WEEK 9

6.2 Perimeter, area and volume

6..2.2 Understand the concept of area (2)

Here we have a range of area problems which can be solved by focussing on an object's base and height.

Monday: This is a nice, general geometry problem where we are not concerned with segments of a specified length.

There are some very neat solutions to this, but it may take a while for a flash of insight to occur!

One way to solve the task is to imagine cutting off the small triangle on the right and joining it to the left hand edge of the slanting rectangle. This produces a parallelogram with the same (vertical) base as the square-on rectangle, and the same (horizontal) height. Another method is to partition both rectangles into the same two small and the same two larger triangles.

Tuesday: Using Monday's relationship and the Pythagorean triple 8, 15, 17, we can write 17x = 8×15, so x ≃ 7.06.

Wednesday: One approach to this task is to find a way of counting squares. However, it is difficult to do this in a satisfactory way as two of the triangle's vertices are not grid-points. How can we get round this?

Students might notice that the midpoint of the lower side of the triangle is a grid-point, and that it lies directly below the top vertex. If we cut the triangle into two parts along the line segment joining these two points, we can use this segment as the base of two triangles each with a (horizontal) height of 3 units (see the diagram, below-left). So the area of the original triangle is 6×3 unit squares.
[We could also regard the segment as part of the invariant line of a shear and thus transform the original triangle into an isosceles triangle with (horizontal) base 6 units and (vertical) height 6 units (see the diagram, below-right).]

Thursday: Here we emphasise the idea that in using the formula Area = ½ × base × height to find the area of a triangle, the base and height don't have to be horizontal and vertical.

Some students might find it rather tedious to find the area using the same method three times. Others might gain a certain satisfaction from verifying that we get the same result each time, especially if they notice how the expressions that they get along the way each time are related.

Using BC as the base, the area formula gives us this expression: ½ × 20 × 15 (which gives us 150 unit squares for the area).

If we use AC as the base, we can find the length of the base and height using Pythagoras' Rule: √(15² + 15²) = 15√2 units and  √(10² + 10²) = 10√2 units, giving us an area of ½ × 15√2 × 10√2 units squares. We can easily transform this expression into our earlier expression ½ × 20 × 15.

If we use AB as the base, we can find the length of the base and height using Pythagoras' Rule: √(5² + 15²) = √250 = 5√10 units and  √(6² + 18²) = 6√10 units, giving us an area of ½ × 5√10 × 6√10 units squares. Again, we can transform this expression into our earlier expression ½ × 20 × 15.

Friday: The first part of this task is fairly trivial, if students are comfortable with the area formula for a triangle, or if they can see that the area of our triangle will always be half the area of the rectangle with BC as one of its sides and part of the x-axis as the opposite side.

It will be interesting to see what students' intuitions are about the perimeter. Some might think it stays the same (at least if the x-coordinate of A is constrained to lie between 0 and 16). Some might surmise that the perimeter is a minimum when A is at (10, 0). This turns out to be right and can be confirmed by measuring or using Pythagoras' Rule. One way to prove it is with a diagram like this (which relates to the more general solution to the classic Heron's Problem). The red segment is shorter than the sum of the two blue segments with which it forms a triangle.