Year 9: Geometry Problems and Proof - PowerPoint Presentation

Year 9: Geometry Problems and Proof www.drfrostmaths.comDr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 18 th January 2016

Geometry Problems: Section 1 – Lengths of Lines Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Starter Find the length of the following lines.         2 2 2             2           3 (In terms of )   tends to result from finding diagonals of squares. ! To get the diagonal of a square, To get the side length of a square from the diagonal,   tends to result from equilateral triangles (or half an equilateral triangle). ! To get the height of an equilateral triangle, half the side length then   ? ? ? ? ? 1 1

Further Examples 1 A regular octagon is inscribed inside a square. Find the fraction of the length of the square that is taken up by the octagon.     1 1 By forming a square, we find the length of the square is . Thus the fraction is   ?     A circle is inscribed inside a square inside a circle. What fraction of the outer circle is occupied by the smaller one? Let radius of smaller circle be 1. By Pythagoras radius of outer circle is . So it’s half.   ?

Test Your Understanding [Kangaroo Pink 2012 Q11] Six identical circles fit together tightly in a rectangle of width 6 cm as shown. What is the height, in cm, of the rectangle? Solution:   1 1     ? Harder Easier [IMC 2006 Q14] A rectangle is cut into two pieces along the dotted line shown. The two pieces are then rearranged to form a right-angled triangle. What is the perimeter of the triangle formed ? Solution: 24   ?

Touching Circles When you have touching circles/semicircles or a circle in contact with other shapes. Connect the centres of any touching circles. Draw a line from the centre of any circle to any points of contact with other shapes. Try to form right-angled triangles from any diagonals (e.g. by adding vertical/horizontal lines). [Based on SMC 2014 Q19] The diagram shows a quarter-circle of radius 2, and two touching semicircles. The larger semicircle has radius 1. Can you form a right-angle triangle and find expressions for all three sides? (You may wish to make the radius of the smaller circle )   ? Identify triangle       ? ? ?

Test Your Understanding Harder 3cm 5cm Using Pythagoras,   ? Easier [IMC 2005 Q21] Two circles with radii 1cm and 4cm touch. The point is on the smaller circle, is on the larger circle and is a tangent to both circles. What is the length of ? A cm B 3cm C cm D cm E 4cm   Suppose the radius of the circle is 1. We wish to find the length of each square. Suppose we called it . Form an appropriate right-angled triangle (using the tips on the previous slide) and find expressions for the three sides.   ? Reveal       (We’d then subsequently use Pythagoras to give us an equation and solve, but we’ll do this later) 1cm 1cm

One final example… An equilateral triangle is inscribed inside a circle of radius 2. Determine the perimeter of the triangle. 2   As per usual advice, draw a line from centre of the circle to a point of contact… If we add this line, can we notice anything special about this triangle formed? If you ever see a 30-60-90 triangle, you should recognise it as half an equilateral triangle . This allows us to use Pythagoras to find the remaining side. 1   Perimeter (Side note: the above diagram also tells us that the centre of an equilateral triangle is a third of the way up the height – very useful!)  

[Kangaroo Pink 2008 Q3] Four unit squares are placed edge to edge as shown. What is the length of the line ? A 5 B C D E 13 Solution: B   Exercise 1 1 Two identical squares are inscribed in a semicircle of radius . Find the length of each square. Solution: A square of length 2 is inscribed in a semicircle. Find the radius of the semicircle . Solution:   2 3 [Kangaroo Pink 2011 Q6] The diagram shows a shape made from a regular hexagon of side one unit, six triangles and six squares. What is the perimeter of the shape? A B C 12 D E 9 Solution: 12         4 ? ? ? ?

[IMC 2009 Q7] Four touching circles all have radius 1 and their centres are at the corners of a square. What is the radius of the circle through the point of contact , , and ?A B C 1 D E   Exercise 1 5 ?

Exercise 1 [Based on IMC 2015 Q24] In square a quarter-circle arc with centre is drawn from to . A point on this arc is 1 unit from and 8 units from . Suppose we wanted to find the the length of the side of square . Let the length of the square be . Draw a line in your diagram horizontally right from the point to form a right-angled triangle. Find the lengths of the three sides of your triangle in terms of (you do not at this stage need to find the value of ).         6 ?

