CBSE Class 10 Mathematics Real Numbers, Polynomials, Triangles and Trigonometry Assignment Set 15

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 those sides in the same ratio.
Proof: In △ABC let DE ∥ BC with D on AB and E on AC. Join BE and CD, and draw EM ⊥ AB and DN ⊥ AC.
1. ar(△ADE) = ½ × AD × EM and ar(△BDE) = ½ × DB × EM, so \( \dfrac{ar(ADE)}{ar(BDE)} = \dfrac{AD}{DB} \).
2. ar(△ADE) = ½ × AE × DN and ar(△CDE) = ½ × EC × DN, so \( \dfrac{ar(ADE)}{ar(CDE)} = \dfrac{AE}{EC} \).
3. △BDE and △CDE lie on the same base DE between the same parallels DE and BC, so ar(△BDE) = ar(△CDE).
From 1, 2 and 3: \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \). Hence proved.

Teacher's Note:
1. The proof compares areas of triangles that share a height.
2. Remember the key step that triangles on the same base and between the same parallels are equal in area.

Answer: Assume √3 is rational. Then √3 = p/q, where p and q are integers with no common factor other than 1 and q ≠ 0.
1. Squaring gives 3q² = p², so 3 divides p², and therefore 3 divides p.
2. Write p = 3m. Then 3q² = 9m², so q² = 3m², which means 3 divides q² and hence q.
3. Now 3 divides both p and q, which contradicts the assumption that they have no common factor.
So √3 is irrational.

Teacher's Note:
1. This is a proof by contradiction.
2. The result that 3 divides p² implies 3 divides p works because 3 is prime.

Answer: Statement: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Proof: Let △ABC be right-angled at B. Draw BD ⊥ AC.
1. △ADB ~ △ABC (angle A is common and ∠ADB = ∠ABC = 90°), so \( \dfrac{AD}{AB} = \dfrac{AB}{AC} \), giving AB² = AD × AC.
2. △BDC ~ △ABC (angle C is common and ∠BDC = ∠ABC = 90°), so \( \dfrac{CD}{BC} = \dfrac{BC}{AC} \), giving BC² = CD × AC.
3. Adding: AB² + BC² = AC(AD + CD) = AC × AC = AC².
Hence proved.

Teacher's Note:
1. The proof uses similarity of the two small triangles with the big one.
2. Draw the altitude to the hypotenuse first.

Answer: Assume 5 + √3 is rational, say equal to r. Then √3 = r − 5.
Since r and 5 are rational, r − 5 is rational, so √3 would be rational.
This contradicts the fact that √3 is irrational. Hence 5 + √3 is irrational.

Teacher's Note:
1. Use the known result that √3 is irrational.
2. A rational number minus a rational number is always rational.

Answer: The required number divides 245 − 5 = 240 and 1029 − 5 = 1024 exactly, so it is HCF(240, 1024).
1024 = 240 × 4 + 64
240 = 64 × 3 + 48
64 = 48 × 1 + 16
48 = 16 × 3 + 0
HCF = 16, and 16 > 5, so the remainder condition is valid. The largest number is 16.

Teacher's Note:
1. Subtract the remainder from each number before taking the HCF.
2. The divisor must be greater than the remainder.

Answer: Let △ABC ~ △PQR with ar(△ABC) = ar(△PQR).
For similar triangles, \( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{CA^2}{RP^2} \).
The left side equals 1, so AB = PQ, BC = QR and CA = RP.
By the SSS criterion, △ABC ≅ △PQR.

Teacher's Note:
1. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides.
2. A ratio of 1 forces all three sides to be equal.

Answer: 12576 = 4052 × 3 + 420
4052 = 420 × 9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = 4 × 6 + 0
The last non-zero remainder is 4, so HCF = 4.

Teacher's Note:
1. Divide the larger number by the smaller and keep dividing the divisor by the remainder.
2. Stop when the remainder becomes zero.

Answer: Sum of zeroes: p + q = −p, so q = −2p.
Product of zeroes: pq = q, so q(p − 1) = 0, giving q = 0 or p = 1.
If q = 0, then p = 0. If p = 1, then q = −2.
So (p, q) = (1, −2) or (0, 0). Check: x² + x − 2 has zeroes 1 and −2.

Teacher's Note:
1. Write both the sum and the product relations and solve them together.
2. Do not miss the case q = 0.

Answer: A cubic with these values is x³ − (sum)x² + (sum of pair products)x − (product).
= x³ − 4x² + 1·x − (−6)
= x³ − 4x² + x + 6 (any non-zero constant multiple is also valid).

Teacher's Note:
1. Use the form x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ.
2. Watch the sign of the constant term.

Answer: Sum of zeroes = \( -\dfrac{1}{6} \) and product = \( \dfrac{k}{6} \).
α² + β² = (α + β)² − 2αβ = \( \dfrac{1}{36} - \dfrac{k}{3} = \dfrac{25}{36} \).
So \( \dfrac{k}{3} = -\dfrac{24}{36} = -\dfrac{2}{3} \), giving k = −2.

