Math Problem Solutions: Equations, Geometry, and Proofs
Math Problem Solutions
Solving Equations
21: Adding the two equations and dividing by 10, we get: x + y = 10. Subtracting the two equations and dividing by -2, we get: x – y = 1. Solving these two new equations, we get, x = 11/2.
Geometry and Circles
30: Let ABCD be the rhombus circumscribing the circle with center O, such that AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively. We know that the tangents drawn to a circle from an exterior point are equal in length.
- AP = AS (1)
- BP = BQ (2)
- CR = CQ (3)
- DR = DS (4)
Adding (1), (2), (3), and (4) we get: AP + BP + CR + DR = AS + BQ + CQ + DS. (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ). Therefore, AB + CD = AD + BC (5). Since AB = DC and AD = BC (opposite sides of parallelogram ABCD), putting in (5) we get, 2AB = 2AD or AB = AD. Therefore, AB = BC = DC = AD. Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus.
Rational and Irrational Numbers
26: Let us assume 5 + 2√3 is rational, then it must be in the form of p/q where p and q are co-prime integers and q ≠ 0. i.e., 5 + 2√3 = p/q. So, √3 = (p – 5q) / (2q) (i). Since p, q, 5, and 2 are integers and q ≠ 0, the RHS of equation (i) is rational. But the LHS of (i) is √3 which is irrational. This is not possible. This contradiction has arisen due to our wrong assumption that 5 + 2√3 is rational. So, 5 + 2√3 is irrational.
Polynomials and Zeros
27: Let α and β be the zeros of the polynomial 2x2 – 5x – 3. Then α + β = 5/2 and αβ = -3/2. Let 2α and 2β be the zeros of x2 + px + q. Then 2α + 2β = -p, 2(α + β) = -p, 2 x 5/2 = -p. So p = -5. And 2α x 2β = q, 4αβ = q. So q = 4 x -3/2 = -6.
Speed, Distance, and Time
28: Let the actual speed of the train be x km/hr and let the actual time taken be y hours. Distance covered is xy km. If the speed is increased by 6 km/hr, then the time of journey is reduced by 4 hours i.e., when speed is (x+6) km/hr, time of journey is (y-4) hours. Therefore, Distance covered = (x+6)(y-4). xy = (x+6)(y-4). -4x + 6y – 24 = 0. -2x + 3y – 12 = 0 (i). Similarly, xy = (x-6)(y+6). 6x – 6y – 36 = 0. x – y – 6 = 0 (ii). Solving (i) and (ii) we get x = 30 and y = 24. Putting the values of x and y in equation (i), we obtain Distance = (30 x 24) km = 720 km. Hence, the length of the journey is 720 km.
OR
Let the number of chocolates in lot A be x and let the number of chocolates in lot B be y. Therefore, the total number of chocolates = x + y. Price of 1 chocolate = ₹ 2/3, so for x chocolates = 2/3x and price of y chocolates at the rate of ₹ 1 per chocolate = y. Therefore, by the given condition 2/3x + y = 400. 2x + 3y = 1200 (i). Similarly, x + 4/5y = 460. 5x + 4y = 2300 (ii). Solving (i) and (ii) we get x = 300 and y = 200. Therefore, x + y = 300 + 200 = 500. So, Anuj had 500 chocolates.
Triangle Similarity
22: In ΔABC, ∠1 = ∠2. Therefore, AB = BD (i). Given, AD/AE = AC/BD. Using equation (i), we get AD/AE = AC/AB (ii). In ΔBAE and ΔCAD, by equation (ii), AC/AB = AD/AE. ∠A = ∠A (common). Therefore, ΔBAE ~ ΔCAD [By SAS similarity criterion].
Angles in Circles
23: ∠PAO = ∠PBO = 90° (angle between radius and tangent). ∠AOB = 105° (By angle sum property of a triangle). ∠AQB = ½ x 105° = 52.5° (Angle at the remaining part of the circle is half the angle subtended by the arc at the center).
Trigonometry
25: sin(A+B) = 1 = sin 90, so A+B = 90 (i). cos(A-B) = √3/2 = cos 30, so A-B = 30 (ii). From (i) & (ii) ∠A = 60° and ∠B = 30°.
Surface Area and Volume
34: Radius of the base of the cylinder (r) = 2.8 m = Radius of the base of the cone (r). Height of the cylinder (h) = 3.5 m. Height of the cone (H) = 2.1 m. Slant height of conical part (l) = √(r2 + H2) = √(2.8)2 + (2.1)2 = √7.84 + 4.41 = √12.25 = 3.5 m. Area of canvas used to make tent = CSA of cylinder + CSA of cone = 2 x π x 2.8 x 3.5 + π x 2.8 x 3.5 = 61.6 + 30.8 = 92.4 m2. Cost of 1500 tents at ₹120 per sq.m = 1500 x 120 x 92.4 = 16,632,000. Share of each school to set up the tents = 16632000 / 50 = ₹332,640.
Trapezium and Proportionality
33: Let ABCD be a trapezium with DC∥AB and EF is a line parallel to AB and hence to DC. To prove: DE/EA = CF/FB. Construction: Join AC, meeting EF in G. Proof: In △ABC, we have GF∥AB. CG/GA = CF/FB [By BPT] (1). In △ADC, we have EG∥DC (EF ∥AB & AB ∥DC). DE/EA = CG/GA [By BPT] (2). From (1) & (2), we get, DE/EA = CF/FB.