Chapter 9      -  Areas of parallelograms and triangles

Figures on the same base and between the same parallels

Figures on the same base and between the same parallels

Two triangles with common base EF.

Two figures are between two parallels AB & CD but have different bases’ length.

Two figures with common base XY and between two parallels XY & PQ.

“So, two figures are said to be on the same base and between same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base”. [As shown in the figure above]

Example: Choose which of the following figures lie on the same base and between the same parallels. For such a figure, write the common base and the two parallels.

Solution:

I. Rectangle ABCD and trapezium EFCD are on the same base CD but they are not between the same parallels.

II. Parallelogram PQRS and ∆ ROQ are on the same base RQ and between the same parallel lines PS and QR.

III. Parallelogram EFGH and pentagon UVWXY are between the same parallel EF and HG but they are not on the same base.

IV. Parallelogram ABCD and parallelogram MNCD lie between the same parallels AB and CD and they have common base CD.

 RECTANGLE AREA AXIOM

If ABCD is a rectangular region such that AB = a units and BC = b units in length, then

CONGRUENT AREA AXIOM

Consider two ∆ ABC and ∆ DEF which are congruent to each other.

Now, we calculate the area of ∆ ABC, then we get


ar(∆ABC) = 1/2× base × height.

                                    = {\color{Blue} \frac{1}{2}} × 3 cm × 4 cm.

                                    = 3 cm × 2 cm.
                                    = 6 cm2.

Also, we calculate the area of ∆ DEF, then we get
ar(∆DEF) = 
{\color{Blue} \frac{1}{2}}  × base × height.

                                 = {\color{Blue} \frac{1}{2}}  × 3 cm × 4 cm.

                                 = 3 cm × 2 cm.
                                 = 6 cm2.

Here, we also see that the area of triangles ABC and DEF are equal.

Therefore, we conclude that if two figures are congruent with each other, they must have equal areas.

Now, consider a rectangle ABCD and square PQRS which are equal in area.

We calculate the area of rectangle ABCD, then we get ar(ABCD) = Length × Breath.

= 9 cm × 4 cm.
= 36 cm2

Also, we calculate the area of square PQRS, then we get
ar(PQRS) = Side × Side.
                 = 6 cm × 6 cm.
                  = 36 cm2.

Here, we see that areas of triangles ABC and DEF are equal. Rectangles ABCD and square PQRS are not congruent, but still, they have equal areas. Therefore, we can conclude that if two figures are congruent, they will have equal areas whereas, if two figures have equal areas, they need not be congruent with each other.

AREA MONOTONE AXIOM:

If R1, R2 are two polygonal regions such that R1 is a part of R2, then

ar(R1 ) ≤ ar(R2)

i.e. In the given figure region, ABCD is the part of region PQRS, that means

In the next figure (below) the region, the vertices A, B, C, D of parallelogram ABCD coincide with the vertices P, Q, R, S of the parallelogram PQRS.

The third possibility for figure ABCD and PQRS is

   

But in this case, ar(ABCD) >ar(PQRS) and PQRS is the part of ABCD which is not held by the statement of the axiom. So, this case shall not be considered.

AREA ADDITION AXIOM:

If a planer region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T)=ar (P)+ar (Q).

In the figure above, ar(ABCD) = ar(∆ADB) + ar(∆BCD)