The relation of parallelograms standing on the same base and lying between the same parallel lines

Two figures are congruent when one can be laid on top of the other and they match exactly, side for side and angle for angle. Two figures have equal area when they cover the same amount of flat surface, whatever their shapes. The two ideas sound alike, but they are different, and this whole lesson rests on that difference.

Study each pair and think about four questions.

• (a) Do the two figures match when one is laid over the other? The two hearts in (a) and the two rhombuses in (f) do. In (b), (c), (d) and (e) the shapes are different, so they do not.
• (b) Is the area of the pair equal? A pair that matches always has equal area. A pair that does not match may or may not have equal area, and only measuring can tell.
• (c) Which pairs are congruent? Pairs (a) and (f) are congruent. The other four pairs are not.
• (d) Are all figures with equal area congruent? No. Equal area tells us only that two figures cover the same amount of surface. It tells us nothing about their shapes.

A pair of rectangles makes point (d) clear. A rectangle 2 cm wide and 6 cm long has area \( 2 \times 6 = 12 \text{ cm}^2 \). A rectangle 4 cm wide and 3 cm long has area \( 4 \times 3 = 12 \text{ cm}^2 \). The areas are equal, but one rectangle is long and thin and the other is almost square, so neither fits on the other.

Many figures in geometry are drawn between a pair of parallel lines. To compare their areas we need two exact ideas: what it means for figures to stand on the same base, and what it means for them to lie between the same parallel lines.

The three figures below put these ideas to the test. In each one, matching arrow marks on two lines show that those lines are parallel.

Now check each figure against both conditions.

• In figure (i), ABDC and ABFE stand on the same base AB but lie between different parallel lines. ABDC reaches up to the line CD, while ABFE reaches only the lower line EF.
• In figure (ii), rectangle ABCD and triangle DEF lie between the same parallel lines AB and DG, but they stand on different bases, DC and DF.
• In figure (iii), triangle ABE and parallelogram ABDC stand on the same base AB and lie between the same parallel lines AB and ED. Both conditions hold.

Only figure (iii) meets both conditions at once. The result in this lesson is about figures of that kind, so both conditions must be checked before it is used.

The height of a parallelogram is the perpendicular distance from its base to the opposite side. It is measured at right angles to the base, never along a slanting side. When the base lies on one line and the opposite side lies on a parallel line, the height is simply the distance between those two lines.

\[ A = b \times h \]

Here \( A \) is the area, \( b \) is the base and \( h \) is the height. Parallel lines stay the same distance apart all along their length, so every perpendicular drawn from one to the other has the same length. That is why every figure standing between the same two parallel lines has the same height.

Take two parallelograms, ABCD and ABEF, drawn on the same base AB between the parallel lines AB and FC. The base AB is 7 cm. A perpendicular DG is drawn from D to AB, and DG is 3 cm. What is the area of each, and how are the two areas related?

Step 1: The area of parallelogram ABCD is base × height, with base AB and height DG:

\[ \text{Area of } ABCD = AB \times DG \]

Step 2: Putting in AB = 7 cm and DG = 3 cm:

\[ 7 \text{ cm} \times 3 \text{ cm} = 21 \text{ cm}^2 \]

Step 3: Parallelogram ABEF stands on the same base AB, and its side FE lies on the same line FC. The perpendicular distance from FC to AB is still DG, so its area is:

\[ \text{Area of } ABEF = AB \times DG \]

Step 4: Putting in the same values:

\[ 7 \text{ cm} \times 3 \text{ cm} = 21 \text{ cm}^2 \]

Step 5: Comparing the two results, both areas are \( 21 \text{ cm}^2 \). So the area of parallelogram ABCD equals the area of parallelogram ABEF.

The two parallelograms lean in opposite directions and do not look alike. Their slant never entered the calculation. Only the base and the distance between the lines did, and those were the same for both.

One measurement is not a proof. Another pair of parallelograms, drawn with different lengths, might seem to behave differently. The theorem below shows that the result holds for every such pair.

Given: parallelograms ABCD and ABEF stand on the same base AB and lie between the same parallel lines AB and FC.

To be proved: area of parallelogram ABCD = area of parallelogram ABEF.

The idea behind the proof comes from the shaded trapezium ABED, which lies inside both parallelograms. Parallelogram ABEF is this trapezium with triangle ADF added on the left. Parallelogram ABCD is the same trapezium with triangle BCE added on the right. So the whole question becomes: are triangles ADF and BCE equal in area? The proof answers it by showing that the two triangles are congruent.

The proof is written as a table of statements, each with its reason.

Statement 5 is where the whole part axiom does its work. Triangle ADF and trapezium ABED together make the whole of parallelogram ABEF, and triangle BCE and trapezium ABED together make the whole of parallelogram ABCD. So parallelograms standing on the same base and lying between the same parallel lines are equal in area. Proved.

In the figure, \( AB \parallel XY \). ABPQ is a rectangle and ABCD is a parallelogram. Prove that the area of parallelogram ABCD = area of rectangle ABPQ.

Solution: parallelogram ABCD and rectangle ABPQ stand on the same base AB and lie between the same parallel lines AB and XY. Every angle of a rectangle is a right angle, so AQ is perpendicular to AB. That makes AQ the height of the parallelogram as well.

To be proved: area of parallelogram ABCD = area of rectangle ABPQ.

A rectangle is a parallelogram whose angles are all right angles, so Theorem 1 gives the same result at once. The example is still worth knowing, because it explains why the area of any parallelogram is base × height: the parallelogram can be swapped for a rectangle on the same base between the same parallels, and a rectangle's area is length × breadth.