# Mind Map

## Area of parallelograms of Class 9

## AREA OF A QUADRILATERAL

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 quadrilateral ABCD = Area of ABC + Area of ADC

= AC BX + AC DY = dp + dq = 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.

Draw AL BD and CM BD.

BD divides area ABCD into two equal parts.

ar(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

ΔALO ≅ ΔCMO (AAS)

That means AO = OC (CPCT)

BD bisects AC.

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

**Sol: **In ΔBCD, CD2 = BD2 + BC2

⇒ (13)2 = (BD)2 + 52 ⇒ BD = 12

In ΔABD, BD2 = AB2 + AD2

⇒ (12)^{2} = (AB)^{2} + 82 ⇒ AB =

ar(quad. ABCD) = ar(ΔABD) + ar(ΔBCD)

= cm2.

**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.

**Sol: **Draw CM AB, then

AM = 7 cm, MB = 13 – 7 = 6 cm

BC = 10 cm

from right triangle MBC

MC = (by Pythagoras theorem)

= 8 cm.

Area of ABCD = (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

OB AC and OD AC.

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

.

**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

ar(ΔAOB) = ar(ΔBOC)

= ar(ΔAOD) = ar(ΔCOD)

To prove: ABCD is a ||gm.

**Construction:** Draw CL AB and DM 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

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 ADL + Area of BMC

=

= h(2AB + DL + MC) = h {AB + (LM + DL + MC)} (as AB = LM)

= h(AB + CD)

Area of trapezium = ×height (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

In ΔADO and CDO

OD = OD, OA = OC

AOD = ∠COD = 90°

ADO ≅ABO (S.A.S)

AD = CD

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.

### CONGRUENCE AREA AXIOM

- If ΔABC≅ΔDEF

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.

### AREA ADDITION AXIOM

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.