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2 Mastering Heuristics Series Handbook for discerning parents Spatial Visualisation A Problem-solving Tool for Challenging Problems in Area and Perimeter (Primary 5 & 6) Sunny Tan Maths Heuristics Private Limited ISBN: Printed in Singapore February 04 Copyrights Sunny Tan Published by Maths Heuristics Private Limited Edited by Karen Ralls-Tan of RE: TEAM Communications Distributed by Maths Heuristics Private Limited 95A Thomson Road, Goldhill Centre, Singapore Tel: For orders and enquiries: enquiry@mathsheuristics.com All rights reserved. No parts of this publication may be reproduced, stored in any retrieval system, or transmitted via any retrieval system or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of the copyright owner. No parts of this publication may be used in the conducting of classes, commercial or otherwise, without the prior permission of the copyright owner. Every effort has been made to contact the holders of any copyright materials found herein. Should there be any oversight, the publisher will be pleased to make any necessary amendment at the first opportunity. No patent liability is assumed with respect to the use of information contained herein. The publisher, author and editor of this publication have taken all reasonable care to ensure that the content herein is error free. However, some typographical, factual or calculation oversights still elude us. We sincerely apologise for these, and would greatly appreciate it if you could highlight these to us via at enquiry@mathsheuristics.com. From time to time, we do provide addenda and errata, in order to ensure appropriate updates to this publication. These may be found at 3

3 CONTENTS 4 Preface 5 How to Use This Book 7 Before You Begin 0 Chapter Rearranging Parts: Non-Circles 7 Chapter Rearranging Parts: Circles 46 Chapter 3 Drawing Lines 69 Chapter 4 Overlapping Parts 80 Chapter 5 Difference in Area 9 Chapter 6 Ratio (Unit Method) 9 Chapter 7 Visual Clues 03 Answers 03 Chapter Rearranging Parts: Circles 03 Chapter Rearranging Parts: Non-circles 04 Chapter 3 Drawing Lines 04 Chapter 4 Overlapping Parts 05 Chapter 5 Difference in Area 05 Chapter 6 Ratio (Unit Method) 05 Chapter 7 Visual Clues 4

4 PREFACE Heuristics in Primary Maths Syllabus Heuristics is a specialised mathematical problem-solving concept. Mastering it facilitates efficiency in solving regular as well as challenging mathematical problems. The Ministry of Education in Singapore has incorporated Problem-solving Heuristics into all primary-level mathematics syllabuses. Learning Heuristics Effectively However, the Problem-solving Heuristics are not taught systematically; they have been dispersed into the regular curriculum. This not only makes it difficult for students to pick up Heuristics skills, but can also make mathematics confusing for them. For us parents, it is difficult to put aside the regular-syllabus mathematical concepts we were brought up on to re-learn Heuristics, much less teach our own children this new concept. Take the Heuristic technique of Algebraic Equations, for instance. Although this technique has never been, and is still not, taught at primary level, parents may attempt to teach their children to solve mathematical questions using Algebraic Equations. This will only confuse their children. According to current primary-level mathematics syllabus, other Heuristics techniques should be used instead. These and other challenges were what I observed firsthand during my years as a mathematics teacher, and provided me the impetus for my post-graduate studies, mathsheuristics programmes and the Mastering Heuristics Series of guidebooks. About Mastering Heuristics Series This series of books is a culmination of my systematic thinking and experience, supported by professional instructional writing and editing, to facilitate understanding and mastery of Heuristics. I have neatly packaged Heuristics into main techniques (series of guidebooks) and mathematical scenarios (chapters within each guidebook). For each mathematical scenario, I offer several examples, showing how a particular heuristics technique may be applied, and then explaining the application in easy-to-follow steps and illustrations without skipping a beat. This particular guidebook in the series deals specifically with Spatial Visualisation for Area and Perimeter the use of visual clues and manipulation to effectively analyse and solve challenging problems in Area and Perimeter. This problem-solving technique is very powerful because it helps students literally see the solution to the problem posed. The Mastering Heuristics Series provides a comprehensive guide to Heuristics. While each guidebook introduces parents to how Heuristics works, students have the opportunity to see the technique applied in different scenarios and to get in some practice. For students enrolled in mathsheuristics programmes, each guidebook serves as a great companion, while keeping parents well-informed of what their children are learning. Sunny Tan February 04 5