Exercise 1 [Kangaroo Pink 2015 Q9] In the grid, each small square has side of length 1. What is the minimum distance from ‘Start’ to ‘Finish’ travelling only on edges or diagonals of the squares? A B C D E 6 Solution: C   7 8 [ IMC 2011 Q19] Harrogate is 23km due north of Leeds, York is 30km due east of Harrogate, Doncaster is 48km due south of York, and Manchester is 70km due west of Doncaster. To the nearest kilometre, how far is it from Leeds to Manchester, as the crow flies? A 38km B 47km C 56km D 65km E 74km Solution: B (Manchester is 40km West and 25km south of Leeds. Use Pythagoras) ? ?

Exercise 1 9 [IMC 2009 Q20] A square, of side two units, is folded in half to form a triangle. A second fold is made, parallel to the first, so that the apex of this triangle folds onto a point on its base, thereby forming an isosceles trapezium. What is the perimeter of this trapezium? A B C D E 5 Solution: D           ?

Exercise 1 10 [Kangaroo Pink 2010 Q7] The diagram shows a square and two equilateral triangles and . has length 1. What is the length of ? A B C D E Solution: A Find the area of a regular octagon of side length 1.               11                     Area: 4 triangles + 4 rectangles + 1 square   ? ?

Exercise 1 12 [Based on Hamilton 2006 Q4] A circle is inscribed in a square and a rectangle is placed inside the square but outside the circle. Two sides of the rectangle lie along sides of the square and one vertex lies on the circle, as shown. The rectangle is twice as high as it is wide. We wish to find the ratio of the area of the square to the area of the rectangle. a) Using the information, give suitable numeric lengths to the width and height of the rectangle. b) As per the usual advice for diagrams involving circles, add a suitable line to your diagram, and form a right-angle triangle. If the radius of your circle is , give expressions for the three lengths of your triangle.             ? Reveal

Exercise 1 13 [IMC 2012 Q21] The parallelogram is formed by joining together four equilateral triangles of side 1 unit, as shown. What is the length of the diagonal ?               By forming the triangle above, we know the base is and the height of the triangle (using the ‘splitting method’) is . Use Pythagoras again gives:   ?

Exercise 1 14 [IMC 2004 Q18] In the triangle , there is a right angle at and angle is . The bisector of the angle meets at , as shown. What is the ratio ? A B C D E                   By using the information, we can work out all the angles as shown. Since is isosceles, then . Suppose we made it 2. We can see is half an equilateral triangle, thus . The answer is therefore E. Alternatively we could have used the Angle Bisector Theorem, which would mean the ratio of is the same as .   ?

Exercise 1 15 [Cayley 2014 Q4] The square lies inside the regular octagon . The sides of the octagon have length 1. Find length .                             because and (as is the interior angle of an octagon). By Pythagoras,   ?

Exercise 1 16 [JMO Mentoring May2012 Q4] A triangle has two angles which measure 30° and 105°. The side between these angles has length 2 cm. What is the perimeter of the triangle?                       By drawing and splitting into two right-angle triangle, we can see we have half a square and half an equilateral triangle. Solution:   ?

Exercise 1 17 In the diagram, the letter is made from two arcs, and , which are each five-eighths of the circumference of a circle of radius 1, and the line segment , which is tangent to both circles. At points and , common tangents to the two circles touch one of the circles. What is the length ? A B C 2 D E           Since the angles in the triangles are , we can see that the lines connecting and to the centre of the diagram are both just 1 (as the triangles are isosceles). So the answer is just 2.   ?

Exercise 1 18 [IMC 1999 Q25] A rectangular sheet of paper with sides 1 and has been folded once as shown, so that one corner just meets the opposite long edge. What is the value of the length ? A B C D E           Since the paper is folded, By Pythagoras, . Thus . Consequently, as , is also an isosceles right-angled triangle and so       ?

Exercise 1 19 [Kangaroo Pink 2008 Q24] In the diagram, is a unit square. Arcs of radius one unit are drawn using each of the four corners of the squares as centres. The arcs centred at and intersect at ; the arcs centred at and intersect at . What is the length of ? A B C D E     is an equilateral triangle of side 1. Its height is (recall that we can halve the side length then ). Therefore, observing that is the overlap of two equilateral triangles,   ?

Geometry Problems: Section 2 – Finding lengths using algebra Dr J Frost (jfrost@tiffin.kingston.sch.uk) (Note to teacher: Please don’t do this lesson until you’ve taught solving quadratic equations to your class)

Introducing variables for unknown lengths We touched upon how it can be useful to introduce variables for unknown lengths . We then hope to be able to form an equation so we can solve ( usually using Pythagoras ).           [Hamilton 2006 Q4] A circle is inscribed in a square and a rectangle is placed inside the square but outside the circle. Two sides of the rectangle lie along sides of the square and one vertex lies on the circle, as shown. The rectangle is twice as high as it is wide.What is the ratio of the area of the square to the area of the rectangle? We formed the following triangle in the previous exercise. Now we can use Pythagoras to solve!   However 1 is not possible as . Thus the area of the rectangle is 2 and area of the square . Ratio is .   ?

