Area of a parallelogram

Base times height, no division by two — a diagonal already did that job.

Every triangle’s area needed a division by two. This lesson’s shape does not.

The same kind of parallelogram this course has built before, A, B, C and D, with the perpendicular height from D to line AB drawn, meeting it at F.

A parallelogram, A, B, C and D, where A, B and D are draggable and C always follows to keep the shape a genuine parallelogram. The perpendicular height from D to line AB is drawn, meeting it at F. A live reading gives the base AB, the height DF, and base times height. This needs JavaScript switched on.

Drag A, B or D. Base times height always gives the whole parallelogram's area.

Compare it to the last lesson’s triangle: same base-times-height product, but nothing here gets halved.

Base and height mean the same thing here as they did for a triangle: any side can be called the base, and the height is the perpendicular distance from that base’s line to the side opposite it — DF, not the slanted side AD.

Two triangles glued along a diagonal

Draw the diagonal AC. It splits the parallelogram into triangles ABC and CDA, and the last chapter’s own argument already showed those two are congruent (ASA, from equal alternate angles on the two pairs of parallel sides). Congruent triangles have equal area, so each is exactly half the parallelogram.

Both triangles share the same base, AB (or its equal and parallel opposite, DC), and the same height, the perpendicular distance between the two parallel sides — which is exactly DF. From the last lesson, each triangle’s area is base times height divided by two:

A_{\triangle ABC} = A_{\triangle CDA} = \frac{AB \times DF}{2}

The two together make the whole parallelogram, so add them:

A_{\text{parallelogram}} = \frac{AB \times DF}{2} + \frac{AB \times DF}{2} = AB \times DF

The two halves cancel the division by two that a single triangle always needed. A parallelogram is two congruent triangles, glued along a diagonal — a fact this course proved last chapter, for a completely different reason, and now gets to reuse for free.

One to watch out for.

The height still has to be measured perpendicular to the base, to the line the base sits on, not along one of the slanted sides — the same care the last lesson’s height needed.

A parallelogram with base 8 units and height 5 units.

8 \times 5 = 40 \text{ square units}

A parallelogram with base 6 units and a slanted side of 10 units — but a height of only 3 units.

6 \times 3 = 18 \text{ square units}

The slanted side plays no part in the calculation at all — only the base and the perpendicular height between the two parallel sides, exactly as with a triangle.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.