295 Oasis School Mathematics-8 295 7. Determine whether the following straight lines have positive slopes, negative slopes, zero slopes or not defined. (a) (b) (c) (d) 23.5 Intercepts of a Straight Line In the given figure, the straight line AB intersects the x–axis at A(a, 0) and y–axis at (0, b). The length OA and OB are called the x–intercept (a) and y–intercept (b) of the straight line AB. To find the intercepts of a straight line from its equation, we can find the x–intercept by putting y = 0 and y intercept by putting x = 0. For example: Find the x–intercept and y–intercept of the line 3x+2y = 12. Putting x = 0, we get, 3x + 2y = 12 or 3× 0 + 2y = 12 or 2y = 12 ∴ y = 12 2 = 6 and putting y = 0, we get, 3x + 2 × 0 = 12 or 3x = 12 ∴ x = 12 3 = 4 Hence, x–intercept = 4 and y–intercept = 6. θ X O O O O Y X' X Y' Y C D X' X Y Y' X' X A B Answer 1. (a) 1 2 (b) -7 8 (c) -1 2 (d) 3 2 (e) 5 3 (f) 2 2. (a) 6 (b) 8 3. (a) 2 and 9 (b) 2 3 and 5 3 (c) -2 and 7 (d) -2 3 and 5 3 (e) -2 and 3 4. (ii) 1, 2 (iii) 2 5. Consult your teacher. 6. Consult your teacher. 7. (a) Positive (b) 0 (c) Negative (d) ∞ X' X Y' Y A(a, o) a b x-intercept y-intercept B(o, b) O Note: At X–axis, y = 0 At Y–axis, x = 0
296 Oasis School Mathematics-8 Worked Out Examples Example: 1 Find the x–intercept and y–intercept from the following graphs. (i) (ii) Solution: (i) From the graph, we get x–intercept (OA) = 2 and y–intercept (OB) = 4. (ii) From the graph, we get x–intercept (OA) = 3, y–intercept (OB) = –5. Example: 2 Draw the graph of the line y = 2x – 4 and find the x–intercept and y–intercept made by the line. Solution: Here, y = 2x – 4 x 0 2 3 y -4 0 2 We see that the line cuts the x–axis at 2 and y–axis at –4. Hence, x–intercept = 2 and y–intercept = –4. X' X Y' Y O X' X Y' Y O X' X Y' Y (0,-4) (2,0) (3,2)
297 Oasis School Mathematics-8 297 Exercise 23.4 1. Find x–intercept and y–intercept in each of the following. (a) (b) (c) (d) 2. Find the x–intercept and y–intercept of the following equations: (a) 2x + 5y = 10 (b) 3x + 2y = 12 (c) 4x + 5y = 20 (d) 4x – y = 12 (e) y = 2x + 4 3. Draw the graphs of the following equations and find x–intercept and y-intercept. (a) y = 2x – 8 (b) 2x + y = 4 (c) 2x + y = 6 (d) y = 2x + 4 (e) y = –3x 2 – 6 4. (a) Find at which point the line 3x – 2y = 6 cuts the x and y axis and draw the graph of the line. (b) Find the points where 2x+3y = 12 cuts X-axis and Y-axis. Answer 1. (a) 2, 4 (b) 3, -5 (c) -3, 2 (d) -4, -3 2. (a) 5, 2 (b) 4, 6 (c) 5, 4 (d) 3, -12 (e) -2, 4 3. Consult your teacher 4. (a) (2, 0), (0, -3) (b) (6, 0) (0, 4) X' X Y' Y X' X Y' Y X' X Y' Y X' X Y' Y
298 Oasis School Mathematics-8 23.6 Simultaneous Equations Let us consider the equations x + y = 8 and 2x–y = 7. These two equations contain two variables 'x' and 'y' and both are linear equations. These equations are simultaneous equations. Hence, a pair of linear equations in two variables are called a system of simultaneous linear equations or simply simultaneous equations. Method of solving simultaneous equations There are many methods of solving simultaneous equations. Here we will discuss only three methods. (i) Graphical method (ii) Elimination method (iii) Substitution method Graphical method of solving simultaneous equation The steps for solving of simultaneous equations graphically