Test Your Understanding           ? Form right-angled triangle (and find expressions for lengths) Find the radius of the smaller circle.   ? Solve your equation ? Use Pythagoras to get equation

Using variables to represent ratio A rectangle whose length is twice its height is inscribed inside a circle of radius 1. Find the height of the rectangle.       Ratio is 2:1, so use and (so that we can easily halve them). So height is   Q ? As per usual advice, connect centre of circle to point of contact. What could we make width and height of rectangle?

Exercise 2 [SMC 2014 Q19] The diagram shows a quadrant of radius 2, and two touching semicircles. The larger semicircle has radius 1. What is the radius of the smaller semicircle? A B C D E [ IMC 2015 Q24] In square a quarter-circle arc with centre is drawn from to . A point on this arc is 1 unit from and 8 units from . What is the length of the side of square ? A 9 B 10 C 11 D 12 E 13               1 2 ? ?

Exercise 2 [IMC 2011 Q23] A window frame in Salt’s Mill consists of two equal semicircles and a circle inside a large semicircle with each touching the other three as shown. The width of the frame is 4m. What is the radius of the circle in metres? A B C D E 1 [IMC 2010 Q23] The diagram shows a pattern of eight equal shaded squares inside a circle of area square units. What is the area (in square units) of the shaded region? A B C D E 2   3             4 ? ?

Exercise 2 5 [Cayley 2010 Q5] A square sheet of paper ABCD is folded along FG, as shown, so that the corner B is folded onto the midpoint M of CD. Prove that the sides of triangle GCM have lengths in the ratio . Let and . Since due to folding, Then: Thus ratio of three sides:         ?

Exercise 2 6 [Cayley 2008 Q5] A kite has sides AB and AD of length 25cm and sides CB and CD of length 39cm. The perpendicular distance from B to AD is 24cm. The perpendicular distance from B to CD is h cm. Find the value of h . Pythagoras on : (in both cases you should be able to spot Pythagorean’s triples and multiples of them) We could continue to use Pythagoras to find . But a smarter way is to realise the area of can be found in two ways, using either CD as the base or BD: The perpendicular height of the triangle using BD as the base is . Thus   E F ?

Exercise 2 [Hamilton 2010 Q5] The diagram shows three touching circles, whose radii are a , b and c , and whose centres are at the vertices Q, R and S of a rectangle QRST. The fourth vertex T of the rectangle lies on the circle with centre S . Find the ratio . By connecting the centres of the circles we get lengths . Using Pythagoras: But we can also see from the diagram that , thus: Ratio is:   7         ?

[Kangaroo Pink 2004 Q24] The shaded area is equal to . What is the length of ? A 1 B 2 C 3 D 4 E impossible to determine   Exercise 2 8 Let radii of two circles be and . Since the diameter of the large circle is , the radius must be . Therefore Drawing the indicated triangle we know two of the lengths, and can use Pythagoras to find the third.           ?

Exercise 2 9 [Maclaurin 2011 Q5] Three circles touch the same straight line and touch each other, as shown. Prove that the radii a , b , and c , where c is smallest, satisfy the equation:               Key here is to form a number of right-angled triangles. Notice the top length of the rectangle is the same as the bottom length – this allows us to express the length in two different ways: Then dividing everything by :   ?

Geometry Problems: Section 3 – Areas and Circles Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Starter In terms of , find the area of the shaded regions. (You may assume any curved line is an arc of a circle)               Hint: You know how to find the area of a right-angled triangle! Could we perhaps split the shape up?   1 2 N ? ? ?  

Common Shapes You’ll See These are particularly common circle-related shapes you will see in Maths Challenge/Olympiad questions.             You can either think of this as double the previous shape, or as adding two quarter circles together and discarding the square to leave the overlap. ? ?

