ID : in-9-full-year-9th-grade-review [1] Class 9 Full Year 9th Grade Review For more such worksheets visit www.edugain.com Answer t he quest ions (1) ABCD is a parallelogram with E and F as the mid-points of side AB and CD. Line segments CE and AF intersect the diagonal BD at point P and Q respectively. If PQ = 5 cm and BC = 12 cm, f ind the length of diagonal BD. (2) Find the value of g. (3) Find the equation of straight line which is parallel to y-axis, and is at a distance of b f rom y-axis is (4) If solutions of a linear equation are (-5, 5), (0, 0) and (5, 5), f ind the equation.
ID : in-9-full-year-9th-grade-review [2] (5) The f ollowing table shows prof its of 5 companies in a business group last year. Company Profit Company 1 Rs. 300,000 Company 2 Rs. 400,000 Company 3 Rs. 250,000 Company 4 Rs. 300,000 Company 5 Rs. 350,000 What is the average prof it of the business group? (6) Vandita measures the heights of 9 of her f riends as f ollows (in centimetres) 127, 99.9, 140, 139, 160, 133, 142, 154, 139 What is the average of their heights? (7) Find the multiplicative inverse of the f ollowing. A) -14 12 B) -9 25 (8) Archana makes the kite using two pieces of paper. 1st piece of paper is cut in the shape of square where one diagonal is of the length 26 cm. At one of the vertex of this square a second piece of paper is attached which is of the shape of an equilateral triangle of length to give the shape of a kite. The length of the sides of triangle is a, such that a 2 = 4 3. Find the area of this kite.
ID : in-9-full-year-9th-grade-review [3] Choose correct answer(s) f rom given choice (9) If in the f igure below AB = 15cm, BC=20cm and CA = 7cm, f ind the area of the rectangle BDCE. a. 192 cm 2 b. 42 cm 2 c. 84 cm 2 d. 230.4 cm 2 (10) A f raction P Q can be expressed as a terminating decimal, if "Q" has no prime f actors other than a. 3, 7, 5 b. 2, 3 c. 2, 5 d. 2, 3, 5 (11) A sphere and a cone have the same radii. If the volume of the sphere is half of the volume of the cone, f ind the ratio of the cone's height and radius. a. 8:1 b. 3:4 c. 1:3 d. 1:2
(12) If AB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ACD + BED. ID : in-9-full-year-9th-grade-review [4] a. 180º b. 360º c. 270º d. 235º (13) Find sum of zeroes of polynomial y 2 + y - 30. a. 0 b. 1 c. -2 d. -1 (14) If AB and PQ are parallel, compute the angle Z. a. 439 b. 254 c. 259 d. 101 (15) The perimeter of a triangular f ield is 216 m and the ratio of the sides is 17:10:9. Which of the f ollowing is the area of the f ield in sq m : a. 1546 b. 1296 c. 864 d. 15552 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : in-9-full-year-9th-grade-review [5] (1) 15 cm (2) f /e Equation of line y = mx + c Since it goes through center, c is 0, hence equation is y = m x For f irst point x = f and y = gf, hence gf = m f m = g (1) For second point x = e and y = f, hence f = m e m = f /e (1) Step 4 On comparing two equations, g = f /e (3) x = b If a line is parallel to the y-axis, then x value of it is constant f or all values of y. Take a look at the image to see this case Further, if the line is distance b away f rom the y-axis, it also means that this constant value of x is b. So the equation f or that line is x=b
(4) x + y = 0 ID : in-9-full-year-9th-grade-review [6] The points (a,b) that solve a linear equation would satisf y ax+by=c, where a,b,c are constants. Substituting (0,0) we see that c = 0. Substituting the other two points (-5,5) and (5,-5), we get the f ollowing equations -5x + 5y=0 and 5x + (-5y) = 0 From these two equations, we can theref ore eliminate the variables and see that the answer is x + y = 0 (5) Rs. 320000 If you look at the question caref ully, you will notice that the table below shows prof its of 5 companies in a business group last year. Company Profit Company 1 Rs. 300,000 Company 2 Rs. 400,000 Company 3 Rs. 250,000 Company 4 Rs. 300,000 Company 5 Rs. 350,000 Total prof its made by the business group = Rs. 300,000 + Rs. 400,000 + Rs. 250,000 + Rs. 300,000 + Rs. 350,000 = Rs. 1600000 Average prof it of the business group = Total prof its made by the business group Number of companies in a group = 1600000 5 = 320000 Now the average prof it of the business group is Rs. 320000.