are as follows: • Draw the graph of the first equation. • Draw the graph of the second equation with the same axes and the same scale. • From the graph, read the point of intersection of the two straight lines drawn. • The required solutions are the x and y coordinates of the point which is the point of intersection of two simultaneous equations. Worked Out Examples Example: 1 Solve graphically: 2x – y = 5 and x – y = 1 Solution: Here, 2x – y = 5 or 2x – 5 = y ∴ y = 2x – 5 …(i) And x – y = 1 or x –1 = y y = x – 1 …(ii) x 0 1 4 y -5 -3 3 x 0 2 3 y -1 1 2 X' X Y' Y O (4,3)
299 Oasis School Mathematics-8 299 Now, plotting these points on the graph, we get two lines. From the graph, Point of intersection of equation (i) and (ii) is (4,3) ∴ x = 4, y =3. Example: 2 Solve graphically: 2x + 3y = 13, y = 3. Solution: Given equations, 2x +3y = 13 ………….(i) or, 3y = 13 – 2x or, y = 13–2x 3 x 2 -1 5 y =13–2x 3 3 5 1 Again, y = 3………… (ii) x 1 2 3 y=3 3 3 3 From the graph, Point of intersection of (i) and (ii) is (2, 3). ∴ x = 2, y = 3 Exercise 23.6 1. Write down the solution of each pair of equations: X' X Y' Y (2, 3) y = 3 2x+3y=13 O X' X Y' Y X' X Y' Y (a) (b) 4 3 3 2 2 1 1
300 Oasis School Mathematics-8 X' X X' X Y' Y' Y Y (c) (d) O O 2. Complete the following table and plot them on the graph and find the solution. x y = 2x+1 x y = 4x–1 (a) x y = 2 x y = x+1 (b) x y = x–6 x y = 2 – x (c) 3. Solve the following equations graphically: (a) x + y = 6, x – y = 2 (b) 2x – y = 5, x + y = 4 (c) x + 2y = 1, x – y = 4 (d) x – 2y = 5, 2x –y =1 (e) 2x = 4 + y, x + 3 = –2y (f) x + 2y + 16 = 0, 2x + 4 – 3y = 0 (g) y = 3-5x 3 , y = –4 (h) x = 3, 2x = 26 – 5y (i) 3x +y = 2, 3x – 2y = 5 4. Convert the following verbal problems into the equations and solve them graphically. (a) Find the two numbers whose sum is 7 and difference is 1. (b) The sum of two numbers is 6 and their difference is 2. Find the numbers. (c) The difference of two numbers is 2. One number is one third of the other. Find the numbers. Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) 4, 2 (b) 3, 1 (c) 3, -1 (d) -1, -3 (e) 1, -2 (f) -8, -4 (g) 3, -4 (h) 3, 4 (i) 1, -1 4. (a) 4,3 (b) 4, 2 (c) 1, 3)
301 Oasis School Mathematics-8 301 Additional Text Elimination method and substitution method of solving simultaneous equations Elimination Method In this method we make the coefficient of one of the variables equal and we add or subtract the given equations to eliminate that variable. We substitute the value of that variable in an equation to get the value of other variable. For example, solve the equations, x +y = 5 and x –y = 3 Solution: Here, x +y = 5 …………… (i) x–y = 3 …………… (ii) Adding equation (i) and (ii), x + y = 5 x – y = 3 2x = 8 or, x = 8 2 = 4 Substituting the value of x in equation (i), x + y = 5 or, 4+y = 5 or, y = 5 – 4 y = 1 ∴ x = 4 y = 1. Substitution method In this method, we express one of the variables in terms of other and we substitute its value in other equation to get the values of the variable. For example solve the equations x +y = 7 and 2x –y = 5 Solution: Given equation, x +y = 7 ……………(i) 2x – y = 5 ……………(ii) From equation (i) y = 7 – x Note: If the co-efficeints of a variable are same in both equations and their sign is opposite add the equations. If their sign is same, subtract the equations.
