**Additional Materials **

**Materials Required**

**Estimated Time Needed**

*(Times are approximate and will depend on the needs of the students.)*

Homework

- Improving Individual Solutions to the Assessment Task (10 minutes)
**Return the original assessment task Shape Statements to the students, together with a second, blank, copy of the task.**

*Look at your original responses, and think about what you have learned in this lesson.**Using what you have learned, try to improve your work.*

**If you have not added questions to individual pieces of work, then write your list of questions on the board. Students should choose from this list only the questions that they think are appropriate to their own work.**

- Solutions: Shape Statements
The solutions below show just one or two strategies for each statement. Students may think of other correct approaches.

- The first statement is
**sometimes true**. The student should give examples of where it is true and also where it is not. For example, it is true if the shapes are two different size circles, but it is not true if one shape is a 3,4,5 triangle and one is a 3 by 2 rectangle. -
The second statement is

**sometimes true**. It is true if the sides joined are the parallel ones.This is shown in the diagram:

Students may use the area of a trapezoid formula:

½ × (sum of parallel sides) × perpendicular distance between them

The sum of the parallel sides for trapezoid X and trapezoid Y are equal. Therefore, Area X = Area Y

Alternatively, students may split the trapezoid as shown:

The two areas marked A

_{1}and A_{2}are equal as they are triangles with the same height and base length. The two areas marked B_{1}and B_{2}are also equal as they are triangles with the same height and base length.Therefore, Area A

_{1}+ Area B_{1}= Area A_{2}+ Area B_{2}, so the trapezoid has been split into two equal areas. -
The third statement is

**sometimes true**. It is true for acute-angled triangles: the triangle's area is one half of the rectangle area in each case. This is shown in the diagram below:Area A = Area B and Area X = Area Y (diagonals of a rectangle cut the rectangle in half.) The area of the rectangle is always twice as big as the area of the triangle.

However, if a triangle has an obtuse angle it is not possible to draw three rectangles in the way described above.

- The first statement is

- Solutions: Always, Sometimes, or Never True?
**A. Cutting shapes: (a) Always true (b) Sometimes true.**This task is intended to help students to overcome the misconception that areas and perimeters are related in some way.

(a) Clearly, it is always true that cutting a piece off a shape will result in a decrease in area.

(b) On the hint cards, examples are shown where cutting a piece off can reduce, leave unchanged, or increase the perimeter.

**B. Sliding a triangle: (a) Always true (b) Always true.**(a) The base is always the same. The height will always be the same, because it is sliding along a line parallel to the base. That means the triangle will always have the same area as the area of a triangle is "half the base times the height".

(b) By sliding the top vertex to the right, we can make the perimeter as large as we like, while preserving the area.

**C. Rectangles: (a) Always true (b) Sometimes true.**(a) This is most easily seen by labeling equal areas:

Area A

_{1}= ½ × r × p and Area A_{2}= ½ × p × r. Therefore, A_{1}= A_{2}Area B

_{1}= ½ × s × q and Area B_{2}= ½ × q × s. Therefore, Area B_{1}= Area B_{2}The diagonal cuts the rectangle in half:

Area A

_{1}+ Area X + Area B_{1}= Area A_{2}+ Area Y + Area B_{2}.Therefore Area X = Area Y.

(b) The perimeters are equal when the two shaded rectangles have a common vertex at the center of the large rectangle:

Perimeter X = 2q + 2r and Perimeter Y = 2s + 2p

The two perimeters are equal when q + r = s + p. This is true when q = p and s = r.

**D. Medians of triangles: Always true.**Since each of the medians bisects one side of the triangle, we can see that pairs of small triangles (labeled X

_{1}, X_{2}; Y_{1}, Y_{2}; and Z_{1}, Z_{2}) have the same base length and height and so have equal areas.Triangles AMB and AMC also have equal areas, therefore:

Area X

_{2}+ Area X_{1}+ Area Z_{1}= Area Y_{1}+ Area Y_{2}+ Area Z_{2}.But we know that Area X

_{1}= Area X_{2}and Area Y_{1}= Area Y_{2}and Area Z_{1}= Area Z_{2}, so:2 × Area X

_{1}= 2 × Area Y_{1}Area X

_{1}= Area Y_{1}Therefore, all 6 triangles having the same area.

**E. Square and circle: Never true.**Suppose the square has a perimeter of 4x units, which implies the area is x

^{2}units^{2}.If the circle also has a perimeter of 4x units,

2πr = 4x

⇒ r = 2x/π

⇒ πr

^{2}= 4x^{2}/π > 1So the circle has a greater area.

**F. Midpoints of a quadrilateral: Always true.**Let ABCD be the quadrilateral and M, N, P, Q be the midpoints of the sides. Draw the diagonals AC and BD.

AMQ and ABD are similar triangles (angle BAD is a common angle A, M = ½ AB, and AQ = ½ AD

Therefore, MQ is parallel to BD, and MQ = ½ BD

Simarly, NP is parallel to BD, and NP = ½ BD

This shows that MQPN is a parallelogram.

Now consider triangle ABO.

If we draw the line XY (as shown), we can show that triangles AMX, MBY, YXM, XYO are all congruent.

For example:

- Triangle MXY is congruent to triangle OYX (Angle, Side Angle: Angle MXY = Angle OYX, the side XY is common, and Angle XYM = Angle YXO)
- Triangle AXM is congruent to triangle MYB (Angle, Side, Angle: Angle XMA = Angle YBM, the side AM = MB, and Angle MAX = Angle BMY)

Therefore, all four triangles have equal area.

Thus the shaded portion of triangle ABO is one half of the triangle.

This argument may be extended to the whole figure.