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GCSE Mathematics – Revision Notes Topic ‘Geometry and Algebra’ © irevise.com 2014. All revision notes have been produced by mockness ltd for irevise.com. Email: [email protected] Copyrighted material. All rights reserved; no part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, reprinting, or otherwise without either the prior written permission of irevise.com or a license permitting copying in the United Kingdom issued by the copyright licensing Agency. AQA – Geometry and Algebra Contents Proportionality ........................................................................................................................................ 3 Map Scales .............................................................................................................................................. 8 Trial and improvement ......................................................................................................................... 10 Quadratic Formula ................................................................................................................................ 13 Sketching Graphs ..................................................................................... Error! Bookmark not defined. 6. Transformations ................................................................................... Error! Bookmark not defined. 7. Area and Perimeter .............................................................................. Error! Bookmark not defined. 8. Points, Lines and Angles and Triangles ................................................ Error! Bookmark not defined. 9. Circles ................................................................................................... Error! Bookmark not defined. 10. Enlargements ..................................................................................... Error! Bookmark not defined. 11. Trigonometry ..................................................................................... Error! Bookmark not defined. Bisector of a given Angle...................................................................... Error! Bookmark not defined. Division of a line segment into equal parts.......................................... Error! Bookmark not defined. Line perpendicular to a given line , passing through a point on ...... Error! Bookmark not defined. Line parallel to a given line, through a point not on ......................... Error! Bookmark not defined. 14. Volume ............................................................................................... Error! Bookmark not defined. 15. Cylinders............................................................................................. Error! Bookmark not defined. 16. Spheres and Cones ............................................................................. Error! Bookmark not defined. 1. Proportionality Proportionality describes the relationship between two variables where the relationship between the two variables can be expressed in terms of a proportionality constant. Direct proportionality The variable y is said to be directly proportional to the variable x if their relationship can be described as: y=cx Where c is a constant. The above can be re-written as , i.e. the ratio y/x is constant. Direct proportionality can be observed when plotting y against x. y = 3x 13 12 11 10 9 y 8 7 6 5 4 3 2 1 0 0 1 2 3 4 x As x increases y increases. The rate of the increase is the proportionality constant, 3. Inverse proportionality The variable y is said to be inversely proportional to the variable x if their relationship can be described as This can be written as yx = c, i.e. the product yx is constant. If y is inversely proportional to x then y decreases as x increases. Proportion to powers Direct and inverse proportionality can also be to the powers of x. For example we could write: y is directly proportional to x y is directly proportional to y is directly proportional to √ y is directly proportional to and y is inversely proportional to x y is inversely proportional to y is inversely proportional to √ y is inversely proportional to Examples Question 1.1 y is directly proportional to x. We know that if y = 28 then x = 7. Write the equation connecting y and x. Find y when x is 12. Answer y is directly proportional to x: Also Therefore So that we can write And if x = 12 then Question 1.2 y is inversely proportional to x. We know that if x = 3 then y = 8. Find the value of y when x = 5. Answer y is inversely proportional to x: Determine c by using known values: therefore So Finally, when x = 5: therefore y = 4.8 Question 1.3 W is inversely proportional to x. If W = 6 then x = 20. Find W for x = 24. Answer W is inversely proportional to x: Again, use known values given: therefore So that and for x = 24 Question 1.4 The variable M is directly proportional to It is known that if r = 5 then M = 200. Determine the value of M when r = 