AREAS OF PARALLELOGRAMS AND TRIANGLES
Main Concepts and Results:
The area of a closed plane figure is the measure of the region inside the figure:
Fig.1
The shaded parts (Fig.1) represent the regions whose areas may be determined by means of simple
geometrical results. The square unit is the standard unit used in measuring the area of such figures.
• If ∆ ABC ≅ ∆ PQR, then ar (∆ ABC) = ar (∆ PQR)
Total area R of the plane figure ABCD is the sum of the areas of two triangular regions R 1 and R 2 ,
that is,
ar (R) = ar (R 1 ) + ar (R 2 )
Fig. 2
• Two congruent figures have equal areas but the converse is not always true,
• A diagonal of a parallelogram divides the parallelogram in two triangles of equal area,
• (i) Parallelograms on the same base and between the same parallels are equal in area
(ii) A parallelogram and a rectangle on the same base and between the same parallels are equal in
area.
• Parallelograms on equal bases and between the same parallels are equal in area,
• Triangles on the same base and between the same parallels are equal in area,
• Triangles with equal bases and equal areas have equal corresponding altitudes,
• The area of a triangle is equal to one-half of the area of a rectangle/parallelogram of the same
base and between same parallels,
• If a triangle and a parallelogram are on the same base and between the same parallels, the area of
the triangle is equal to one-half area of the parallelogram.
Exercise:
1) What do you mean by the term Congruent?
2) Two figures are called congruent, if they have the same shape and the same size. The
given statement is true or false?
3) If two figures A and B are congruent, they must have equal areas. The given statement
is true or false?
4) Prove ar(T) = ar(P) + ar(Q) with the help of a suitable diagram.
5) Two figures are said to be on the same base and between the same parallels, if they
have a common base (side) and the vertices (or the vertex) opposite to the common
base of each figure lie on a line parallel to the base. True or false? Also give explanation
if the statement is true?
6) Which of the following figures lie on the same base and between the same parallels. In
such a case, write the common base and the two parallels.
Fig. 3
7) Parallelograms on the same base and between the same parallels are equal in area.
True or false? Also prove the same if it is true.
8) Parallelograms on the same base and between the same parallels are equal in area.
Prove the theorem.
9) In Fig. 4, ABCD is a parallelogram and EFCD is a rectangle
Also, AL ⊥ DC.
Prove that (i) ar (ABCD) = ar (EFCD)
(ii) ar (ABCD) = DC × AL
Fig .4
10) If a triangle and a parallelogram are on the same base and between the same parallels,
then prove that the area of the triangle is equal to half the area of the parallelogram.
11) In Fig.5, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and
CF = 10 cm, find AD.
Fig.5
12) If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD,
show that
ar (EFGH) = ar(ABCD).
13) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram
ABCD. Show that ar (APB) = ar (BQC)
14) In Fig. 6, P is a point in the interior of a parallelogram ABCD. Show that
Fig.6
15) In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
Fig.7
16) A farmer was having a field in the form of a parallelogram PQRS. She took any point A
on RS and joined it to points P and Q. In how many parts the fields is divided? What are
the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions
of the field separately. How should she do it?
17) Prove that: Two triangles on the same base (or equal bases) and between the same
parallels are equal in area.
18) Prove that: Two triangles having the same base (or equal bases) and equal areas lie
between the same parallels.
19) Show that a median of a triangle divides it into two triangles of equal areas.
20) In Fig. 8, ABCD is a quadrilateral and BE || AC and also BE meets DC produced at E.
Show that area of ∆ ADE is equal to the area of the quadrilateral ABCD.
Fig. 8
21) In Fig.9, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
Fig.9
22) In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = ar(ABC).
23) Show that the diagonals of a parallelogram divide it into four triangles of equal area.
24) In Fig. 10, ABC and ABD are two triangles on the same base AB. If line- segment CD is
bisected by AB at O, show that ar(ABC) = ar (ABD)
Fig. 10
25) D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ ABC. Show
that
26) In Fig. 11, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Fig. 11
27) D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar
(EBC). Prove that DE || BC.
28) XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E
and F respectively, show that ar (ABE) = ar (ACF)
29) The side AB of a parallelogram ABCD is produced to any point P. A line through A and
parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see
Fig. 12). Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Fig. 12
30) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
Prove that ar (AOD) = ar (BOC).
31) In Fig. 13, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
Fig. 13
32) A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat
of the village decided to take over some portion of his plot from one of the corners to
construct a Health Centre. Itwaari agrees to the above proposal with the condition that
he should be given equal amount of land in lieu of his land adjoining his plot so as to
form a triangular plot. Explain how this proposal will be implemented.
33) ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that ar (ADX) = ar (ACY).
[Hint : Join CX.]
34) In Fig.14, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Fig. 14
35)
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar
(AOD) = ar (BOC). Prove that ABCD is a trapezium.
36) In Fig.15, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals
ABCD and DCPR are trapeziums
Fig. 15
37) Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal
areas. Show that the perimeter of the parallelogram is greater than that of the
rectangle.