Teacher's Note:
1. Express α² + β² through the sum and product of zeroes.
2. Then solve the simple linear equation for k.

Answer: Subtract the second equation from the first: 7y = 35, so y = 5.
Then 2x + 15 = 11, so x = −2.
Put x = −2 and y = 5 in y = mx + 3: 5 = −2m + 3, so m = 1.
Solution: x = −2, y = 5, m = 1.

Teacher's Note:
1. Elimination works well because the coefficients of x are equal.
2. Substitute the solution into the third equation to find m.

Answer: Split the middle term: 4√3x² + 8x − 3x − 2√3 = 4x(√3x + 2) − √3(√3x + 2) = (√3x + 2)(4x − √3).
Zeroes: \( x = -\dfrac{2}{\sqrt{3}} \) and \( x = \dfrac{\sqrt{3}}{4} \).
Sum = \( \dfrac{-8 + 3}{4\sqrt{3}} = -\dfrac{5}{4\sqrt{3}} \), which equals −b/a ✓.
Product = \( -\dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{4} = -\dfrac{1}{2} \), and c/a = \( \dfrac{-2\sqrt{3}}{4\sqrt{3}} = -\dfrac{1}{2} \) ✓.

Teacher's Note:
1. Split the middle term using two numbers whose product is (4√3)(−2√3) = −24.
2. Then compare the sum and product with −b/a and c/a.

Answer: Since cot A = 1/tan A, \( \dfrac{\tan A + 4}{\dfrac{4}{\tan A} + 1} = \dfrac{\tan A(\tan A + 4)}{4 + \tan A} = \tan A \).
With sin A = m/n, cos A = \( \dfrac{\sqrt{n^2 - m^2}}{n} \), so tan A = \( \dfrac{m}{\sqrt{n^2 - m^2}} \).
The value is \( \dfrac{m}{\sqrt{n^2 - m^2}} \).

Teacher's Note:
1. Simplify the expression first before putting in values.
2. It reduces neatly to tan A.

Answer: x² + y² + z² = r² sin²A cos²C + r² sin²A sin²C + r² cos²A
= r² sin²A (cos²C + sin²C) + r² cos²A
= r² sin²A + r² cos²A = r²(sin²A + cos²A) = r².
Hence proved.

Teacher's Note:
1. Group the first two terms and use sin²C + cos²C = 1.
2. Apply the same identity once more for angle A.

Answer: LHS = \( \left(1 + \dfrac{\cos A}{\sin A} - \dfrac{1}{\sin A}\right)\left(1 + \dfrac{\sin A}{\cos A} + \dfrac{1}{\cos A}\right) \)
= \( \dfrac{(\sin A + \cos A - 1)(\cos A + \sin A + 1)}{\sin A \cos A} \)
= \( \dfrac{(\sin A + \cos A)^2 - 1}{\sin A \cos A} = \dfrac{1 + 2\sin A \cos A - 1}{\sin A \cos A} = 2 \) = RHS.

Teacher's Note:
1. Convert everything to sine and cosine.
2. The product has the form (x − 1)(x + 1) with x = sin A + cos A.

Answer: Since sec²θ − tan²θ = 1, we get sec θ − tan θ = 1/p.
Adding: 2 sec θ = p + 1/p, so sec θ = \( \dfrac{p^2 + 1}{2p} \). Subtracting: 2 tan θ = p − 1/p, so tan θ = \( \dfrac{p^2 - 1}{2p} \).
sin θ = tan θ ÷ sec θ = \( \dfrac{p^2 - 1}{p^2 + 1} \). Hence proved.

Teacher's Note:
1. Find sec θ − tan θ from the identity sec²θ − tan²θ = 1.
2. Then use sin θ = tan θ ÷ sec θ.

Answer: Let the building be AB = 7 m and the tower be CD, with the horizontal line from A meeting the tower at E.
In the right triangle formed by the foot of the tower: tan 45° = AB/BD, so BD = 7 m. Hence AE = BD = 7 m.
In the upper right triangle: tan 60° = CE/AE, so CE = 7√3 m.
Height of tower = CE + ED = 7√3 + 7 = 7(√3 + 1) = 7 × 2.732 ≈ 19.12 m.

Teacher's Note:
1. The horizontal distance is the same in both right triangles.
2. Remember to add the building height to the part of the tower above the observer.

Answer: α + β = 6 and αβ = a.
3α + 2β = 2(α + β) + α = 12 + α = 20, so α = 8.
Then β = 6 − 8 = −2, and a = αβ = 8 × (−2) = −16.

Teacher's Note:
1. Rewrite 3α + 2β using the known sum of the zeroes.
2. Find both zeroes before computing the product.

Answer: Let the smaller number be x, so the larger is 3x.
3x − x = 26, so 2x = 26 and x = 13.
The numbers are 13 and 39. Check: 39 − 13 = 26 and 39 = 3 × 13 ✓.

Teacher's Note:
1. Express both numbers using one variable.
2. Verify the answer against both conditions.