h.

Teacher's Note:
1. Upstream speed is 18 − x and downstream speed is 18 + x.
2. Reject the negative root.

Answer: Height above eye level = 28.5 × tan 60° = 28.5√3 ≈ 49.36 m.
Height of chimney = 49.36 + 1.5 ≈ 50.86 m.

Teacher's Note:
1. Do not forget to add the boy's height.
2. Use √3 = 1.732.

Answer: Equating squared distances: (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)².
−14x − 2y + 50 = −6x − 10y + 34, so 8x − 8y − 16 = 0, i.e. x − y = 2.

Teacher's Note:
1. Equate the squares of the two distances to avoid roots.
2. The result is the perpendicular bisector of the segment.

Answer: Write 1 = sec²θ − tan²θ in the numerator: 1 + secθ − tanθ = (secθ − tanθ)(secθ + tanθ) + (secθ − tanθ) = (secθ − tanθ)(secθ + tanθ + 1).
Cancelling the common factor with the denominator, LHS = secθ − tanθ = \( \dfrac{1-\sin\theta}{\cos\theta} \) = RHS.

Teacher's Note:
1. Replace the 1 in the numerator using sec²θ − tan²θ = 1.
2. Then factorise and cancel.

Answer: For collinear points the area of the triangle is 0: k(2k − 6 + 2k) + (1 − k)(6 − 2k − 2 + 2k) + (−4 − k)(2 − 2k − 2k) = 0.
This gives 8k² + 4k − 4 = 0, i.e. (2k − 1)(k + 1) = 0. So k = 1/2 or k = −1.

Teacher's Note:
1. Set the area formula of a triangle to zero.
2. Both roots are valid.

Answer: Divide p(x) by (x − √2)(x + √2) = x² − 2: the quotient is 2x² + 7x − 15.
2x² + 7x − 15 = (2x − 3)(x + 5), so the other zeroes are 3/2 and −5.

Teacher's Note:
1. Divide by the quadratic factor made from the known zeroes.
2. Factorise the quotient.

Answer: Let the centre be O(x, y). Equating OA = OB for (3, −7) and (3, 3): (y + 7)² = (y − 3)², so 20y = −40 and y = −2.
Equating distances to (6, −6) and (3, −7): (x − 6)² + 16 = (x − 3)² + 25, so x = 3. The centre is (3, −2) and the radius is 5.

Teacher's Note:
1. The centre is equidistant from all three points.
2. Verify with the radius 5.