5 HOW TO USE THIS BOOK :BEFORE YOU BEGIN: This chapter instills basic but important steps and truths in the heuristics technique that must be applied across every question in the guidebook. This helps to standardise the given information for easy application of the technique being taught. In this guidebook on Spatial Visualisation, the steps include being familiar with parts and formulae of basic and derivative shapes. :CHAPTERS AND SECTIONS: Various scenarios are neatly separated into different chapters and sections. This allows the heuristics technique to be learnt and applied in a focused manner. :EXAMPLES: Each example of heuristics application comes with Working and Explanation, and includes Confusion Alert and Alternative boxes. WORKING: Working shows heuristics application in action (how quick it is to solve a question). 6

6 EXPLANATION: Explanation shows the thought process (the detailed steps) behind the heuristics application. It takes readers through the solution in the following manner: step-by-step method so readers can follow what happens at every stage. systematic approach so readers begin to see a pattern in applying the technique. easy-to-follow steps so readers can quickly understand the technique minus the frustration. For Spatial Visualisation, readers will see that its application always involves: identifying basic shapes, derivative shapes and parts of shapes, analysing how these are inter-related, and how these may be manipulated to arrive at the answer, and the application of formulae (as listed in the Before You Begin Chapter) to carry out the actual solution. This quickly helps readers see and understand how to approach each question, including picking out hints often provided in the questions. Confusion Alert boxes highlight areas where students are likely to falter or make mistakes in. It also gives the rationale to help clarify their doubts. Alternative boxes show other approaches to the solution process. This acknowledges the different views that students may have to the problem. :LET S APPLY: Learning is only effective with practice. Hence, at the end of each chapter/section is a list of questions to hone readers skills in the heuristics technique. :ADDITIONAL TIPS: For on-going sharing and discussions on the use of Spatial Visualisation, visit: For detailed workings to all the Spatial Visualisation Let s Apply sections, visit: 7

7 BEFORE YOU BEGIN :THINGS TO NOTE:. You need to be familiar with the various formulae for different basic shapes rectangle, square, triangle, circle. However, the topic on circles is only covered at Primary 6.. These formulae will also be needed when working with derivative shapes semi-circle (half circle), quadrant (quarter circle), etc. 3. Challenging higher-order questions involve a combination of shapes. The combination sometimes forms a larger composite shape. Examples of combination of shapes. A semi-circle, a circle and 4 a triangle within a square. A square within a circle. Examples of composite shape. triangles, forming a square. Various squares and rectangles of different sizes, forming a square. 4. Basic shapes and their formulae. a) Rectangle Area = Length x Breadth = L x B Perimeter = Length + Breadth = L + B = (L + B) b) Square Length = Breadth Area = Length x Length = L x L Perimeter = 4 x Length = 4 x L 8

8 c) Triangle Area = x Base x Height = x b x h ( bh) Perimeter = Sum of 3 sides d) Circle Diameter = x radius (r) Circumference = x diameter = x x radius ( d) ( r) Area = x radius x radius ( r ) can assume various forms. Read the question carefully to determine which form of to apply. = 3.4 = 7 Calculate 9

9 CHAPTER REARRANGING PARTS: NON-CIRCLE :EXAMPLES:. In the figure below, AB = BC = CD = DE = EF = FA. The area of the whole figure is 0 cm². Find the area of the shaded part. :WORKING: 6 parts = 0 parts shaded = x 0 = 6 3 = 40 x 0 Area of the shaded part is 40cm. 0

10 :EXPLANATION: Draw lines as shown. Notice how each shaded part has a mirror part that forms sectors of the hexagon. BAF = BOF BCD = BOD Similarly, DEF = DOF All the triangles listed above are identical. Hence, there are 6 identical triangles. 6 identical triangles = 0 identical triangles shaded = x 0 = 6 3 = 40 Area of the shaded part is 40 cm. x 0

11 . A rectangular garden measuring m by 6 m has a 5-m walking path running through it. Find the area of the walking path. :WORKING: Garden and path Garden only = L x B = 6 x = 546 = L x B = (6 5) x = x = 44 Path = = 05 Area of the walking path is 05 m