302 Oasis School Mathematics-8 Substituting this value in equation (ii) we get, 2x –y = 5 or, 2x – (7–x) = 5 or, 2x – 7 + x = 5 or, 3x = 5 + 7 or, 3x = 12 or, x = 12 3 = 4. Again, substituting the value of x in (i) x + y = 7 or, 4 + y = 7 or, y = 7–4 = 3 ∴ x = 4 and y=3. Worked Out Examples Example: 1 Solve the given equations by elimination method: 3x – 2y = 5 and 2x +y = 8 Solution: Given equation, 3x –2y = 5 ………………….. (i) 2x +y = 8 ……………………(ii) Multiplying equation (ii) by 2 and adding these equations, 3x –2y = 5 4x +2y = 16 Adding, 7x = 21 or, x = 21 7 or, x = 3. or, 2y = 4 or, y = 4 2 or, y = 2. ∴ x = 3 and y = 2 Substituting the value of x in (i) 3x – 2y = 5 or, 3×3–2y = 5 or, 9 –2y = 5 or, –2y = 5 – 9 or, –2y = –4 Cancel (–) sign from both sides.
303 Oasis School Mathematics-8 303 Example: 2 Solve the given equations by substitution method: 2x +y =5 and 3x –2y = 4 Solution: Given equations, 2x +y = 5 ………………(i) 3x – 2y = 4 ……………..(ii) From equation(i), y = 5 – 2x Substituting this value in equation (ii) 3x –2y = 4 or, 3x – 2(5–2x) = 4 or, 3x – 10+4x = 4 or, 7x – 10 = 4 or, 7x = 4+10 or, 7x = 14 or, x = 14 7 or, x = 2 Substituting the value of x in (i) 2x +y = 5 2×2+y = 5 or, 4 + y = 5 or, y = 5 –4 or, y = 1 ∴ x = 2 and y = 1. Exercise 23.7 1. Solve the following simultaneous equations (by elimination method). (a) x + y = 3, x – y = 1 (b) 2x + y = 7 and 3x – y = 8 (c) 3x – 2y = 4, 3x + y = 7 (d) 2x + y = 8, 2x + 3y = 12 (e) 3x – y = 15, 3x + 2y = 24 (f) 2x + 3y = 15, x –2y = –3 (g) 2x + y = 1, 3x – 2y = 5 2. Solve the following simultaneous equations (by substitution method). (a) –y = –x + 4, 2x – 3y = 6 (b) x – y = 2, x + y = 6 (c) x + 2y = 7, 2x + 3y = 12 (d) 3x + 2y = 5, 3x – y = 2 (e) 2x – y = 3, 3x + 2y = 8 (f) 2x + y = 6, 2y = 2 + x (g) y = 2x, 4x + 3y = 20 Answer 1. (a) 2, 1 (b) 3,1 (c) 2,1 (d) 3, 2 (e) 6, 3 (f) 3,3 (g) 1, -1 2. (a) 6, 2 (b) 4, 2 (c) 3, 2 (d) 1,1 (e) 2, 1 (f) 2,2 (g) 2,4
304 Oasis School Mathematics-8 23.7 Quadratic Equation Let's take an equation x² + 2x + 5 = 0. It is the second degree equation with one variable which is in the form of ax² + bx + c = 0. This type of equation is the quadratic equation where, a = 1, b = 2 and c = 5 Hence an algebraic equation having degree two is the quadratic equation. It is in the form of ax2 + bx + c = 0. where a, b, c are constants. Examples: x2 – 5x + 6 = 0 x2 – 2x = 0 x2 – 8 = 0, etc. All these are in the form of ax2 + bx + c = 0. In the first equation, a = 1, b = –5, c = 6 In the second equation, a = 1, b = –2, c = 0 and in the third equation, a = 1, b = 0, c = –8 Solution of Quadratic Equations A quadratic equation can be solved by different methods. Here we are discussing the solution of quadratic equation by factorisation method. • Simplify the given equation and convert it in the form of ax2 +bx+c = 0 • Factorise the expression on the left hand side. • Write each factor is equal to zero to get the values of x. Steps Worked Out Examples Example: 1 Solve: (x+1) (x–3) = 0 Solution: Here, (x + 1) (x – 3) = 0 Either, (x + 1) = 0 or, x – 3 = 0 or, x = 3, or, x = –1 Hence, x = –1, 3.
