This problem solving activity has a measurement focus.
Bill is making a badge. He uses a ruler and a compass, and he colours 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 uses a radius of 10cm for his circles.
What is the area of the red piece?
In this problem students apply their knowledge of circles to a construction task. They should understand the significance of the two centres being on the two circumferences and know how to construct a rhombus.
As they work on the problem, students may notice that the two triangles formed by joining the centres of the circles, are equilateral. They may also approach the problem by applying their knowledge of Pythagoras’ theorem, or by using trigonometric ratios.
A similar problem is: Dan’s Badge, Measurement, Level 6.
Bill is making a badge. He uses a ruler and a compass, and he colours 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 uses a radius of 10cm for his circles. What is the area of the red piece?
Construction: Since both circles have the same radius, the centre of the second circle is anywhere on the circumference of the first circle. 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
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/bill-s-badge at 8:58pm on the 26th February 2024