CBSE Class 9 Mathematics Areas of Parallelograms and Triangles Notes Set 16

Figures on the Same Base and Between the Same Parallels

Two figures are on the same base and between the same parallels when they share a base and their other sides or vertices lie on a line parallel to the base. Such figures have the same height.
1. Parallelograms on the same base and between the same parallels are equal in area.
2. Triangles on the same base and between the same parallels are equal in area.
3. If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half that of the parallelogram.
4. Triangles with equal areas on the same base (and on the same side of it) lie between the same parallels. In particular if two such triangles have equal areas, their corresponding altitudes are equal.

Area Formulae

1. Parallelogram: base x corresponding height.
2. Triangle: \( \tfrac12 \times \) base \( \times \) corresponding height.
3. Trapezium: \( \tfrac12 \times \) height \( \times \) (sum of the parallel sides).
4. Rhombus: \( \tfrac12 \times \) product of the diagonals.
5. Triangles of the same height have areas in the ratio of their bases, and triangles on the same base have areas in the ratio of their heights.

Useful Results

1. A median divides a triangle into two triangles of equal area.
2. The diagonals of a parallelogram divide it into four triangles of equal area.
3. The lines joining the midpoints of the sides of a triangle divide it into four congruent triangles, each with one quarter of the area.
4. The centroid joins with the vertices to give three triangles of equal area, each one third of the whole.
5. A line through the point of intersection of the diagonals of a parallelogram divides it into two equal areas.

Worked Example

Example. The diagonals of a parallelogram ABCD meet at O. A line through O meets AB at X and CD at Y. Show that area(AXYD) = \( \tfrac12 \) area(ABCD).
Triangles AOX and COY are congruent by ASA (AO = OC, vertically opposite angles, alternate angles OAX and OCY), so their areas are equal.
Area(ACD) = \( \tfrac12 \) area(ABCD). Also area(ACD) = area(AOYD) + area(COY) = area(AOYD) + area(AOX) = area(AXYD). Hence area(AXYD) = \( \tfrac12 \) area(ABCD).