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Simple Problems In Mathematics

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Published in: Mathematics
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Simple Problems in Mathematics.

The L / Kolkata

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  1. 1. 2. A. The Learning Hall Mathematics Class IX-X sep 16, 2018 + 3x + m, the value of m is; If x+4 is a factor of the polynomial b) Since, x+4 is a factor of + 3x + m p(-4) = 0 Or, (-4)2 + = 0 or, 16-12 +m=O Or, 4+m=0 Or, m=-4 The perimeter of a triangle is 120 m and its sides are in the ratio 5:12:13; Find the area of the triangle Now, Let a,b,c be the sides of the triangle Hence, a+b+c = 120 Now sides ratio = 5:12:13 Therefore, 5X+12X+13x = 120 Therefore, 30x = 120
  2. Therefore, sides are 5x = 5x4 = 20 = = 48 = 13x4 = 52 Since, 5:12:13 is a Pythagorean Triplet, Therefore, the triangle is right angled. Therefore, area of the triangle = 1/2 x Base x Height = 1/2 x x 12x = 1/2 x 20x 48 = 480 rn2 Area of the Triangle = 480 m2 3. A. Ram has two rectangles in which their areas are given a) 25a2-35a+12 b) 35y2 + 13y-12 i) ii) Give possible expression for the length & breadth of each rectangle Which mathematical concept is used in the problem l) Possible length and breadth of the rectangle are the factors of the given area 25a2- 35a + 12 = 25a2 -(20+ 15)a + 12 = 25a2 -20a- 15a + 12 = 5a(5a-4) - 3(5a-4) So possible length and breadth are (5a-3) and (5a-4) respectively Area = 35y2 + 13y-12 = 35y2 + (28-15)y-12 = 35y2 + 28y-15y-12 = 7y(5Y+4) - 3(5Y+4)
  3. 4. A. 5. So possible length and breadth are (7y-3) and (5Y+4) respectively Factorization of Polynomials A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule ? Radius of the sphere = r= 3.5/2 = 1.75 mm Volume of the sphere = 4/3 x pi x r3 = 4/3 x 22/7 x = 22.46 mm3 The volume of medicine required to fill one capsule is 22.46 mm3 p x z O In the given figure, OP is perpendicular to AB; If x:y:z = 1:3:5, then find the degree measures of x, y & z A. Since OP is perpendicular to AB, / AOP = 900
  4. or, or, or, Hence It is given that, y = 3a From the figure, we can see that, = a = 100 — 3a = 300 z = 5a = 500 z = 500 x + y + z = 900 a +3a + 5a = 90 9a = 900 a = 100 x = 100 X y = 300

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