12 :EXPLANATION: Remove the path and join the garden parts. This forms a shorter rectangle measuring m by (6 5) m, that is m by m. Longer rectangle = L x B = 6 x = 546 Shorter rectangle = L x B = x = 44 Path = Longer rectangle Shorter rectangle = = 05 Area of the walking path is 05 m. 3

13 3. The figure below shows a rectangle WXYZ with two identical right-angled isosceles triangles cut out at diagonally-opposite corners. The area of the shaded figure is 64 cm². Find the area of one of the triangles. :WORKING: Rectangle Rectangle = 6 x u = 6u = 4 x u = 4u Rectangle 3 = 4 x 6 = 44 Shaded parts = u + 6u = u u = 64 30u = u = 0 u = 4 Unshaded triangle = = = = 8 x B x H x u x u x 4 x 4 Area of one of the triangles is 8 cm. 4

14 :EXPLANATION: Shift one triangle so that both triangles join to form a square. This is what you will get. Draw lines as shown. There are 3 shaded rectangles. Rectangle Rectangle = 6 x u = 6u = 4 x u = 4u Rectangle 3 = 4 x 6 = 44 Shaded parts = u + 6u = u We know that the shaded area is 64 cm. Equate both values u = 64 30u = u = 0 u = 4 Now, calculate the area of one unshaded triangle. Unshaded triangle = x B x H = = = 8 x u x u x 4 x 4 Area of one of the triangles is 8 cm. 5

15 4. Find the area of the trapezium shown in the figure below. :WORKING: The triangles = x B x H = = 4 x 6 x 8 Base of the triangles together = 0 0 = 0 = Base of rectangle Rectangle = x The triangle = x 4 = 48 Trapezium = The triangles + Rectangle = = 7 Area of the trapezium is 7 cm. 6

16 :EXPLANATION: Draw lines as shown. Separate the rectangle, and join the triangles. There are now figures the rectangle and a larger triangle. Let s work on the larger triangle. In the larger triangle, the angle at the top is 80 o 53 o 37 o, that is 90 o. So, B = 6 and H = 8. Larger triangle = x B x H = x 6 x 8 = 4 Let s work on the rectangle. Note that both the rectangle and the larger triangle have same base and height. Rectangle = x Larger triangle = x 4 = 48 Let s combine. Trapezium = Larger triangle + Rectangle = = 7 Area of the trapezium is 7 cm. 7

17 5. The figure below is made up of a small square, big square and a rectangle. The perimeter of the shaded rectangle is cm and the sum of the area of the squares is 78.5cm². Find the area of the shaded rectangle. :WORKING: Side of composite square = ( ) = Composite square = x = rectangles = 78.5 = 4.5 rectangle = 4.5 =.5 Area of the shaded rectangle is.5 cm. 8

18 :EXPLANATION: Draw lines as shown. Note that the rectangles are identical, and that the 4 shapes form a composite square. Let s look at the composite square. Side of composite square = Length of rectangle + Side of small square = Length of rectangle + Breadth of rectangle = L + B We know that perimeter of the rectangle is cm or (L + B). Equate the values of the perimeter. (L + B) = (L + B) = So, side of composite square = L + B = Composite square = x = Now, let s look at the rectangles. Original squares = 78.5 rectangles = Composite square Original squares = 78.5 = 4.5 rectangle x rectangles = = x 4.5 =.5 Area of the shaded rectangle is.5 cm. 9

19 :LET S APPLY: Problems Involving Rearranging Parts: Circle. A square ABCD is cut into 3 triangles. The area of the square is 80 cm². Given that M is the mid-point of AD, find the area of the shaded figure.. A rectangular plot of land measuring m by 4 m has a 4-m walking path running through it. Find the area of the walking path. 3. A, B, C and D are mid-points of the sides of a square as shown. If the area of the square is 96 cm², what is the area of the shaded triangle? 4. The figure shows a rectangle WXYZ. Some parts of it have been cut off, resulting in the shaded shape. The area of the shaded shape is 60 cm. Find the area of rectangle WXYZ. 0

20 5. The figure shows a big square of 90 cm² area. A, B, C and D are mid-points along the sides of the square. Find the area of the shaded parts. 6. The figure shows a big square, two identical right-angled triangles and a small square. The perimeter of the shaded region is 44 cm, and the total area of the two squares is 9 cm. Find the area of one triangle. 7. The figure consists of two rectangles, two big squares and two small squares. The total area of the four squares is 480 cm². The perimeter of each rectangle is 40 cm. Find the area of one rectangle.

