CBSE Class 9 Mathematics Areas of Parallelograms and Triangles Notes Set 08

Parallelogram: Properties to Remember

1. A quadrilateral with both pairs of opposite sides parallel is a parallelogram. Opposite sides are equal, opposite angles are equal and adjacent angles add up to 180°.
2. Its diagonals bisect each other, and each diagonal cuts the parallelogram into two congruent triangles, so each triangle has half the area.
3. If one pair of opposite sides of a quadrilateral is both equal and parallel, the quadrilateral is a parallelogram. If the diagonals of a quadrilateral bisect each other, it is also a parallelogram.
4. A parallelogram with one right angle is a rectangle. The diagonals are equal only in a rectangle and perpendicular only in a rhombus.

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 one side (or have equal bases on the same line) and their opposite vertices or sides lie on a single line parallel to that base. The distance between the two parallel lines is the common altitude.

1. Parallelograms on the same base and between the same parallels are equal in area. Reason: area = base × altitude, and both factors are equal.
2. Triangles on the same base (or on equal bases) and between the same parallels are equal in area.
3. If a triangle and a parallelogram stand on the same base and between the same parallels, the triangle has exactly half the area of the parallelogram.
4. Converse: parallelograms (or triangles) on the same base with equal areas, lying on the same side of it, lie between the same parallels.

Area Formulas

1. Parallelogram: area = base × corresponding altitude. Any side may be used as the base, so base1 × altitude1 = base2 × altitude2.
2. Triangle: area = \( \frac{1}{2} \) × base × corresponding altitude.
3. Trapezium: area = \( \frac{1}{2} \) × (sum of the parallel sides) × distance between them.
4. Rectangle: length × breadth. The rectangle has the smallest perimeter among all parallelograms with the same base and the same altitude, because the perpendicular is the shortest distance between two parallel lines.

Median and Mid-point Results

1. A median divides a triangle into two triangles of equal area (equal bases, same apex, same altitude).
2. The three medians meet at the centroid G, and △AGB, △BGC and △CGA each have area \( \frac{1}{3} \) of the triangle.
3. Mid-point theorem: the segment joining the mid-points of two sides of a triangle is parallel to the third side and half as long. Converse: a line through the mid-point of one side parallel to a second side bisects the third side.
4. Joining the mid-points of the consecutive sides of any quadrilateral gives a parallelogram whose area is half the area of the quadrilateral. It is a rectangle if the diagonals of the quadrilateral are perpendicular and a rhombus if they are equal.
5. The segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the bases and equals half their sum.
6. The mid-point of the hypotenuse of a right triangle is at equal distances from the three vertices.

How to Attack a Proof on Areas

1. Look for two triangles on the same base between the same parallels, or a triangle and a parallelogram sharing a base.
2. Draw the helpful construction: join a diagonal, draw a line through a point parallel to a given line, or drop the altitudes.
3. Add or subtract common regions to turn a known equality into the wanted one, as in ar(△ACQ) − ar(△ABQ) = ar(△AQP) − ar(△ABQ).
4. Write each step with its reason; the reasons are the theorems listed above.

Solved Examples

Example 1. In a parallelogram ABCD, AB = 10 cm and the altitudes corresponding to the sides AB and AD are 7 cm and 8 cm. Find AD.
Answer:
Area of the parallelogram = AB × 7 = 10 × 7 = 70 cm².
The same area = AD × 8, so AD = 70 ÷ 8 = 8.75 cm.

Example 2. Show that a median of a triangle divides it into two triangles of equal area.
Answer:
Let AD be the median of △ABC and draw AL ⊥ BC.
Because BD = DC, ar(△ABD) = \( \frac{1}{2} \) × BD × AL = \( \frac{1}{2} \) × DC × AL = ar(△ADC).
So a median halves the area of the triangle.

Example 3. In △ABC, AD is a median and DE is a median of △DAC. Show that ar(△AED) = \( \frac{1}{4} \) ar(△ABC).
Answer:
AD is a median of △ABC, so ar(△ACD) = \( \frac{1}{2} \) ar(△ABC).
DE is a median of △DAC, so ar(△AED) = \( \frac{1}{2} \) ar(△ACD) = \( \frac{1}{4} \) ar(△ABC).

Example 4. The diagonals AC and BD of a quadrilateral ABCD meet at O with BO = OD. Prove that ar(△ABC) = ar(△ADC).
Answer:
In △ABD, O is the mid-point of BD, so AO is a median and ar(△AOB) = ar(△AOD).
In △CBD, CO is a median, so ar(△COB) = ar(△COD).
Adding, ar(△AOB) + ar(△COB) = ar(△AOD) + ar(△COD), that is ar(△ABC) = ar(△ADC).

Example 5. A point O lies inside a parallelogram ABCD. Show that ar(△AOB) + ar(△COD) = \( \frac{1}{2} \) ar(ABCD).
Answer:
Draw a line through O parallel to AB meeting AD at E and BC at F.
△AOB and parallelogram ABFE stand on the same base AB between the same parallels, so ar(△AOB) = \( \frac{1}{2} \) ar(ABFE).
Likewise ar(△COD) = \( \frac{1}{2} \) ar(DEFC).
Adding, ar(△AOB) + ar(△COD) = \( \frac{1}{2} \) ar(ABCD).

Example 6. ABCD is a trapezium with AB ∥ DC. E is the mid-point of AD and EF ∥ DC meets BC at F. Prove that EF = \( \frac{1}{2} \) (AB + DC).
Answer:
Let EF meet the diagonal AC at G. In △ADC, E is the mid-point of AD and EG ∥ DC, so G is the mid-point of AC and EG = \( \frac{1}{2} \) DC.
In △ABC, G is the mid-point of AC and GF ∥ AB, so F is the mid-point of BC and GF = \( \frac{1}{2} \) AB.
EF = EG + GF = \( \frac{1}{2} \) (AB + DC).

Example 7. D is the mid-point of side AB of △ABC, P is any point on BC, and CQ ∥ PD meets AB at Q. Show that ar(△BPQ) = \( \frac{1}{2} \) ar(△ABC).
Answer:
Join CD. CD is a median, so ar(△BCD) = \( \frac{1}{2} \) ar(△ABC).
△PDQ and △PDC lie on the same base PD between the parallels PD and QC, so ar(△PDQ) = ar(△PDC).
ar(△BPQ) = ar(△BPD) + ar(△PDQ) = ar(△BPD) + ar(△PDC) = ar(△BCD) = \( \frac{1}{2} \) ar(△ABC).

Example 8. The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.
Answer:
Let P, Q, R, S be the mid-points of AB, BC, CD, DA. In △ABC and △ADC, PQ ∥ AC ∥ SR and PQ = SR = \( \frac{1}{2} \) AC, so PQRS is a parallelogram.
In △ABD, PS ∥ BD. Since AC ⊥ BD, PQ ⊥ PS, so one angle of the parallelogram is 90°.
Hence PQRS is a rectangle.