NCERT Exemplar Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles

NCERT Exemplar Class 9 Maths Solutions Chapter 9: Exercise 9.1
Page: 85-87, Total Questions: 10

Question: 1
The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles

Answer: [A]
Solution:
Let AN be the median of △ABC
`
We have drawn AM perpendicular to BC
We know that median bisects a line into 2 equal parts.
Hence BN = NC …(i)
Now, area of a triangle is given as =12 (Base) (Height)
Area of △ABN=12(BN)(AM) …(ii)
(AM is the perpendicular distance between point A and BN. Hence AM is the height of △ABN )
Area of △ACN=12(CN)(AM)
(AM is the perpendicular distance between point A and CN. Hence AM is the height of △ACN )
From (i), we have BN = NC
Area of △ACN=12(BN)(AM) …(iii)
Comparing equations (ii) and (iii), we get
Area of △ABN = Area of △ACN
Hence the median of a triangle divides it into two triangles of equal area.
Hence, option (A) is the correct answer.

Question: 2
In which of the following figures (Fig. 9.3), you find two polygons on the same base and between the same parallels?

Answer: [d]
Solution.
(a) We have a triangle and a parallelogram but we can easily see that they are not between the same parallel lines. Hence it is incorrect.
(b) We have a triangle and a parallelogram on the same base BC, but we can easily see that they are not between the same parallel lines. Hence it is also incorrect.
(c) Same as option a, this is incorrect.
(d) Here, the parallelograms, PQRA and BQRS are on the same base QR and between the same parallels QR and PS.
Hence, option (d) is the correct answer.

Question: 3
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 cm ²
(B) a square of area 25 cm ²
(C) a trapezium of area 24 cm ²
(D) a rhombus of area 24 cm ²

Answer: [D]
Solution.
Let ABCD be a rectangle.
Given sides 8 cm and 6 cm.

Let the length of the rectangle = 8 cm
Breadth of rectangle = 6 cm
Let E, F, G and H be the mid-points of the sides of the rectangle
AE = EB = DG = GC = 3 cm
AH = HD = BF = FC = 4 cm
Then we have to find out about the figure EFGH.
In △HAE, using Pythagoras' theorem
EH2=AE2+AH2
EH2=32+42=9+16=25
So, EH = 5 cm.
Similarly we can find out HG = GF = FE = 5 cm
Hence, EFGH is a rhombus.
Now, the area of the rhombus = 12d1d2
Where d1,d2 are the diagonals of the rhombus
area of rhombus = 12(EG)(HF)
H ere, EG = AD = 8 cm
and HF = AB = 6 cm
Area of rhombus EFGH = 8×62=4×6=24cm2
Therefore, option (D) is correct.

Question:4
In Fig. 9.4, the area of parallelogram ABCD is:


(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL

Answer: [C]
Solution.
We know that
Area of parallelogram = Base × corresponding altitude
Area of parallelogram ABCD = AB × DL … (1)
Now, AB = DC (opposite sides of a parallelogram are equal)
Substituting in (1), we get
Area of parallelogram ABCD = DC × DL
Hence, option (C) is the correct answer.
We can check for other options as well.
In option (A), AB × BM; BM is not the corresponding altitude to the side AB
In option (B), BC × BN; BC is not the corresponding altitude to the side BN
In option (D), AD × DL; AD is not the corresponding altitude to the side DL
Therefore, option (C) is correct.

Question: 5
In Fig. 9.5, if parallelogram ABCD and rectangle ABEM 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)

Answer: [C]
Solution:
In a rectangle opposite sides are equal.
So in ABEM,
AB = EM … (1)
In a parallelogram also the opposite sides are equal.
So in ABCD,
CD = AB …(2)
Adding eq. (1) and (2)
We get AB + CD = EM + AB
So, CD = EM …(3)
We know that, the shortest distance between any two parallel lines is the perpendicular distance between them. So, perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
BE < BC and AM < AD
On adding these inequalities, we get
BE + AM < BC + AD
or
BC + AD > BE + AM
On adding AB + CD on both sides, we get
AB + CD + BC + AD > AB + CD + BE + AM
⇒ AB+BC+CD+AD>AB+CD+BE+AM (rearranging LHS)
⇒ AB+BC+CD+AD>AB+BE+EM+AM[ Q CD = EM (from eq. 3)]
⇒ Perimeter of ABCD> Perimeter of ABEM
Therefore, option (C) is correct.