8. Find r when M = 3125. . =5 Answer M is directly proportional to : Again, if: we have that So we can write the equation Now, if r = 8 we have: Similarly, if M = 3125: and Question 1.5 The cost of a bucket of ice cream, C, depends on its weight w. The price of ice cream, p, is 16 euro/kg. Describe the relationship between C, w and p. Write the equation that describes this relation. What would you spend if you wanted to buy a bucket of 8kg of ice cream? Answer The cost of a bucket of ice cream C is directly proportional to the weight w. The constant of proportionality is the price of 1 kg of ice cream, namely 16 euro/kg. We then have: The cost of a bucket og 8kg would then be: Question 1.6 The circumference C of a circle is given by the formula Where d is the diameter of the circle. Describe in words the proportionality relation, specify the constant of proportionality. Find the diameter of a circle with circumference C=35. Answer The circumference of a circle is directly proportional to the diameter d. The constant of proportionality is , the constant ratio between the circumference and the diameter of a circle. If the circumference C is 35 then: 2. Map Scales Map scales are ratios used to represent distances. They can be used as an alternative representation of distances when the original set of distances cannot be conveniently displayed. When drawing a map of a large area the original distances cannot be represented as they are. We use map scales to relate original distances to new, more convenient distances so that it is possible to draw a map. A certain unit in our map represents a certain unit of land or sea. For example we can decide that a distance of 1 meter = 100 cm corresponds to 1cm in our map. We would write that as 1:100, meaning that 1 unit in the map represents 100 units of territory. Examples Question 2.1 The scale on a map is 1:250000. What is the actual distance corresponding to 1cm in the map? Express the answer in kilometres. Answer 1 unit in the map represents 250000 units. 1cm in the map represents 250000cm. 250000cm = 2.5km. Question 2.2 The scale on a map is 1:500000. Two towns are 8cm apart on the map. What is the actual distance between the towns in kilometres? Answer 1cm on the map corresponds to 500000cm. 8cm on the map corresponds to 8*500000cm = 4000000cm. 4000000cm = 40km Question 2.3 The scale on a map is 1:300000. What is the actual distance in kilometers represented by 11cm in the map. Answer 1cm on the map corresponds to 300000cm. 11cm on the map corresponds to 11*300000cm = 3300000cm. 3300000cm = 33km. Question 2.4 The scale on a map is 1:200000. What distance in the map represents 5km? Answer 5km = 500000cm. Also: Therefore: And Question 2.5 In a map 1 inch represents 4 miles. A road on the map measures 6 inches to the nearest inch. What is the shortest possible distance of the road? Answer The road measures 6 inches to the nearest inch. This implies that the shortest measure could be 5.5 inches before rounding up. 1 inch represents 4 miles; therefore 5.5 inches represent 5.5*4 miles = 22 miles. 3. Trial and improvement Trial and improvement involves testing an unknown variable with different values; the value that produces the result closest to the solution is selected. When solving by trial and improvement the first value you test should be one that you think is close to the answer. Using this as a starting point you then test different values until you find the one that is closest. Examples Question 3.1 Use trial and improvement to find a solution to the equation Give your solution to 1 decimal place. Answer Guess a value for x and evaluate We want the result to be close to 25. x 2 3 2.5 2.7 2.8 2.9 +x 10 30 18.125 22.38 24.75 27.29 . Comment Result is too small, try with a bigger number Too big, try a number between 2 and 3 Too small, try a larger value Still too small, try a larger value Quite close to 25. Try next value, then choose the best Too big. Previous result was closer to 25 To 1 decimal place the best solution to the equation is given by Question 3.2 Use trial and improvement to find a solution to the equation Give your answer to 1 decimal place. Answer Re-arrange the expression as Again guess values for x, then improve. We want to find the value of x that yield results closest to 26. – 5x 3 4 3.5 3.6 12 44 25.37 28.65 Comment Too small. Try larger value Too big. Try a value between 3 and 4 Still small, but quite close to 26. Try next value and compare Result too big. Previous value was closer to 26 To 1 decimal place the best solution to the equation is given by Question 3.3 The perimeter of a square is given by the formula Where l is the length of the side of the square. If the perimeter of a square is P = 49, find by trial and improvement the value of l when l is a whole number (i.e. no decimals). Answer It is known that 4l = P = 49. We want the whole number that is closest to the solution. Comment 15 14 13 12 60 56 52 48 Result too large. Try a smaller value