Example Let radius of the circle be 1 for simplicity. Then we can see a quarter of the shaded region can be formed by starting with a square and cutting out a quarter circle. This has area . Area of shaded region Fraction shaded =   ? [IMC 2012 Q19] The shaded region shown in the diagram is bounded by four arcs, each of the same radius as that of the surrounding circle. What fraction of the surrounding circle is shaded? A B C D E it depends on the radius of the circle  

Exercise 3 1 [JMO 2015 A5] Two circles of radius 1 cm fit exactly between two parallel lines, as shown in the diagram. The centres of the circles are 3 cm apart. What is the area of the shaded region bounded by the circles and the lines ? Solution: ) cm 2 (note how the answer is written ) [Kangaroo Grey 2005 Q9] In the diagram, the five circles have the same radii and touch as shown. The square joins the centres of the four outer circles. The ratio of the area of the shaded parts of all five circles to the area of the unshaded parts of all five circles is A B C D E Solution: B   2 ? ?

Exercise 3 3 [IMC 2014 Q20] The diagram shows a regular pentagon and five circular arcs. The sides of the pentagon have length 4. The centre of each arc is a vertex of the pentagon, and the ends of the arc are the midpoints of the two adjacent edges. What is the total shaded area? A B C D E Solution: D [IMC 2014 Q10] The diagram shows five touching semicircles, each with radius 2. What is the length of the perimeter of the shaded shape? A B C D E Solution: B   4 ? ?

Exercise 1 Answers 5 [Hamilton 2011 Q4] A square just fits within a circle, which itself just fits within another square, as shown in the diagram. Find the ratio of the two shaded areas.       Let width of inner square be 2. Then radius of circle is and length of outer square . Area on left: (cut inner square out of circle then quarter) Area bottom-right: Ratio:   ?

Exercise 3 6 [ IMC 1997 Q23] The square is inscribed in a circle of radius 1. Semicircles are drawn with diameters as shown, and the parts of these semicircles which lie outside the original circle are shaded. What is the total area of these four shaded regions? A B 2 C D 1 E Solution: B (We could start with the area of everything using the four semicircles and the square, then cutting out the circle in the middle)   ?

Exercise 3 7 [Cayley 2009 Q2] The boundary of a shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line AB, which is 10cm long. What is the length of the perimeter of the figure. Solution: cm [IMC 2008 Q17] The shaded region is bounded by eight equal circles with centres at the corners and midpoints of the sides of a square. The perimeter of the square has length 8. What is the length of the perimeter of the shaded region? A B C D E Solution: D   8 ? ?

Exercise 3 9 [Kangaroo Pink 2007 Q18] A coin with diameter 1 cm rolls around the outside of a regular hexagon with edges of length 1cm until it returns to its original position. In centimetres, what is the length of the path traced out by the centre of the coin? A B C D E Solution: C   ?

Exercise 1 Answers 10 [IMC 2006 Q21] The diagram shows two semicircular arcs, and . The diameters, and , of the two semicircles are parallel; is of length 4 and is a tangent to semicircular arc . What is the area of the shaded region ? A B C D 4 E       If radius for big semicircle is 2, then radius of smaller one is . The shaded region consists of two shapes: a semicircle, of area , and the bit at the top. The bit at the top is a segment. We can get the area of a segment by starting with a quarter circle and cutting out the triangle. So   ?

Exercise 3 [STMC Regional 2007/08 Q5] The square in the diagram below has sides of length two units. The shaded sections are enclosed by 4 semi-circles. Calculate the exact value of the total area of the unshaded regions . Solution: (Note that if we added four semicircles and subtracted a square, this would give us the shaded region)   11 ?

[Hamilton 2005 Q2] The region shown shaded in the diagram is bounded by three touching circles of radius 1 and the tangent to two of the circles. Calculate the perimeter of the shaded region. Exercise 1 Answers 12   The circles meet at angles. So each of the three central arcs are a sixth of a circle. We have an additional two quarter arcs at the bottom, making a total of one full circle! Perimeter:   ?

Exercise 3 13 [Hamilton 2010 Q4] The diagram shows a quarter-circle with centre O and two semicircular arcs with diameters OA and OB. Calculate the ratio of the area of the region shaded grey to the area of the region shaded black. Solution: [Kangaroo Grey 2005 Q18] Two rectangles and are shown in the diagram. What is the area of the rectangle ? A 10cm 2 B 12cm 2 C 13cm 2 D 14cm2 E 16cm2Solution: B   14 ? ?

Exercise 3 15 [IMC 2003 Q19] What is the area of the pentagon shown? A B C D E Solution: B [IMC 2005 Q18] Three-quarters of the area of the rectangle has been shaded. What is the value of ? A 2 B 2.4 C 3 D 3.6 E 4 Solution: E   16 ? ?

