WEEK 6

 

6.1 Geometrical properties

6.1.3 Understand and use Pythagoras’ theorem

This week's tasks involve a variety of problems that can be solved using Pythagoras' Theorem.

 

Monday: Here we use the coordinates (x, y) of points to calculate the distance (d) of the points from the origin, using the Pythagorean relation x² + y² = d².

Students who are familiar with the Pythagorean triple 5, 12, 13 might spot, without calculating, that D lies on the desired circle. 

Given that 13² = 169, it is fairly straightforward to determine whether the other points are inside or outside the desired circle:
for point A we have 9² + 9² = 81 + 81 = 162 < 169
for point B we have 10² + 8² = 100 + 64 = 164 < 169
for point C we have 11² + 7² = 121 + 49 = 170 > 169
for point D we have 12² + 5² = 144 + 25 = 169.

Tuesday: The red and the blue marble are almost the same distance from the green marble: it is very difficult to tell which distance is shorter just by eye.

For the red marble, we can use the grid to construct a 2 units by 7 units right angled triangle whose hypotenuse is the line segment that joins the red and the green marble. So the square of this distance is 4 + 49 = 53 square units. Similarly the square of the distance from the blue to the green marble is 36 + 16 = 52. So the blue marble is (very slightly) closer to the green marble than is the red marble.

Wednesday: We can see from the grid that BC = BD, so triangle BCD is isosceles. It looks as though the other two given triangles might be isosceles too, though it turns out that this is only the case for triangle ABC. Here AB² = 1² + 7² = 50 and AC² = 5² + 5² = 50. BD and AD are not equal, since BD² = 2² + 4² = 20, while AD² = 3² + 3² = 18.

Thursday: These diagrams are based on the 5, 12, 13 right-angled triangle. In the first diagram there are two isosceles triangles, which are fairly easy to spot. In the second diagram, it turns out that VY = VW, which is perhaps less easy to spot and not that easy to verify - it hinges on the fact that the 13 units hypotenuse VX is divided in the ratio 7.5 : 12 or 5 : 8 by point Y.

Students might find it interesting to explore similar pairs of diagrams based on other Pythagorean triples. The situation shown in the first diagram is relatively trivial: for any right angled triangle like ABC but involving a different Pythagorean triple, it is a simple matter to find a point D such that CB = CD, so that BCD is isosceles. In turn, the triangle ADE will also be isosceles as it is similar to triangle BCD. 

The situation shown in the second diagram is likely to be more challenging. For a right angled triangle like VWX but again involving a different Pythagorean triple, it is again a simple matter to find a point Y such that VY = VW; however, for this to happen, the length of VZ is likely to involve a fairly complicated fractional number.

Friday: Finally, a 'real life' problem....!