This problem solving activity has a geometry focus.
The High Peak Jewellery Company wants a new logo. Martin has come up with one based on a ring and a diamond. It is constructed by inscribing a rectangle in a circle as shown below.
The diamond is then drawn inside the rectangle.
Can Martin construct the shape using only ruler and compasses? What is the area of the diamond shape?
This geometrical problem involves the application of standard constructions. This should support students to see how to solve the area aspect of the problem. At first it appears that there is insufficient information for a solution to be found, however, in such cases a diagram can be a useful tool to enable students to proceed.
Note: A number of problems appear to have insufficient information, including the Level 6 Number and Algebra (Equations and Expressions) problem Pigs, Goats and Sheep. In this case, the fact that whole numbers were involved meant that extra necessary information could be derived.
The High Peak Jewellery Company wants a new logo. Martin has come up with one based on a ring and a diamond. It is constructed by inscribing a rectangle in a circle as shown below.
The diamond is then drawn inside the rectangle.
Can Martin construct the shape using only ruler and compasses?
What is the area of the diamond shape?
A circle is inscribed in the rhombus. What is its area?
Invent your own logo for The High Peak Jewellery Company.
To construct the logo Martin has first to draw in the circle. That is straightforward. Now draw in any diameter through the centre of the circle. To get the point which is 3 cm from the centre and the circumference use the normal method of producing the perpendicular bisecting a line segment. Then extend the perpendicular bisector if necessary to cut the circle at two points. Repeat the process on the other side of the centre. Now join the four points on the circle to make the rectangle. Finally join up the points to make the rhombus.
The key to finding the area is to see that the side of the rhombus equals the radius of the circle, since the diagonals of a rectangle are equal. Thus, using Pythagoras’ Theorem, the other side of the triangle is
62 = a2 + 32
27 = a2
√27 = a
3√3 = 27
Hence the area is half the base times the height = 0.5 x 3 x 3√3. This is approximately 7.79 cm2.
To find the radius of the circle inscribed in the rhombus, note that the radius of this circle is perpendicular to the side of the rhombus.
Now Sin A = 3/6 using the larger triangle and Sin A = R / (3√3).
Therefore R / (3√3) = 3/6
Giving R = (9√3)/ 6 or 2.6.
Printed from https://nzmaths.co.nz/resource/rings-and-diamonds at 12:08am on the 26th April 2024