In the figure A and B are points on PQ and PR
such that AB || QR.
AR and BQ are drawn.
Answer the following questions, which refer to
a theorem. You need to redraw the sketch.
1.1 Complete: ΔΔ area APB area AQB = . . . . . . . .
1.2 Complete: ΔΔ area APB area ABR = . . . . . . . .
1.3 What can you say about the area of ΔAQB and the area of ΔABR, and why?
1.4 What can you deduce from 1.1, 1.2 and 1.3?
1.5 Give the wording of the theorem which is under consideration here.
1.1 Complete: ΔΔ area APB area AQB = area AQB
1.2 Complete: ΔΔ area APB area ABR = area ABR
1.3 The area of ΔAQB is equal to the area of ΔABR. This is because both triangles have the same base AB and are between the same parallel lines AB and QR.
1.4 From 1.1, 1.2, and 1.3, we can deduce that the area of triangle APB is equal to the sum of the areas of triangles AQB and ABR.
1.5 The theorem under consideration here is the Parallelogram Theorem, which states that the area of a parallelogram is equal to the sum of the areas of the two triangles formed by one of its diagonals.