305 Oasis School Mathematics-8 305 Example: 2 Solve: x2 = 5x Solution: Here, x2 = 5x or, x2 – 5x = 0 or, x(x – 5) = 0 Either, x = 0 or, x – 5 = 0 x = 5 Hence, x = 0, 5 Example: 3 Solve: x2 – 49 = 0 Solution: Here, x2 – 49 = 0 or (x)2 – (7)2 = 0 or, (x + 7)(x – 7) = 0 Either, x + 7 = 0 ⇒ x = –7 or, x – 7 = 0 ⇒ x = 7 Hence, x = –7, 7 Example: 4 Solve: 2x2 – 5x – 12 = 0 Solution: Here, 2x2 – 5x – 12 = 0 or, 2x2 – (8 – 3)x – 12 = 0 or, 2x2 – 8x +3 x – 12 = 0 or, 2x(x – 4) + 3 (x – 4) = 0 or, (x – 4) (2x + 3) = 0 Either, x–4 = 0, or, 2x + 3 = 0 or, x = 4, or , x = -3 2 Hence, x = 4, -3 2
306 Oasis School Mathematics-8 Example: 5 Solve: 3x – 7 2x – 5 = x + 1 x – 1 Solution: Here, 3x – 7 2x – 5 = x + 1 x – 1 or (3x – 7) (x – 1) = (x + 1) (2x – 5) or 3x2 – 3x – 7x + 7 = 2x2 – 5x + 2x – 5 or 3x2 – 10x + 7 = 2x2 – 3x – 5 or 3x2 – 2x2 – 10x + 3x + 7 + 5 = 0 or x2 – 7x + 12 = 0 or x2 – 4x – 3x + 12 = 0 or x(x – 4) – 3(x – 4) = 0 or (x – 4) (x – 3) = 0 Either, x – 4 = 0 or, x – 3 = 0 ∴ x = 4 x = 3 Hence, x = 4, 3. Example: 6 If 5 times of a number is added to the square of the number, the result is 66. Find the numbers. Solution: Let the required number be x. Then by the given condition, x2 + 5x = 66 or, x2 + 5x – 66 = 0 or, x2 + 11x – 6x – 66 = 0 or, x(x + 11) – 6(x + 11) = 0 or, (x + 11) (x – 6) = 0 Either, x + 11 = 0 or, x – 6 = 0 ∴ x = –11 ∴ x = 6 Hence, the required number = 6, –11.
307 Oasis School Mathematics-8 307 Exercise 23.7 Solve the following: 1. (a) (x–3) (x+2) = 0 (b) (x+6) (x–6) = 0 (c) (x+3) (x+4) = 0 (d) (x+1) (x–5) = 0 2. (a) x2 –2x=0 (b) 2x2 +5x=0 (c) 3x2 = 6x (d) x2 = 5x 3. (a) x2 – 25 = 0 (b) x2 – 81 = 0 (c) x2 – 16 =0 (d) 4x2 –81 = 0 4. (a) x2 + 5x + 6 = 0 (b) x2 + x – 20 = 0 (c) x2 – 3x – 10 = 0 (d) 2x2 – x– 6 = 0 (e) x2 –9x = 70 (f) x2 –10x = 39 (g) 3x2 = 2x + 8 (h) 2x2 = 3x – 1 5. (a) x + 1 x = 5 2 (b) x + 1 3x–7 = x – 1 2x–5 6. Solve the following problems: (a) If 6 is added to the square of a number, the result is 31. Find the number. (b) If 2 is subtraced from the half of the square of a number, the result is 30. Find the numbers. (c) If 5 times a number is added to its square, the result is 6. Find the numbers. (d) If 6 times of a number is added to the square of the number, the result is 55. Find the numbers. (e) One number is 5 less than another and their product is 36. Find the numbers. (f) If the sum of two numbers is 6 and the product is 8, find the two numbers. Answer 1. (a) 3, -2 (b) -6, 6 (c) -3, -4 (d) -1, 5 2. (a) 0, 2 (b) 0, -5/2 (c) 0, 2 (d) 0, 5 3. (a) 5, -5 (b) 9, -9 (c) 4, -4 (d) 9/2, -9/2 4. (a) -3, -2 (b) -5, 4 (c) 5, -2 (d) 2, -3/2 (e) -5, 14 (f) -3, 13 (g) 2, -4/3 (h) 1, 1/2 5. (a) 2, 1/2 (b) 3, 4 6. (a) 5, -5 (b) 8, -8 (c) -6, 1 (d) -11, 5 (e) 4, 9 and -4,-9 (f) 4, 2