21 CHAPTER REARRANGING PARTS: CIRCLE :EXAMPLES:. The figure below shows a square, a quadrant and a semi-circle. The area of the square is 40 cm². Find the area of the shaded parts. (Take = 3.4) :WORKING: 3 4 square = 3 4 = 80 x 40 Area of the shaded parts is 80 cm².

22 :EXPLANATION: HINT Clues are always in the question itself. Focus only on shapes mentioned in the question. In this example, we can see triangles. However, since the question does not mention anything about triangles, we ignore the triangles. Instead, we focus on the square, quadrant and semi-circle which are mentioned in the question. Draw half a diagonal line as shown. Pay attention to the parts shaded pink. Each pink-shaded part has a mirror unshaded part. Switch the corresponding parts. This is what you will get. 3

23 Extend the half diagonal into a full diagonal. The square is now cut into 4 equal parts. 3 parts are shaded, while part is unshaded. of the square is shaded. 3 4 Whole square = 40 square = x = 80 Area of the shaded parts is 80 cm². 4

24 . The figure below shows a big semi-circle, with a circle and identical smaller semi-circles within it. Find the area of the shaded parts. (Take =3.4) :WORKING #: shaded parts = Quadrant Triangle = ( 4 x r ) ( x B x H) = ( 4 x 3.4 x x ) ( x x ) =.4 Shaded parts =.4 x =.8 Area of the shaded parts is.8 cm². :WORKING #: Shaded parts = Semi-circle Triangle = ( x r ) (B x H) = ( =.8 x 3.4 x x ) (4 x ) Area of the shaded parts is.8 cm². :EXPLANATION: HINT Clues are always in the question itself. Focus only on shapes mentioned in the question. In this example, we focus on the big semi-circle, circle and identical smaller semi-circles which are mentioned in the question. Note how the semi-circle is made up of quadrants which are mirror image of each other. 5

25 Let s look at one of the quadrants. Draw lines as shown. Pay attention to the parts shaded pink. Each pink-shaded part has a matching unshaded part. Switch the corresponding parts. This is what you will get. Shaded parts = Quadrant Triangle = ( x r ) ( x B x H) = ( 4 4 =.4 x 3.4 x x ) ( Remember there are identical quadrants. Total shaded parts =.4 x =.8 ALTERNATIVE Shaded parts = Semi-circle Triangle = ( x r ) (B x H) = ( =.8 x x ) x 3.4 x x ) (4 x ) Area of the shaded parts is.8 cm². 6

26 3. The diagram below shows an isosceles triangle ABC with semi-circles within it. The semi-circles have a radius of 5 cm. Find the area of the shaded part. (Take as 3.4) :WORKING: Square = L x B = 30 x 30 = 900 Circle = r = (3.4 x 5 x 5) = corners = Square Circle = = corner = 4 = x The shaded area is cm. 7

27 :EXPLANATION: HINT Clues are always in the question itself. Focus only on shapes mentioned in the question. In this example, we focus on the triangle and semi-circles which are mentioned in the question. Note how the triangle is made up of smaller triangles which are mirror image of each other. Rotate the left smaller triangle clockwise on Point A, until Point B touches Point C. This is what you will get. Square = L x B = 30 x 30 = 900 Circle = r = (3.4 x 5 x 5) = corners = Square Circle = = corner = 4 = x The shaded area is cm. 8

28 4. The figure below is formed by overlapping 7 circles of 8 cm diameter. The inner circle passes through the centre of the 6 outer circles. Find the area of the shaded region. (Take = ) 7 :WORKING: circle 6 circles 3 3 circle shaded x 6 circles shaded = circles shaded = x r = ( x x 4 x 4) = Area of the shaded region is 498 cm. :EXPLANATION: There are 6 circles with identical shading as shown above. Look at one of the circles. 9