Question:6
The mid-point of the sides of a triangle, along with any of the vertices as the fourth point, makes a parallelogram of area equal to
(A) 12ar(ABC)
(B) 13ar(ABC)
(C) 14ar(ABC)
(D) ar(ABC)

Answer: [A]
Solution.
Let the given triangle ABC be shown as below, with medians AE, BF, and CD.

D is the mid-point of AB, E is the mid-point of BC, F is the mid-point of AC
Mid-point theorem states that the line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side
So, DF=12BC
DF = BE = EC …(i)
DE=12AC
DE = AF = FC …(ii)
FE=12AB
FE = AD = DB …(iii)
In △ADF and △FEC
AD = FE (from iii)
DF = EC (from i)
AF = FC (given)
△ADF≅△FEC(SSS congruence)
Similarly we can prove that,
△ADF≅△FEC≅△DBE≅△EFD
So the bigger triangle is divided into 4 smaller congruent triangles.
So area of △ABC=4 (area of one smaller triangle) …(iv)
If we consider the parallelograms formed, then we can see that one ∥gm consists of two such triangles.
∥BDFE = △DBE + △EFD
∥DFCE = △FEC + △EFD
∥ADEF = △ADF + △EFD
So area of every parallelogram=2 (area of one smaller triangle) ....…(v)
Using (iv) and (v),
area of every ∥gm= 12ar(ABC)
Therefore, option (A) is correct.

Question: 7
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

Answer: [B]
Solution:
Area of a parallelogram is given as the product of its base and corresponding altitude.
Altitude is the perpendicular distance between the two parallel sides.
Now, parallelogram on the same or equal bases and between the same parallels will have the same area. This is because the base is the same, and also as they are between the same bases so the altitude will be the same for them.
So, the ratio of their area is 1 : 1.
Hence, (B) is the correct answer.

Question: 8
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)

Answer: [D]
Solution.
Since the diagonal of a parallelogram divides it into two triangles of equal area, and rectangle and a rhombus are also parallelograms. It may be a kite as the diagonals of a kite divide it into two triangles of equal areas. Then ABCD need not be any of (a), (b) or (c).
Hence, (d) is the correct option.

Question: 9
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 the parallelogram is
(A) 1 : 3
(B) 1 : 2
(C) 3 : 1
(D) 1 : 4

Answer: [B]
Solution.
We know that a triangles 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. Let us prove the same. We have a parallelogram ABCD and triangle BEC on the same base BC.


Construct EM perpendicular to BC
Area of a triangle is given as: 12 (Base) (Height)
Area of △BEC=12(BC)(EM) …(i)
Area of a parallelogram is given as: (Base) (corresponding altitude)
Area of a parallelogram ABCD= (BC) (EM) …(ii)
Ratio of the area of the triangle to the area of parallelogram = (i) : (ii)
= 12(BC)(EM)/(BC)(EM)
= 12
Hence, the ratio of the area of the triangle to the area of parallelogram is 1 : 2.
Hence, (B) is the correct answer.

Question:10
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is


(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)

Answer: B
Given: ABCD is a trapezium in which AB∥DC,
E and F are the midpoints of AD and BC, respectively

(Mid-segment Theorem on Trapezium: the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them.)
So, EF=12(a+b)
Now, ABFE and EFCD are also trapeziums.
ar(ABFE)=12(EF+AB)h
=12[12(a+b)+a]×h
=h4(3a+b)
ar(EFCD)=12(CD+EF)h
=12[b+12(a+b)]×h
=h4(a+3b)
Required ratio
∴ar(ABEF)ar(EFCD)=h4(3a+b)h4(a+3b)=(3a+b)(a+3b)
So, the required ratio is (3a+b):(a+3b)
Hence, (B) is the correct answer.

NCERT Exemplar Class 9 Maths Solutions Chapter 9: Exercise 9.2
Page: 88, Total Questions: 5

Question:1
Write true or false and justify your answer.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm².

Answer: [False]
Solution.
We have a parallelogram ABCD, and X is the midpoint of AB.