308 Oasis School Mathematics-8 23.8 Inequality (Inequation) Inequality: Let us consider the mathematical statement x + 3 = 7. Value of x determines whether the given statement is true or false. If x = 2, 2+3 = 7 (false statement) If x = 4, 3+4 = 7 (true statement) Such open statement containing (=) sign is the equation. Open statement containing trichotomy signs such as <, ≤,>,≥ are called inequality. For example: x < 5 : x is less than 5 x ≤ 4 : x is less than or equal to 4 x > 6 : x is greater than 6 x ≥ 7 : x is greater than or equal to 7. S.N. Some mathematical notations: Meanings 1. a > b → a is greater than b. 2. a < b → a is smaller than b. 3. a ≥ b → a is greater than or equal to b. 4. a ≤ b → a is smaller than or equal to b. Note: <means less than, > means greater than, ≤ means less than or equal to ≥ means greater than or equal to Properties of inequalities 1. If same number is added to both sides or subtracted from the both sides, the sign of inequality remains the same. i.e. a < b ⇒ a + c < b + c and a – c < b – c. 2. If both sides of an inequality are multiplied or divided by a positive number, the inequality sign remains the same. i.e. a < b ⇒ ac < bc and a c < b c if c > 0. 3. If both sides of an inequality are multiplied or divided by a negative number, the inequality sign alters. i.e. a < b ⇒ ac > bc and a c < b c if c < 0. and -a > -b Solution sets on a number line Suppose we have to show all the numbers between any two given numbers say –2 and 5. See the following number line. Here the line represents all the real numbers. The right arrow shows the direction for increasing order of numbers while the left arrow shows the direction for decreasing -5 -4 -3 -2 -1 0 1 2 3 4 5
309 Oasis School Mathematics-8 309 order of numbers. Hence the line represents a real number line. Inequality in the number line: Let's observe the following number line. Worked Out Examples Example: 1 Show the following inequalities in the number line. (a) x ≥ –3 (b) x < 2 (c) 1 ≤ x < 5 Solution: Example: 2 Write the inequality represented by the given graph. -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Solution: This number line represents all the numbers less than or equal to 4. ∴ Required inequality is x ≤ 4. -5 -4 -3 -2 -1 0 1 2 3 4 5 x ≥ 2 6 7 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 x ≥ 1 6 7 (i) (ii) -5 -4 -3 -2 -1 0 1 2 3 4 5 –1< x ≤ 6 6 7 (iii) -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 -2 ≤ x ≤ 3 6 7 (iv) -4 -3 -2 -1 0 1 2 3 4 x ≥ -3 (a) -4 -4 -3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3 4 4 5 5 6 x < 2 1 ≤ x < 5 (b) (c)
310 Oasis School Mathematics-8 Example: 3 Find the solution set for the given inequality in number line. 2x + 5 ≥ x + 8 Solution: 2x+5 ≥ x+8 or, 2x+5-5 ≥ x+8–5 [Subtracting 5 from both sides] or, 2x ≥ x+3 or, 2x–x ≥ x–x+3 [Subtracting x from both sides] or, x ≥ 3 Example: 4 Solve the following inequality and find its solution set. 2x – 4 ≤ 6 ∈ N Solution: 2x – 4 ≤ 6 or, 2x – 4 + 4 ≤ 6 + 4 or, 2x ≤ 10 or, 2x x ≤ 10 x or, x ≤ 5. Since x is less than or equal to 5, and x ∈ N. ∴ Required solution set is {5, 4, 3, 2, 1} Example: 5 Solve the given inequalities and represent in a number line –3 < x –2 ≤ 4 Solution: Here, –3 < x – 2 ≤ 4 Now, –3 < x – 2 ≤ 4 or, –3+2<x–2+2 ≤ 4 + 2 [adding 2 in all sides] or –1 < x ≤ 6 -3 -2 -1 0 1 2 3 4 5 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