Still too large. Try smaller again Still large but close to 49. Try next value and compare Closer to 49. Best solution Our answer is then l = 12. Question 3.4 The area A of a circle is given by the formula Where r is the radius of the circle. Find r to 1 decimal place when A = 786. Answer Write Then again look for the value for which the result is closest to 786. Comment 14 15 16 15.9 15.8 615.75 706.85 804.25 794.23 784.27 Result is small. Try larger Still small. Try larger again Result bigger than 786. Try slightly smaller Closer to 786. Check next smaller value Best solution To 1 decimal place the solution is r = 15.8 Question 3.5 Find the solution to the equation Correct to 1 decimal place by trial and improvement. Answer First we re-arrange the equation: Again we look for the value of w for which decimal place. is closest to 145 when w has 1 Comment 4 3 3.3 3.2 352 111 164.07 144.79 Result is too large. Try smaller Result too small. Try a value between 3 and 4 Result bigger than 145. Try slightly smaller Closest to 145. Best solution The solution to 1 decimal place is w = 3.2 4. Quadratic Formula A quadratic equation has standard form. The numbers a, b and c are the coefficients of the equation, while x is said to be a variable. Note that in the standard form all non-zero terms appear in one side, and the other side is always zero. The solutions of an equation are the values of the variable x for which the equation is true. Therefore, to find the solutions of a quadratic equation we must find the values for which the expression is equal to 0. If solutions of the equations exist we usually find them by factorization. However, when factorization is not an option we can solve the equation by using the quadratic formula. √ The quantity is called the discriminant of the quadratic equation, and is usually represented by the symbol Δ (greek uppercase delta). We can rewrite the above as: √ Where The discriminant Δ can be positive, negative or zero. If Δ > 0 There are two distinct roots (solutions) If Δ = 0 There is 1 root If Δ < 0 There are no real roots Examples Question 4.1 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer We have: a=1 b=5 c=3 And the determinant is The determinant is positive; therefore there are two distinct roots. Question 4.2 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer We have: a = 12 b=-4 c=2 And the determinant is ( ) The determinant is negative; therefore there are no real roots. Question 4.3 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer We have: a=2 b=-4 c=2 And the determinant is ( ) The determinant is zero, therefore there is 1 real root. Question 4.4 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer We have: a=-3 b=3 c= And the determinant is ( ) ( ) The determinant is zero; therefore there is 1 real root. Question 4.5 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer Note that the term containing x is not present. To say that there is no x is the same of saying that its coefficient is zero (0x). We have: a = 26 b=0 c= And the determinant is ( ( ) ( ) ) The determinant is positive; therefore there are two distinct roots. Question 4.6 In the equation Indicate a, b and c. Determine Δ and indicate whether the equation has 1, 2 or no solutions. Answer Here the constant term c does not appear. We write that as c = 0 . We have: a=-1 b=-5 c= And the determinant is ( ) ( ) ( ) The determinant is positive; therefore there are two distinct roots. We now use the quadratic formula to find roots (if any) of quadratic equations Question 4.7 Find the roots (if any) of the equation Answer Here we have: a= b=-5 c= Calculate the determinant ( ) The determinant is zero so we know that there is one root. √ To check if the solution is correct, just verify that it satisfies the equation: The solution x = 2 does indeed satisfy the equation. Question 4.8 Find the roots of the equation given in Question 4.6 Answer The equation given in question 4.6 is The determinant of the equation, calculated above, is Δ = 25 and there are two solutions. We call the solutions and . √ ( ) √ So we get and Check that and . satisfy the equation: For ( ) ( ) For ( ) ( ) Question 4.9 Find the roots (if any) of the equation Answer First of all we put the equation in standard form, so that all non-zero terms appear in one side of the equation: Then we continue as before and calculate the discriminant: a=1 b= c= ( ) ( ) ( ( ) ) The determinant is positive, therefore there are two distinct roots √ So we get and Check that and ( ) and √ . satisfy the equation: For 0=0 For ( ) ( 0.0676 + 3.9 4 = 0 0=0 ) Question 4.10 Solve the equation Give your answer in the form Answer Again, first re-arrange into standard form, then calculate the discriminant: a=1 b= c= ( ) . The determinant is positive; therefore there are two distinct roots. We are asked to present the answer in the form ( √ ) √ √ We can simplify the above expression by noting that √ √ √ √ And dividing across by 2 √ So our final answer is √ √