Exercise 3 17 [Cayley 2015 Q5] The diagram shows a right-angled triangle and three circles. Each side of the triangle is a diameter of one of the circles. The shaded region R is the region inside the two smaller circles but outside the largest circle. Prove that the area of R is equal to the area of the triangle. Let lengths of triangle be . As the triangle is right-angled, . The area of the triangle is . We can find R by adding the two smaller semicircles and the triangle in the middle, before subtracting the large semicircle. Thus the areas are the same.         ?

Exercise 3 18 [Kangaroo Grey 2012 Q17] In the diagram, is a square, is the midpoint of and is perpendicular to . What is the ratio of the area of the shaded triangle to the area of the square? A B C D E Solution: D [IMC 1998 Q24] Each of the sides of this regular octagon has length 2cm. What is the difference between the area of the shaded region and the area of the unshaded region (in cm 2 )? A 0 B 1 C 1.5 D 2 E Solution: A (we saw how to split up the octagon in Exercise 1)   19 ? ?

Geometry Problems: Section 4 – Cutting a nd ReformingDr J Frost (jfrost@tiffin.kingston.sch.uk)

Question: The diagram shows three semicircles, each of radius 1. What is the size of the total shaded area? A: π + 2 B: 5 Frequently, we can chop a shape into bits and put them back together in some other way, clearly with the same area. The key is cutting this into shapes we know the area of. We have a 2x1 rectangle, and the remaining bits can combine to form a circle of area π . This avoids us having to calculate the complicated area in the middle. Skill #4: Cutting and Reforming C:   D:   E:  

Test Your Understanding The figure shows two shapes that fit together exactly. Each shape is formed by four semicircles of radius 1. What is the total shaded area? We can cut up the shape and reform it to get a rectangle with area 8.   ?

Test Your Understanding The diagram shows a design formed by drawing six lines in a regular hexagon. The lines divide each edge of the hexagon into three equal parts. What fraction of the hexagon is shaded ? A B C D E   (Hint: Can you add lines to the shape such that it’s broken up into smaller shapes all of the same area?) After adding the lines to break up the shape into small triangles, we could just count the total triangles and shaded triangles. But since the pattern in the blue region is repeated, we could just find the fraction shaded in this, which is .   ?

Exercise 2 Answers Q1 The diagram contains six equilateral triangles with sides of length 2 and a regular hexagon with sides of length 1. What fraction of the whole shape is shaded? A B C D E     ?

Exercise 2 Answers Q2 The diagram shows an equilateral triangle with its corners at the mid-points of alternate sides of a regular hexagon. What fraction of the area of the hexagon is shaded? A B C D E     ?

Exercise 2 Answers Q3 Find the fraction of the shape which is shaded.   ?

The diagram on the right shows a rectangle with sides of length 5cm and 4cm. All the arcs are quarter-circles of radius 2cm. What is the total shaded area in cm 2.A B 8 C D 10 E   Exercise 2 Answers ? Q4

Exercise 2 Answers Q5 The diagram shows a ceramic design by the Catalan architect Antoni Gaudi. It is formed by drawing eight lines connecting points which divide the edges of the outer regular octagon into three equal parts, as shown. What fraction of the octagon is shaded? A B C D E   Here we really do have to focus on just part of the shape to keep things simple. of the triangles are shaded. Note that we ignored some of the lines.   ?

Exercise 2 Answers Q6 The diagram shows a square with two lines from a corner to the middle of an opposite side. The rectangle fits exactly inside these two lines and the square itself. What fraction of the square is occupied by the shaded rectangle? A B C D E   We needn’t break up to non-shaded large triangles as we know they make up half the square. Of the other half, 8 tenths are shaded. So is shaded.   ?

Geometry Problems: A few bonus problems… Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Bonus Questions The radius of the circle is 1. The arc is formed by a circle whose centre is the point A. We wish to find the area shaded.       Find the length of , i.e. the radius of the circle centred at . Hence find the area of segment area formed by the chord . Hence find the area of the shaded region.     a b c ? ? ?

Bonus Questions Q2 Two unit circles are inscribed inside a rectangle, such that each circle passes through the centre of the other. Determine the shaded area. (Hint: Think what the angle of spread of each of the two segments is)       Since , and are all the radius of the circles, they form an equilateral triangle. If we split the shaded region in half and look at the right half, this segment has an angle of . Segment area is third of a circle minus indicated triangle (whose base and height we can find by Pythagoras). So area of shaded   ?

Bonus Questions Q N Four quarter circles join the corners of a square with side length 2. Determine the area of the shaded region. The key is to split the central area into 4. Each area can be found by starting with the area of the arc, and subtracting the two triangles. Since is an equilateral triangle, Area of sector: Area of triangle: Area of quarter of shaded region: Area of shaded:           ?