Now, ar(ABCD)=ar(AXCD)+ar(△XBC) …(1)
Draw XG perpendicular to CD
Area of ∥gm ABCD = (base) (corresponding altitude)

= (AB) (GX) …(2)

Area of trapezium AXCD = 12 (Sum of parallel sides) (Distance between them)

= 12(DC+AX)(GX)

Now, AX=12AB and AB=CD (Given)

Area of trapezium AXCD=12(AB+12AB)(GX)…(3)

Area of △ABC = 12 (base) (height)

=12(AB)(GX)

=12(AB)(GX) …(4)

Now,
Area of trapezium AXCD=12(AB+12AB)(GX)=24cm² (given)

34(AB)(GX)=24cm²
14(AB)(GX)=8cm²
12(AB)(GX)=16cm²

From equation (4),
12(AB)(GX)= Area of △ABC=16cm²

Hence, the given statement is false.

Question:2
Write true or false and justify your answer.
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 cm².

Answer: [False]
Solution.
It is given that PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm.
A is any point on PQ; therefore, PA < PQ.
It is given that A is any point on PQ; therefore, PA < PQ.

PS = 5 cm, PR = 13 cm
In D PSR, using Pythagoras theorem,
PS²+SR²=PR²
5²+SR²=13²
SR²= 169−25 =144
SR = 12 cm = PQ (opposite sides of a rectangle are equal)
Area of a triangle is given as 12×base×height
Now, ar(△PQR)=12×PQ×QR=12×12×5=30cm2
As PA<PQ

ar(△PAS)<ar(△PQR)
or ar(△PAS)<30cm2
But it is given that ar(PAS)=30cm2

Hence, the given statement is false.

Question: 3
Write true or false and justify your answer.
PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS. The area of DASR = 90 cm².

Answer: [False]
Solution.
Given, PQRS is a parallelogram whose area is 180 cm²

We know that the diagonal of a parallelogram divides it into two triangles of equal area, so
∴ar(△QRS)=12ar(∥gm PQRS)
=12×180=90cm²

Given that A is any point on SQ
So, ar(△ASR)<ar(△QRS)
Hence,ar(△ASR)<90cm²
But given that area of △ASR = 90 cm²

Hence, the given statement is false.

Question:4
Write true or false and justify your answer.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 14 ar ( ABC).

Answer: [True]
Solution.
Given △ABC and △BDE are two equilateral triangles.
D is the mid-point of BC, so BD = CD = 12 BC


Let each sides of △ABC be x.
So each side of △BDE is x2.
Area of an equilateral triangle is given as 34(side)2
Now, ar(△BDE)ar(△ABC)=34(x2)234x2=1x24x2=14

Hence, ar(△BDE)=14ar(△ABC)

Therefore, the given statement is true.

Question:5
Write true or false and justify your answer.
In Fig., ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 12ar ( EFGD).

Answer:[False]
Solution.
Given that ABCD and EFGD are two parallelograms and G is midpoint of CD.
Let the perpendicular distance between AB and CD be h
ar (∥gm) = (base) (corresponding altitude)
ar (∥EFGD) = (GD) (h)
ar (∥ABCD) = (CD) (h)
CD = 2 GD (Given that G is midpoint of CD)
So, ar(∥ABCD)=2ar(∥EFGD)
\ar(∥EFGD)=12ar(∥ABCD) ....(i)
Now, DPDC and parallelogram ABCD are on the same base BC and between the same parallels AB and CD.
So, ar(△DPC)=12ar(∥ABCD) …(ii)
From (i) and (ii)
⇒ar(△DPC)=ar(∥EFGD)
Therefore, the given statement is false.

NCERT Exemplar Class 9 Maths Solutions Chapter 9: Exercise 9.3
Page: 89-92, Total Questions: 9

Question:1

In Fig., 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).

Answer:
Solution.
Given: a parallelogram PSDA with points Q and R taken on PS such that PQ = RS = QR
and PA∥QB∥RC
To prove: ar(PQE) = ar(CFD)
Proof:
PQ = QR = RS
PS = PQ + QR + RS = 3 PQ
Also, PA∥QB∥RC and PS∥AD
So PABQ, QBCR, RCDS – all are parallelograms.
Hence, AB=BC=CD=13AD
We know that opposite sides of a parallelogram are equal, so in ∥gm APSD:
PS = AD
⇒ 13PS=13AD
⇒ PQ=CD …(1)
Consider △PEQ and △CFD
PS∥AD and PD is transversal
∠QPE=∠CDF [Alternate interior angles] …(2)
Now, QB∥RC and AD is a transversal
∠QBD=∠RCD [Corresponding angles] …(3)
PS∥AD and PD is transversal
∴∠PQB=∠QBC[Alternate interior angles] …(4)
From (3) and (4)
∠PQE=∠FCD…(5)
From (1), (2), (5)
△PEQ≅△CFD [ASA congruence rule]
Hence, ar(△PQE)=ar(△CED)[Congruent figures have equal area]
Hence proved.