311 Oasis School Mathematics-8 311 Exercise 23.7 1. Show the following inequalities in the number line. (a) x < 2 (b) x ≥ -5 (c) x ≤ -3 (d) x > 4 (e) -1 ≤ x < 3 (f) 3 < x < 7 2. Write down the inequalities represented by each of the following graphs. 3. If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, find the solution set of the following inequations. (a) x ≥ 6 (b) x – 5 > 2 (c) x – 3 < 5 (d) 3x – 1 ≤ 5 4. Solve the following: (a) x + 3 > 6 (b) 2x – 5 ≥ 5 (c) 3y + 2 ≤ 8 (d) 3x – 10 > 5x + 4 (e) 2x + 8 ≤ 3x + 4 (f) 2(x + 3) ≤ x + 2 (g) 2x+3 (x+1) > 2x + 9 (h) x+2 4 ≤ x+3 5 (i) 2x-1 3 > 3x+1 2 5. Solve the following inequalities and find the solution set. (a) 2x – 3 < 5; x ∈ N (b) 2x + 16 ≤ x + 18, x ∈ W (c) 6 – 3x > 9; x ∈ I (d) – 1 ≤ x + 1 < 3, x ∈ I 6. Solve the following inequalities and show them in the number line. (a) 2(x+5) < 4 + 5x (b) x –2 3 > 3x–1 4 (c) 2x+8 ≥ x + 5 (d) 2x+1 4 ≤ x+2 3 (e) -2 ≤ x – 4 ≤ 2 (f) – 1 2 ≤ 2x– 5 2 ≤ 3 2 (g) –6 ≤ 2x + 4 < 10 -4 -3 -2 -1 0 1 2 3 4 5 -4 -4 -4 -3 -3 -3 -2 -2 -2 -1 -1 -1 0 0 0 1 1 1 2 2 2 3 3 (f) (b) (e) -4 -3 -2 -1 0 1 2 3 (d) (c) -4 -3 -2 -1 0 1 2 3 4 5 (a)
312 Oasis School Mathematics-8 Attempt all the questions. Group 'A' [4 × 1 = 4] 1. (a) Factorise : x3 – 8 (b) Solve : 52x = 1 (c) Find the slope of the straight line 3y = 6x + 5 (d) Solve : 2x ≤ –4 Group 'B' [5 × 2 = 10] 2. (a) Simplify : a2 a–b + b2 b–a (b) Simplify : 128x6 y-3 64x-3y3 (c) Simplify : 5x+1–5x 4×5x (d) Draw a graph of the line 2x + y = 5. (e) Solve the given inequality and show it in the number line. 7 – 3x > 10. Group 'C' [4 × 4= 16] 3. (a) If a + 1 a = 5, find the value of a2 + 1 a2 . (b) Find the H.C.F. of 8a3 –27, 4a2 –12a+9. (c) Simplify: 1 a+b + 1 a–b – 1 a2 –b2 (d) Solve graphically: y = x+2 and y = 3x Answer 1. Consult your teacher. 2. (a) x > 2 (b) x ≤ 2 (c) x<1 (d) x>-2 (e) -3 < x ≤ 2 (f) -1<x<4 3. (a) {6, 7, 8, 9} (b) {8, 9} (c) {0, 1, 2, 3, 4, 5, 6, 7} (d) {0, 1, 2} 4. (a) x > 3 (b) x ≥ 5 (c) y ≤ 2 (d) x < -7 (e) x ≤ 4 (f) x ≤ -4 (g) x>2 (h) x ≤ 2 (i) x<-1 5. (a) {1,2,3} (b) {0, 1, 2} (c) {..., -4, -3, -2} (d) {-2,-1, 0,1} 6. (a) x > 2 (b) x<-1 (c) x ≥ -3 (d) x ≤ 5/2 (e) 2 ≤ x ≤ 6 (f) 1≤ x ≤ 2 (g) -5 ≤ x < 3 Assessment Test Paper Full marks : 30
313 Oasis School Mathematics-8 313 BLE New Model Question Time : 3 hrs. F.M. 100 (Compulsory Mathematics) (Based on New Curriculum and Specification Grid) (Group 'A'] [10×1=10] 1. a. What is the corresponding angle of ∠BGH in the given figure? [Ans: ∠DHF] b. Find the area of the given semi-circle. [Ans: 77cm²] 2. a. Find the length of AB in the given right angled triangle ABC. [Ans: 3cm] b. Draw the compass bearing of 060°. 