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Level Five > Geometry and Measurement

Achievement Objectives:

Specific Learning Outcomes:

Construct two intersecting circles where the radius of the second circle is on the circumference of the first.

Use Pythagoras’ theorem to find the area of a rhombus

Devise and use problem solving strategies to explore situations mathematically (be systematic, draw a diagram, use a model)

Description of mathematics:

The construction part of this problem is a straightforward application of ideas involving circles. Students will need to know what the significance of the two centres being on the two circumferences is as far as locating the two circles is concerned. The construction of the rhombus should be relatively easy.

There are a couple of ways of tackling the area. One way is to notice that the two triangles formed by joining the centres of the circles, are equilateral. Another way is to draw in the two diagonals of the rhombus and work out the lengths of the sides of the triangles obtained. This can be done by Pythagoras’ theorem or by using trigonometric ratios.

Area is an important quantity in mathematics. There are obvious applications of plane figures such as rectangles to fields and house sections. In fact the whole business of geometry may well have started in Egypt when fields had to be surveyed after the annual flood in the Nile Delta. Newton and Leibnitz, of course took the whole thing further when they introduced the integral calculus and so opened up a new range of areas to calculation. They also opened up areas other than those of plane figures that were of great use.

Students in the last two years of high school will come across the integral calculus and its applications.

For a similar problem at a more difficult level, see Dan’s Badge, Measurement, Level 6.

Required Resource Materials:
Rulers
Compasses
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:

### Problem

All the students in the class decided to make a badge. Bill did his with ruler and compass and coloured it red and white. His drawing is below. The centre of one circle lies on the circumference of the other. Both circles have the same radius.

Use ruler and compass to draw your own copy of Bill’s Badge.

Bill used a radius of 10cm for his circles. What is the area of the red piece?

### Teaching sequence

1. Plan a way of capturing the interest of the students in the problem. One possibility is to begin with a brainstorm about circles. Draw a circle on the board and ask the students to tell you everything that they know about circles.
2. Pose the problem to the class. Check that everyone understands what the problem is about before asking them to work on it individually (or in pairs).
3. As the students construct the intersecting circles, circulate asking questions that focus on the construction:
What is the length this (pointing)? How do you know?
What shape is formed in the centre? How do you know?
4. As the students move on to working out the area of the rhombus ask questions that focus their thinking on the area of triangles.
How are you working out the area?
How do you work out the area of a triangle? Does that apply to all triangles?
How do you know the "height" of the triangle?
5. Ask the students to record their work so that it can be shared with the class.
6. Share solutions.

#### Solution

Construction: Since both circles have the same radius, the centre of the second circle is anywhere on the circumference of the first circle. So after drawing the first circle, place the point of the compass anywhere on the first circle and draw another circle with the same radius. This will give the two centres. The other two points are at the intersection of the two circles so it is just a matter of joining up the four points to make the rhombus.

Area: Draw in the two diagonals to form the figure below.

Then AC = 10 since AC is the radius of both circles. Hence AX = 5. Similarly AB = 10. Now the angle of AXD = 90 degrees, so we can find XD using Pythagoras.

XD2 = AX2 – AD2 = 102 – 52 = 100 – 25 = 75.

Hence XD = √75 = 5√3 ≈ 8.660

The area of the shaded figure is:
= 4 x the area of triangle AXD
= 4 (half base times altitude)
= 4( 0.5 x 5√3 x 5)
= 50√3
≈86.6 cm 2

AttachmentSize

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