29 CONFUSION ALERT Each of the 6 circles with identical shading is NOT shaded as shown on the right. This is because the bottom shaded areas belong to the circles on the left and on the right. Draw lines as shown. Pay attention to the parts shaded pink. Each pink-shaded part has a matching unshaded part. Switch the corresponding parts. This is what you will get. of the circle is shaded. 3 circle 6 circles 3 3 circle shaded x 6 circles shaded = circles shaded = x r = ( x x 4 x 4) = Area of the shaded region is 498 cm. 30

30 5. The diagram below shows a rectangle and 3 identical circles within it. Given that the length of the rectangle is 7 cm, find the total area of the shaded parts. Use a calculator to obtain the value of. (Leave your answer correct to decimal places) :WORKING: Height of rectangle = Diameter of circle = 7 3 = 9 Shaded area = Triangle Circle = ( x B x H) ( r ) = ( x 7 x 9) ( x 4.5 x 4.5) = = Area of the shaded parts is 57.88cm. 3

31 :EXPLANATION: HINT Clues are always in the question itself. Focus only on shapes mentioned in the question. In this example, we can see triangles. However, since the question does not mention anything about triangles, we ignore the triangles. Instead, we focus on the rectangle and 3 circles which are mentioned in the question. Pay attention to the part shaded pink. The pink-shaded part has a matching unshaded part. Switch the parts. This is what you will get. The unshaded top segments of the circles on the left and on the right form a complete unshaded circle. Height of rectangle = Diameter of circle = 7 3 = 9 Shaded area = Triangle Circle = ( x B x H) ( r ) = ( x 7 x 9) ( x 4.5 x 4.5) = = Area of the shaded parts is 57.88cm. 3

32 6. The diagram below is made up of 3 circles of 8 cm diameter, with centres, X, Y and Z. Find the area of the shaded regions. (Take = ) 7 :WORKING: circle = = ( = 308 x r x 7 x 4 x 4) Area of shaded regions is 308 cm. 33

33 :EXPLANATION: Draw lines shown. Pay attention to the 3 parts shaded pink. Each pink-shaded part has a matching unshaded part. Switch the corresponding parts. This is what you will get. of a circle is shaded. circle = = ( = 308 x r x 7 x 4 x 4) Area of shaded regions is 308 cm. 34

34 :LET S APPLY: Problems Involving Rearranging Parts: Circle. The figure shows a window design. The radius of the semi-circle is 3 cm. Find a) the perimeter of the figure. b) the area of the figure. (Leave your answers correct to decimal places. Take π = 3.4). The figure is made up of a circle, 4 identical semi-circles and a square of 4 cm sides. O is the centre of the circle. Find the area of the shaded regions. (Leave your answer correct to decimal places. Use a calculator to obtain the value of ) 3. The figure shows a big square ABCD with sides measuring 8 cm, and 8 identical sectors. Find the total area of the shaded parts. (Take = 7 ) 4. The figure is made up of identical small circles, identical medium circles and identical quadrants. What is the total area of the shaded parts? (Leave your answer correct to decimal places. Take = 7 ) 35

35 5. The figure below is formed by a circle, and identical rectangles with one end rounded off into a quadrant. Each rectangle has an area of 56 cm before one end is rounded off. The length of the rectangle is 4 times its width. Find the area of the shaded parts. (Take = 7 ) 6. The figure shows identical quadrants and an equilateral triangle. Find the area of the shaded parts. (Take as 7 ) 7. The figure shows sets of 3 quadrants with radii of cm, cm and 3 cm. Find the area of the shaded parts. (Take as 3.4) 8. The figure below consists of a rectangle and six similar circles of 0 cm radius. Find the area of the shaded regions. (Take = 3.4) 36

36 9. The figure is made up of a square and 3 identical quadrant of 8 cm radius. Find the area of the shaded part. (Take = 7 ) 0. The diagram shows a circle with cm radius and centre O. AB and CD are diameters of the circle. Find the area of the shaded parts. (Take = 3.4). The diagram shows 4 identical circles of 4 cm radius, with centres A, B, C and D. Find the area of the shaded parts. (Take = 7 ). The figure is formed by overlapping 7 circle of 4 cm diameter. The inner circle passes through the centres of the 6 outer circles. Find the area of the shaded regions. (Take = 7 ) 37