38) In Fig. 16, D and E are two points on BC such that BD = DE = EC. Show that
ar (ABD) = ar (ADE) = ar (AEC).
Fig. 16
39) In Fig. 17, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
Fig. 17
40) In Fig. 18, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ.
If AQ intersect DC at P, show that ar (BPC) = ar (DPQ)
(Hint: Join AC)
Fig. 18
41) In Fig.19, ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
If AE intersects BC at F, show that
(Hint : Join EC and AD. Show that BE || AC and DE || AB, etc )
Fig. 19
42) Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
43) P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is
the mid-point of AP, show that
44) In Fig. 20, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares
on the sides BC, CA and AB respectively. Line segment AX⊥DE meets BC at Y. Show
that:
Fig. 20
Multiple Choice Questions
45) The area of the figure formed by joining the mid-points of the adjacent sides of a
rhombus with diagonals 12 cm and 16 cm is
(A) 48
(B) 64
(C) 96
(D) 192
46) The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangle
47) In which of the following figures (Fig. 21), you find two polygons on the same base and
between the same parallels?
Fig. 21
48) The figure obtained by joining the mid-points of the adjacent sides of a rectangle of
sides 8 cm and 6 cm is :
(A) a rectangle of area 24 cm2
(B) a square of area 25 cm2
(C) a trapezium of area 24 cm2
(D) a rhombus of area 24 cm2
49) In Fig. 22, the area of parallelogram ABCD is :
(A) AB × BM
(B) BC × BN
(C) DC × DL
Fig.22
(D) AD × DL
50) In Fig. 23, if parallelogram ABCD and rectangle ABEF are of equal area, then :
(A) Perimeter of ABCD = Perimeter of ABEM
(B) Perimeter of ABCD < Perimeter of ABEM
(C) Perimeter of ABCD > Perimeter of ABEM
(D) Perimeter of ABCD = (Perimeter of ABEM)
Fig. 23
51) The mid-point of the sides of a triangle along with any of the vertices as the fourth
point make a parallelogram of area equal to
52) Two parallelograms are on equal bases and between the same parallels. The ratio of
their areas is
(A) 1 : 2
(B) 1 : 1
(C) 2 : 1
(D) 3 : 1
53) ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then
ABCD
(A) is a rectangle
(B) is always a rhombus
(C) is a parallelogram
(D) need not be any of (A), (B) or (C)
54) If a triangle and a parallelogram are on the same base and between same parallels,
then the ratio of the area of the triangle to the area of parallelogram is
(A) 1 : 3
(B) 1 : 2
(C) 3 : 1
(D) 1 : 4
55) ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig. 24). E and F are
the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is
Fig 24
(A) a : b
(B) (3a + b) : (a + 3b )
(C) (a + 3b) : (3a + b )
(D) (2a + b) : (3a + b )
Short Answer Questions with Reasoning
Write True or False and justify your answer
I. If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP).
II. If in Fig. 25, PQRS and EFRS are two parallelograms, then
III.
IV.
V.
VI.
VII.
VIII.
Fig. 25
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24
,
then ar (ABC) = 24
.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any
point on PQ. If PS = 5 cm, then ar (PAS) = 30
.
PQRS is a parallelogram whose area is 180
. and A is any point on the
diagonal QS. The area of ∆ ASR = 90
.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
Then ar (BDE) = ar (ABC).
In Fig. 26 ABCD and EFGD are two parallelograms and G is the mid-point of CD.
Then
ar (DPC) = ar ( EFGD).
Fig. 26
Short Answer Questions
1. PQRS is a square. T and U are respectively, the mid-points of PS and QR (Fig.
27). Find the area of ∆ OTS, if PQ = 8 cm, where O is the point of intersection of
TU and QS.
Fig. 27
2. ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ
(Fig. 28). If AQ intersects DC at P,
show that ar (BPC) = ar (DPQ)
Fig. 28
3. In Fig.29, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ
= QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).
Fig. 29
4. X and Y are points on the side LN of the triangle LMN such that LX = XY = YN.
Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 30). Prove
that
ar (LZY) = ar (MZYX)
Fig. 30
5. The area of the parallelogram ABCD is 90
(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)
(see Fig.31). Find
Fig. 31
6. In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets
AB in Q (Fig. 32), then prove that
ar (BPQ) = ar (ABC)
Fig .32
7. ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is
the mid-point of EF (Fig. 33), prove that
ar (AER) = ar (AFR)
Fig. 33
8. O is any point on the diagonal PR of a parallelogram PQRS (Fig. 34). Prove that
ar (PSO) = ar (PQO).
Fig. 34
9. ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig.
35). AE intersects CD at F. If ar (DFB) = 3
, find the area of the
parallelogram ABCD.
Fig. 35
10. In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ
|| AD has been drawn which meets AB in P and DC produced in Q (Fig. 36).
Prove that
ar (ABCD) = ar (APQD)
Fig. 36