JEE Mathematics Areas of Parallelograms and Triangles Notes Set 02

Areas of Parallelograms and Triangles: Key Results

Two figures are said to lie on the same base and between the same parallels when they share one side lying on one of two parallel lines and their opposite vertices lie on the other parallel line. The perpendicular gap between the parallels is then the common height of both figures.

Theorem 1: Parallelograms on the same base

Parallelograms standing on the same base and lying between the same parallel lines have equal areas.

Idea of the proof for parallelograms ABCD and ABEF on base AB: the triangles CBE and DAF have BC = AD, BE = AF and the included angles CBE and DAF equal (BC || AD and BE || AF). By SAS they are congruent, so they have equal area. Now
Area of ABCD = Area of ABED + Area of triangle CBE
Area of ABEF = Area of ABED + Area of triangle DAF
The two added pieces are equal, hence Area of ABCD = Area of ABEF.

Theorem 2: Triangles on the same base

Triangles standing on the same base and lying between the same parallels have equal areas.

Idea of the proof for triangles ABC and PBC with AP || BC: complete two parallelograms on BC, namely ABCQ (with side BA) and PBCR (with side BP), so that both lie between the same parallels. By Theorem 1 they are equal in area. A diagonal cuts a parallelogram into two equal triangles, so Area of ABC = half of Area of ABCQ = half of Area of PBCR = Area of PBC.

Theorem 3: Equal areas and equal bases

If two triangles have equal bases and equal areas, their corresponding altitudes are equal. This follows from Area = (1/2) × base × altitude: with the same base and the same area the altitude cannot differ.

Useful Consequences

1. A parallelogram and a triangle on the same base and between the same parallels: Area of triangle = (1/2) × Area of parallelogram.
2. A median divides a triangle into two triangles of equal area, because they have equal bases (BD = DC) and the same altitude from the opposite vertex.
3. The diagonals of a parallelogram divide it into four triangles of equal area.
4. If two triangles on the same base (on the same side of it) have equal areas, their apexes lie on a line parallel to the base.
5. If D is any point on side BC of triangle ABC, then Area(ABD) : Area(ADC) = BD : DC, since the height from A is common.
6. Joining the midpoints of two sides of a triangle cuts off a triangle whose area is one quarter of the whole.

Formulas to Remember

Area of parallelogram = base × corresponding height
Area of triangle = \( \frac{1}{2} \) × base × corresponding height

Example 1: Show that a median of a triangle divides it into two triangles of equal area.
Solution: In triangle ABC let AD be the median to BC and let AE be the altitude from A. Area(ABD) = \( \frac{1}{2} \times BD \times AE \) and Area(ACD) = \( \frac{1}{2} \times DC \times AE \). Since BD = DC, the two areas are equal.

Example 2: 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) = (1/2) Area(ABCD).
Solution: In triangles AOX and COY: AO = OC (diagonals bisect each other), angle AOX = angle COY (vertically opposite) and angle OAX = angle OCY (alternate angles). By ASA they are congruent, so Area(AOX) = Area(COY). Now Area(ACD) = Area(AOYD) + Area(COY) = Area(AOYD) + Area(AOX) = Area(AXYD). Since the diagonal AC halves the parallelogram, Area(ACD) = (1/2) Area(ABCD). Hence Area(AXYD) = (1/2) Area(ABCD).

Example 3: In triangle ABC, D is the midpoint of BC and E is the midpoint of AD. Find Area(BEC) if Area(ABC) = 48 square units.
Solution: Area(ABD) = Area(ACD) = 24 (median). In triangle ABD, BE is a median, so Area(BED) = 12. Similarly Area(CED) = 12. Hence Area(BEC) = 12 + 12 = 24 square units, which is half of Area(ABC).