37 CHAPTER 3 DRAWING LINES :EXAMPLES:. The figure below shows 6 identical circles of cm radius. Their centres are placed at the sides and corners of a rectangle. Find the area of the shaded figure. (Take = 3.4) :WORKING: Rectangle = 8 x 4 = 3 4 circles = 4 x 3.4 x x = 50.4 Shaded area = = 8.4 Area of the shaded figure is 8.4 cm. 38

38 :EXPLANATION: Draw lines as shown to form the rectangle mentioned in the question. Length of rectangle = 4 x = 8 Breadth of rectangle = x = 4 Rectangle = L x B = 8 x 4 = 3 Let s look at the remaining shaded areas. Together, the remaining shaded areas are equivalent to 4 circles of cm radius. 4 circles = 4 x r = 4 x 3.4 x x = 50.4 Shaded area = Rectangle + 4 circles = = 8.4 Area of the shaded figure is 8.4 cm. 39

39 . The figure below, not drawn to scale, shows a quadrant with a semi-circle within it. Find the area of the shaded part. (Take = 3.4) :WORKING: Semi-circle = Triangle = = 57 = 00 x 3.4 x 0 x 0 x 0 x 0 segments = = 57 segment = 57 = 8.5 sector = 8 = 57 x 3.4 x 0 x 0 Shaded part = sector Triangle segment = = 8.5 Area of the shaded part is 8.5 cm. 40

40 :EXPLANATION: CONFUSION ALERT Note that the question mentions that the figure is not drawn to scale. Hence, never assume shapes and sizes of shapes based on what you see. Let s look at the semi-circle and the triangle within the semi-circle. Radius of semi-circle = 0 = 0 Semi-circle = = = 57 x r x 3.4 x 0 x 0 The triangle is within the semi-circle. Base of triangle = 0 Height of triangle = Radius of semi-circle = 0 Triangle = x B x H x 0 x 0 = = 00 Note the parts shaded pink. segments = Semi-circle Triangle = = 57 segment = 57 = 8.5 4

41 Let s look at the quadrant ( 4 circle). A line divides the quadrant at 45, so the quadrant is divided into equal sectors ( circle). 8 Let s look at sector ( 8 circle). sector = = 8 8 = 57 x r x 3.4 x 0 x 0 Note how the triangle and segment are contained within the sector. Shaded part = sector Triangle segment = = 8.5 Area of the shaded part is 8.5 cm. 4

42 3. The figure below shows a rectangle that is cut into 4 triangles A, B, C and D. Find the area of triangle D. :WORKING: A + C = B + D = rectangle rectangle B + D = A + C 40 + D = D = 00 D = D = 60 Area of triangle D is 60 cm. 43

43 :EXPLANATION #: A + C = x B x H rectangle = B + D = x B x H rectangle = That means B + D = A + C 40 + D = D = 00 D = D = 60 Area of triangle D is 60 cm. 44

44 :EXPLANATION #: Draw lines as shown. The rectangle is divided into 4 quadrants of different sizes.. Within each quadrant is a diagonal line. Each diagonal line cuts its respective quadrant into equal parts. That means, any part is its respective quadrant. Logically, if we take of every quadrant and combine those areas, we get the rectangle. You can take of every quadrant as follows. A + C = rectangle B + D = rectangle That means B + D = A + C 40 + D = D = 00 D = D = 60 Area of triangle D is 60 cm. 45

45 4. Find the area of a square whose diagonal is cm. :WORKING: Triangle = = 36 Square = x 36 = 7 x x 6 Area of the square is 7 cm. 46

46 :EXPLANATION: The diagonal divides the square into identical triangles. Draw line as shown. This line is the height of one triangle. It is also a diagonal of the square. Height of triangle = Triangle = = = 36 = = 6 x B x H x x 6 diagonal Square = x Triangles = x 36 = 7 Area of the square is 7 cm. 47

47 5. The figure below shows a square of 0 cm sides. The ratio of lengths AB to BC and AD to DE is :. Find the area of the shaded region. :WORKING: XI : YI = AB : EF = : 3 XY: 4u 0 XI: u (0 4) = 5 YI: 3u (3 x 5) = 5 EFI = = 50 x 0 x 5 Shaded area = x 50 = 300 Area of shaded region is 300 cm. 48