3. a. If A = {a, e, i, o, u} and B = {e, a, r}, find A∩B. [Ans: {e, a}] b. Convert 6.201 × 104 into normal form. [Ans: 62010] 4. a. Find the range: (Temperature): 25°C, 28°C, 30°C, 32°C, 33°C. [Ans: 8°C] b. Factorise: 4x² – 1 [Ans: (2x + 1) (2x – 1)] 5. a. If mx=ax, find the value of x. [Ans: 0] b. Find the slope of the line: y – 2x = 6 [Ans: 2] (Group 'B') [17×2=34] 6. a. Find the value of x in the given figure. [Ans: 40°] b. Show that ∠A = 90° in the given figure. c. In the figure, if ∆ABC ~ ∆ADE, find the length of AB. [Ans: 37.5cm] 7. a. Prove that ∆ADB ≅ ADC in the figure. b. In the given semi-circle if the length of ACB is 11cm, find the length of AB. [Ans: 7 cm] A B C D F E H G A B 7cm O A B C 5cm 4cm E A C B D H 1250 3x+5 A B a 3a 2a 25cm B C A D E 10cm 15cm 20cm A D B C A O B C
314 Oasis School Mathematics-8 c. Draw a net of a cube. 8. a. Find the length of AB in the given figure. [Ans: 5 units] b. If the length and area of a rectangle are 6m and 30m² respectively, find its perimeter. [Ans: 22m] c. If U = {1, 2, 3, 4, 5} and A = {2, 4}, find A. [Ans: {1, 3, 5}] 9. a. If a, m and b are in proportion show that m² = ab. b. Convert 1002 into decimal. [Ans: 4] c Find the median: 12, 13, 17, 10, 18, 15, 25 [Ans: 15] 10. a. Factorise: x² – 5x + 6 [Ans: (x – 2) (x – 3)] b. If a = 2 and b = 2, find the value of ab × ba . [Ans: 16] c. Simplify: 3x+1–3x 3x × 2 [Ans: 1] 11. a. Show the equation y = x in graph. b. Solve and show in number line: -5x ≤ 10. [Ans: x ≥ -2] (Group'C') [14×4=56] 12. If U = {1, 2, 3, ....10}, A = {prime numbers less than 10}, B = {even numbers up to 10} and C = {1, 2, 3, 4, 5,}, find A∩B∩C by using Venn-diagram. [Ans: {2}] 13. If x – y = 5, find the value of x3 –y3 – 15xy. [Ans: 125] 14. Find H.C.F. : x² – 4, x² + 3x + 2 [Ans: x + 2] 15. Simplify: 3 x+3 4 x–3 24 x2 –9 + – [Ans: 7x + 3 ] 16. Solve graphically: x + y = 5 and x – y = 1 [Ans: 3, 2] 17. Find the mean from the given data: [Ans: 40] (Marks obtained) 10 20 30 40 50 60 (No. of student) 2 3 5 4 5 6 18. The length of a hall is 20m and breadth is one fourth of length. If the volume of the hall is 400m³, find the height of the room. [Ans: 4m] 19. Simplify: 110112 + 111112 – 100012 [Ans: 1010012 ] 20. A shopkeeper bought a watch for Rs.1,000 and fixed its price 25% above it. If he sold it by allowing 10% discount, find the selling price. [Ans: Rs.1125] 21. 20 men can complete a piece of work in 36 days. How many men should be added to complete the work in 24 days? [Ans: 10] 22. Find the rate of interest if Rs. 4,000amounts to Rs.6,000 in 2 years. [Ans: 25%] 23. Draw a triangle ABC with vertices A(-1, 3), B(3, 6) and C(5, -2) on a graph paper. Reflect it in X-axis and show the image A'B'C' on the same graph paper. [Ans: (-1, -3), B'(3, -6), C' (5, 2)] 24. Verify experimentally that diagonals of a parallelogram bisect each other. [Two figures of different measures are necessary] 25. Construct a rectangle ABCD where diagonal AC =BD = 10 cm and angle between them is 60°. Y Y' A B O X' X