A complete presentation for class 9 ch areas of traingles and parallelograms

2.  The study of Geometry, originated with the
measurement of earth (lands) in the process of
recasting boundaries of the fields and dividing them into
appropriate parts. For example, a farmer Budhia had a
triangular field and she wanted to divide it equally
among her two daughters and one son. Without actually
calculating the area of the field, she just divided one
side of the triangular field into three equal parts and
joined the two points of division to the opposite vertex.
In this way, the field was divided into three parts and
she gave one part to each of her children. Do you think
that all the three parts so obtained by her were, in
fact, equal in area? To get answers to this type of
questions and other related problems, there is a need
to have a relook at areas of plane figures, which you
have already studied in earlier classes.

3.  You may recall that the part of the plane
enclosed by a simple closed figure is called a
planar region corresponding to that figure.
The magnitude or measure of this planar
region is called its area.
Planar
Region

4.  You are also aware of some formulae for
finding the areas of different figures such as
rectangle, square, parallelogram, triangle
etc. Here, attempt shall be made to
consolidate the knowledge about these
formulae by studying some relationship
between the areas of these geometric figures
under the condition when they lie on the
same base and between the same parallels.
This study will also be useful in the
understanding of some results on ‘similarity
of triangles’.

5.  (1) (2)
 P Q A B
T U
S R c D
(1) PQRS and RSTU lies on same base(SR) and
between same parallel lines (PQ parallel SR).
(2) ABCD and ECD lies on same base (CD) and
between same parallels ( AB parallel CD)

6. Theorem 9.1 : Parallelograms on the same base and between the same parallels
are equal in area.
 GIVEN: Two parallelograms ABCD and EFCD, on the same base DC and
between the same parallels AF and DC are given
 TO PROVE: ar (ABCD) = ar (EFCD).
 PROOF: In tri. ADE and tri. BCF,
 angle DAE = angle CBF (Corresponding angles from AD || BC and transversal
AF) (1)
 angle AED =angle BFC angles from ED || FC and transversal AF) (2 )

 There fore,
 angle ADE = angle BCF (Angle sum property of a triangle) (3)
 Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
 So, tri. ADE cong. Tri. BCF [By ASA rule, using (1), (3), and (4)]
 Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
 Now, ar (ABCD) = ar (ADE) + ar (EDCB) = ar (BCF) + ar (EDCB) [From(5)] = ar
(EFCD)
 So, parallelograms ABCD and EFCD are equal in area

7. Let tri. ABP and parallelogram ABCD be on same base AB and between same parallels AB and PC
To prove : ar(PAB)= ½ ar (ABCD)
Cont.: Draw BQ parallel AP to obtain another parallelogram ABQP
 Proof: Now parallelograms ABQP
 and ABCD are on the same base AB and between the
same parallels AP and PC
P D Q C
 .
A B
Therefore, ar (ABQP) = ar (ABCD) (By
Theorem 9.1) (1)
But
tri.PAB cong. tri.BQP (Diagonal PB divides
parallelogram ABQP into two congruent
triangles.)
So, ar (PAB) = ar (BQP) (2)
Therefore, ar (PAB) = ½ ar (ABQP)
[From (2)] (3)
This gives ar (PAB)= ½ ar (ABCD)

8.  Let ∆ABC and ∆ABD be on the same base AB and between the same parallel AB and CD. It is
require to prove that ∆ABC = ∆ABD.
 Construction: A parallelogram ABPQ is constructed with AB as base and lying between the same
parallels AB and CD.
 Q P C D




B
 A B
 Proof: Since ∆ABC and parallelogram ABPQ are on the same base AB and between the same
parallels AB and Q,
 Therefore, ∆ABC = ½(Parallelogram ABPQ)
 But ar(ABPQ)= ar(ABCD) [Same base=AB Same parallels= AB and QD
 Similarly, ∆ABD = ½(Parallelogram ABPQ)
 Therefore, ∆ABC = ∆ABD.

10.  Given- tri. ABC in which AD is median
 TO prove- ar(ABD) = ar(ACD)
 Cont.- Draw AN perpendicular to BC
Proof- ar (ABD)=1/2 × base × altitude (of tri. ABD)
 = ½ BD * AN A
 = ½ CD * AN (As BD = CD)
 = ½ × base × altitude (of tri. ACD)
 = ar(ACD) B N D C


11.  Q. 1: In Fig., ABCD is a parallelogram. Points P and
Q on BC trisects BC in three equal parts. Prove that
ar (APQ) = ar (DPQ) = 1/6 ar(ABCD).
 A D
 B C
 P Q
O

12.  Through P and Q, draw PR and QS parallel to AB. Now PQRS is a
parallelogram and its base PQ = 1/3BC.
 ar (APD) = ½ ar (ABCD) [Same base BC and BC || AD] (1)
 ar (AQD) = ½ ar (ABCD) (2)
 From (1) and (2), we get ar (APD) = ar (AQD) (3)
 Subtracting ar (AOD) from both sides, we get
 ar (APD) – ar (AOD) = ar (AQD) – ar (AOD) A (4) D
 ar (APO) = ar (OQD),
 Adding ar (OPQ) on both sides in (4), we get
 ar (APO) + ar (OPQ) = ar (OQD) + ar (OPQ)
 ar (APQ) = ar (DPQ) B
C
Since, ar (APQ) = ½ ar (PQRS), therefore P Q
 ar (DPQ) = ½ ar (PQRS) Now
 , ar (PQRS) = 1/3 ar (ABCD) Therefore, ar (APQ) = ar (DPQ)
 = ½ ar (PQRS) = ½ × 1/3 ar (ABCD)
 = 1/6 ar (ABCD)
O

13.  1. A point E is taken on the side BC of a
parallelogram ABCD. AE and DC are produced to
meet at F . Prove that ar (ADF) = ar (ABFC).
 2. The diagonals of a parallelogram ABCD
intersect at a point O. Through O, a line is
drawn to intersect AD at P and BC at Q.
Show that PQ divides the parallelogram into
two parts of equal area.
 3. The medians BE and CF of a triangle ABC
intersect at G . Prove that the area of
∆ GBC = area of the quadrilateral AFGE.

14.  4. If the medians of a ∆ABC intersect at G,
show that ar (AGB) = ar (AGC) = ar (BGC)=
1/3 ar (ABC).
 5. In Fig., X and Y are the mid-points of AC
and AB respectively, QP||BC and CYQ and
BXP are straight lines. Prove that ar (ABP) =
ar (ACQ).

 Q A P
 Y X
 B C

15.  AKSHIT (11)
 HARPUNEET SINGH (10)
 DIGVIJAY (41)
 GURAMRIT SINGH (21)
 SAHAJ (47)
 DEEPANSH (32)