48 :EXPLANATION: Draw lines and label additional points as shown. ABI and EFI are similar triangles. That means, the ratio of their bases equals the ratio of their heights. XI : YI = AB : EF = : ( + ) = : 3 XY = (u + 3u) = 4u XY: 4u 0 XI: u (0 4) = 5 YI: 3u (3 x 5) = 5 EFI = = = = 50 CFI = EFI x B x H x EF x YI x 0 x 5 Shaded area = x EFI = x 50 = 300 Area of shaded region is 300 cm. 49

49 6. The figure below, not drawn to scale, is made up of identical rectangles, identical big squares and small square. Find the area of the shaded small square. :WORKING: 0 x 0 = 00 Area of the shaded small square is 00 cm. 50

50 :EXPLANATION: Side of big square = Side of small square + Breadth of rectangle Draw lines as shown. Note how the leftmost vertical line divides the side of the left big square into side of small square and breadth of rectangle. Note how the rightmost vertical line also divides the side of the right big square into side of small square and breadth of rectangle. Draw more lines as shown. Note how the squares shaded pink are identical to the shaded square. Let s look at the breadth of the overall figure. Side of small square + ( x Breadth of rectangle) = 40 Let s look at the length of the overall figure. Mark out Side of small square + ( x Breadth of rectangle) = 40 5

51 We know that Length of the overall figure = 60 This leaves ( x Side of small square) = = 0 Side of small square = 0 = 0 Small square = 0 x 0 = 00 Area of the shaded small square is 00 cm. 5

52 :LET S APPLY: Problems Involving Drawing Lines. The diagram, not drawn to scale, shows a figure EFGH, 3 semi-circles and a right-angled triangle. If AC = 4 cm, BC = 4 cm and AB = 8 cm, find the area of the rectangle EFGH.. A square is drawn in a circle with centre O as shown. The area of the square is 46 cm². Find the area of the circle. (Take = 3.4) 3. The diagram, not drawn to scale, shows a right-angled triangle and a quadrant. Find the area of the shaded parts. (Take = 3.4) 4. A rectangle WXYZ is divided into four triangles. The areas of triangle WZO and triangle YZO are 6 cm² and 7 cm² respectively. The area of triangle XYO to the area of rectangle WXYZ is 3:0. Find the area of the triangle WXO. 53

53 5. The figure is made up of squares of sides 4 cm and 9 cm. Find the area of the shaded figure. 6. WXYZ is a square with an area of 90 cm². The shaded parts A and B are squares with different areas. Find the area of the shaded parts. 7. The figure shows square ABCD with diagonals measuring 40 cm. Given that the shaded parts are identical, find the total area of the shaded parts. (Take = 7 ) 8. ABCD is a square. AOC and BOE are straight lines. Given that E is the mid-point of AD, find the shaded area of the figure. 54

54 9. The figure below shows a square ABCD with 30 cm sides. The ratio of lengths BE to EC is :. Find the area of the shaded region. 55

55 ANSWERS Answers to questions in the prior chapters Let s Apply sections are listed in this chapter. Detailed workings may be downloaded at: CHAPTER REARRANGING PARTS: CIRCLE :LET S APPLY: Problems Involving Rearranging Parts: Circle a) 8.64 cm b) 338 cm ) cm 3) 68 cm 4) cm 5) 384 cm 6) 46 cm 7). cm 8) 58 cm 9) 4 cm 0) cm ) 3693 cm ) 308 cm 56

56 CHAPTER REARRANGING PARTS: NON-CIRCLE :LET S APPLY: Problems Involving Rearranging Parts: Non-circle ) 60 cm ) 44 m 3) 60 cm 4) 64 cm 5) 36 cm 6) 40 cm 7) 80 cm 57

57 CHAPTER 3 DRAWING LINES :LET S APPLY: Problems Involving Drawing Lines ) 59 cm ) 7. cm 3) 53.5 cm 4) 8 cm 5) 30.5 cm 6) 4.5 cm 7) 5600 cm 8) 9 cm 9) 330 cm 58