Mind Map

Area of parallelograms of Class 9


Let ABCD be any quadrilateral. Draw a diagonal AC. From B and D, draw perpendiculars BX and DY to AC.

Let AC = d, BX = p, DY = q

.Area of parallelograms

Area of parallelograms Area of quadrilateral ABCD = Area of Area of parallelogramsABC + Area of Area of parallelogramsADC

= Area of parallelograms AC Area of parallelograms BX + Area of parallelogramsArea of parallelogramsAC Area of parallelograms DY = Area of parallelogramsdp + Area of parallelogramsdq = Area of parallelograms d (p + q)

Ex:A quadrilateral ABCD is such that the diagonal BD divides its area into two equal parts.

Prove that BD bisects AC.

Sol: Let ABCD be a parallelogram having diagonals AC and BD. We have to prove that AC and BD bisect each other at O. i.e. AO = OC.

Area of parallelograms

Draw AL Area of parallelograms BD and CM Area of parallelograms BD.

BD divides area ABCD into two equal parts.

Area of parallelogramsar(ΔABD) = ar (ΔBDC) (given)

Thus, ΔABD and ΔBDC are on the same base AB are of equal area. Therefore, their, corresponding altitudes are equal.

i.e. AL = CM.

inΔALO and ΔCMO,

∠1 = ∠2 (vertically opposite angles)

∠ALO = ∠CMO (each equal to 90o)

and AL = CM


That means AO = OC (CPCT)

Area of parallelograms BD bisects AC.

Ex: Find the area of quadrilateral ABCD (given √5 = 2.23)

Area of parallelograms

Sol: In ΔBCD, CD2 = BD2 + BC2

⇒ (13)2 = (BD)2 + 52 ⇒  BD = 12

In ΔABD, BD2 = AB2 + AD2

⇒ (12)2 = (AB)2 + 82 ⇒  AB = Area of parallelograms

Area of parallelogramsar(quad. ABCD) = ar(ΔABD) + ar(ΔBCD)

= Area of parallelogramscm2.

Ex: In the adjoining figure, AB || DC, DA is perpendicular to AB. Further DC = 7 cm, 

CB = 10 cm, AB = 13 cm. Find the area of quadrilateral ABCD.

Area of parallelograms

Sol: Draw CM Area of parallelograms AB, then

AM = 7 cm, MB = 13 – 7 = 6 cm

BC = 10 cm

Area of parallelogramsfrom right triangle MBC

MC = Area of parallelograms (by Pythagoras theorem)

 Area of parallelograms = 8 cm.

Area of ABCD = Area of parallelograms(13 + 7) 8 = 80 cm2.

Ex: Show that the area of a rhombus is half the product of the lengths of its diagonals.

Sol: Let ABCD be a rhombus whose diagonals AC and BD intersect at O.

As the diagonals of a rhombus intersect at right angles

Area of parallelograms OB Area of parallelograms AC and OD Area of parallelograms AC.

Area (ABCD) = area (ΔABC) + area (ΔADC)

Area of parallelograms

Area of parallelograms


Area of parallelograms.

Ex: If the diagonals AC and BD of a quadrilateral ABCD intersect at O and separate the quadrilateral ABCD into four triangle of equal area, show that the quadrilateral is a parallelogram.

Sol: Given: a quadrilateral ABCD, whose diagonals AC and BD intersect at O in such a way that

Area of parallelograms

ar(ΔAOB) = ar(ΔBOC)

= ar(ΔAOD) = ar(ΔCOD)

To prove: ABCD is a ||gm.

Construction: Draw CL Area of parallelograms AB and DM Area of parallelograms AB

Proof: ar(ΔAOD) = ar(ΔBOC)

⇒ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC) + ar(ΔAOB)

⇒ar(ΔABD) = ar(ΔABC)

⇒ DM = CL

⇒ DC || AB

Similarly, AD || BC

Area of parallelograms ABCD is a parallelogram.

Area of Trapezium:

ABCD is a trapezium in which AB || CD (AB < CD).

AL and BM are perpendiculars on CD from vertices A and B respectively.

Let AL = BM = h.

Area ABCD = Area of rectangle ABML + Area of Area of parallelogramsADL + Area of Area of parallelogramsBMC


Area of parallelograms

= Area of parallelogramsh(2AB + DL + MC) = Area of parallelogramsh {AB + (LM + DL + MC)} (as AB = LM)

= Area of parallelogramsh(AB + CD)

Area of parallelograms Area of trapezium = Area of parallelograms×height Area of parallelograms(the sum of parallel sides) = h/2(a + b)

Where a = AB, b = CD, h = AL or BM.

Ex: Find the sum of the length of the bases of a trapezium whose area is 4.2 m2 and whose height is 280 cm.

Sol: Area = 4.2 m2

Height = 280 cm = 2.8 m

b1 + b2 = m.

Ex. If ABCD is a quadrilateral in which diagonals bisect at right angles and ACBD, then it is a

(a) Isosceles trapezium (b) Trapezium

(c) Rhombus (d) None

Sol. C


OD = OD, OA = OC

AOD = ∠COD = 90°



Similarly AD = BC = CD = AB

As digonals AC, BD are not equal, ABCD is a rhombus.

Important Types of Quadrilaterals Rectangles Squares Parallelograms Rhombus Trapezoid (trapezium) Kites The definitions are important.

 Area of parallelograms



Then, ar. (ΔABC) = ar. (ΔDEF)

  • Two Congruent figures have equal area but the Converse is not always true.
  • Area of a parallelogram is the product of its base and the corresponding altitude.
  • Area of a triangles is half the product of its base and the corresponding altitude. 
  • Triangles on the same base and between the same parallels are equal in area.
  • A median of a triangle divides it into two triangles of equal area.


Total area R of the plane figure ABCD is the sum of two polygonal regions R1 and R2.

i.e. ar. (R) = ar. ar.

  • Parallelogram on the same base and between same parallel are equal in area.
  • Parallelogram and a rectangle on the same base and between the same parallels are equal in area.
  • If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
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