(6) 137.1 ID : in-9-full-year-9th-grade-review [7] If you look at the question caref ully, you will notice that Vandita measures the heights of 9 of her f riends as f ollows (in centimetres) 127, 99.9, 140, 139, 160, 133, 142, 154, 139 Sum of heights of her f riends = 127 + 99.9 + 140 + 139 + 160 + 133 + 142 + 154 + 139 = 1233.9 Total number of her f riends = 9 Average height of Vandita f riends = Sum of heights of her f riends Total number of her f riends = 1233.9 9 = 137.1 Now the average height of Vandita f riends is 137.1. (7) A) 12-14 Multiplicative Inverse: When we multiply a number by its "Multiplicative Inverse", we get 1. Mathematically, n 1 n = 1 The multiplicative inverse of -14 12 is 12-14, since -14 12 12-14 = 1.
B) 25 ID : in-9-full-year-9th-grade-review [8] -9 Multiplicative Inverse: When we multiply a number by its "Multiplicative Inverse", we get 1. Mathematically, n 1 n = 1 The multiplicative inverse of -9 25 is 25-9, since -9 25 25-9 = 1.
(8) 341 cm 2 ID : in-9-full-year-9th-grade-review [9] Following f igure shows the kite, made by two pieces of paper, Now, we can see that, this kite consists of a square ABCD and a equilateral triangle BEF. The area of the equilateral triangle ΔBEF can be calculated using Heron's f ormula f or equilateral triangle, Area = 3 a2 4 = 3 (4 3) 4 = 3 cm 2 The diagonal of the square = 26 cm. The area of the square ABCD = ( 1 2 ) (26) 2 = 338 cm 2. Step 4 Thus, the area of the kite = Area(ABCD) + Area(BEF) = 3 + 338 = 341 cm 2.
(9) a. 192 cm 2 ID : in-9-full-year-9th-grade-review [10] The area of the triangle ABC can be calculated using Heron's f ormula, since all sides of the triangle are known. S = (7 + 15 + 20)/2 = 21 cm. The area of the ΔABC = [ S(S - AB) (S - BC) (S - CA) ] = [ 21(21-15) (21-20) (21-7) ] = 42 cm 2 The height(bd) of the ΔABC = 2 Area(ΔABC) AC = 2 42 7 = 12 cm In right angled ΔBDC, DC 2 = BC 2 - BD 2 DC = [ BC 2 - BD 2 ] = [ (20) 2 - (12) 2 ] = 16 cm The area of the rectangle BDCE = BD DC = 12 16 = 192 cm 2 (10) c. 2, 5
(11) a. 8:1 ID : in-9-full-year-9th-grade-review [11] We know that the volume of a cone with radius r and height h = 1/3 π r 2 h. We also know that the volume of a sphere with radius r = 4/3 π r 3. We have been told that the volume of the sphere in question is half of the volume of the cone in question. Theref ore, 4/3 π r 3 = 1/2 x (1/3 π r 2 h) h/r = 8:1 Step 4 Thus, the ratio of the cone's height and radius is 8:1. (12) c. 270º ACB is the angle subtended by the point C by the diameter. The angle in a semicircle is 90º. Theref ore ACB = 90º If we join points B and C, BECD is a quadrilateral f ormed by 4 points on a semicircle. BECD is theref ore a cyclic quadrilateral The opposite angles of a cyclic quadrilateral add up to 180º BCD + BED = 180º We need to f ind ACD + BED We can see that ACD = ACB + BCD ACD + BED = ( ACB + BCD) + BED = ACB + ( BCD + BED) = 90º + 180º = 270º (13) d. -1
(14) c. 259 ID : in-9-full-year-9th-grade-review [12] Draw a line MN which is parallel to line PQ and line AB If you look at the given f igure caref ully, you will notice that angle Z1 + angle Z2 + angle Z makes f ull angle and the f ull angle is exactly 360. Theref ore angle Z1 + angle Z2 + angle Z = 360 angle Z = 360 - angle Z1 - angle Z2 Angle Z1 and angle Q are alternate angles. Theref ore angle Z1 = angle Q angle Z1 = 52 Similarly angle Z2 and angle B are alternate angles. Theref ore angle Z2 = angle B angle Z2 = 49 Step 4 Theref ore angle Z = 360 - angle Z1 - angle Z2 = 360-52 - 49 = 360-101 = 259
(15) b. 1296 ID : in-9-full-year-9th-grade-review [13] Since we know the perimeter, we can use Heron's f ormula to help us compute the area The f ormula states that the area of a triangle with sides a, b and c, and perimeter 2S = Let us assume the 3 sides are of length a=17x, b=10x and c=9x (we know this because the ratio of the sides is given as 17:10:9) We also know that a+b+c = 216. 17x + 10x + 9x = 216 (17 + 10 + 9)x = 216 36x = 216 x = 216 = 6 36 Step 4 From this we see that a = 102, b = 60 and c=54. Also S=108 Step 5 Putting these values into Heron's f ormula, Area = = Solving, we f ind the area = 1296