Question:2
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.). Prove that ar (LZY) = ar (MZYX)

Answer:

Solution.
Given: △LMN with X and Y points on the side LN
LX = XY = YN and XZ∥LM
To prove: ar(△LZY)=ar(MZYX)
Proof:
Since XZ∥LM and △XMZ and △XLZ are on the same base XZ and between the same parallel lines LM and XZ. Then,
ar(△XMZ)=ar(△XLZ) …(1)
Adding ar(△XYZ) both sides of eq. (1), we get
ar(△XMZ)+ar(△XYZ)=ar(△XLZ)+ar(△XYZ)
⇒ ar(XMZY) = ar(LZY)
This can be rearranged as:
ar (LZY) = ar (MZYX)
Hence proved.

Question:3

The area of the parallelogram ABCD is 90cm2 (see Fig.9.13). Find

(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)

Answer:
(i) 90cm²
(ii) 45cm²
(iii) 45cm²
Solution.
(i) Here, ABCD is a parallelogram with area 90cm2
∴AB∥CD
So, AB∥CF [Extended part of CD is CF]
∴AB∥EF
Hence, ABEF is a parallelogram.
We know that a parallelogram that lies between the same parallel lines and has the same base has equal areas.
Now, parallelogram ABCD and ABEF lie between the same parallel lines AB and CF have the same base AB. So these have equal area.
∴ar(ABCD)=ar(ABEF)
∴ar(ABEF)=90cm²
(ii)△ABD and parallelogram ABCD lie between the same parallel lines and have the same base AB. So we know that
∴ar(ABD)=12×ar(ABCD)
Now, ar(ABCD)=90cm²
⇒ar(ABD)=12×90cm²
∴ar(ABD)=45cm²
(iii) Here, △BEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF. So we know that,
ar(△BEF)=12ar(ABEF)=90cm²
ar(△BEF)=12×90=45cm²

Question:4

In △ABC, D is the midpoint of AB, and P is any point on BC. If CQ∥PD meets AB in Q (Fig.), then prove that ar(BPQ)=12ar(ABC).

Answer:

Solution.
In △ABC, D is the midpoint of AB, and P is any point on BC.
CQ∥PD
Join D and C

Consider △ABC,
Since D is the midpoint of AB
So CD is the median of △ABC
Using the mid-point theorem,
ar(△BCD)=12ar(△ABC)…(i)
Now, CQ∥PD
Since △PDQ and △PDC are on the same base, PD and between the same parallel lines PD & QC. So we have:
∴ar(△PDQ)=ar(△PDC) …(ii)
From (i) & (ii)
ar (△BCD)=12ar(△ABC)
ar(△BPD)+ar(△PDC)=12ar(△ABC)
Now, area of △PDC = area of △PDQ
ar(△BPD)+ar(△PDQ)=12ar(△ABC)
⇒ar(△BPQ)=12ar(△ABC)
Hence proved.

Question:5
ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig.), prove that ar (AER) = ar(AFR)

Answer:

Solution.
Given that ABCD is a square
E and F are the midpoints of BC and CD
R is the midpoint of EF
Proof: Consider △ABE and △ADF
We know that BC = DC (All sides of the square are equal)
⇒ BC²=DC²
⇒ EB=DF(Since E and F are mid points of BC and CD)
∠ABE =∠ADF=90°(All angles in a square are right angles)
⇒AB=AD(All sides in a square are equal)
∴△ABE≅△ADF(By SAS criterion)
Hence, AE = AF (by C.P.C.T)
Now, ER = RF ( R is the midpoint of EF)
AR = AR (Common)
∴△AER≅△AFR(By SSS criterion)
Hence, AR divides the triangle into two triangles of equal area.
⇒ ar(△AER)=ar(△AFR)
Hence proved.

Question:6

O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO).

Answer:

Solution.
Given: O is any point on the diagonal PR of a parallelogram PQRS
Construction: Join QS
Let diagonals PR and QS intersect each other at T.

We know that the diagonals of a parallelogram bisect each other.
T is the mid-point of QS and PR
Since the median of a triangle divides it into two triangles of equal area.
In △PQS, PT is the median of side QS
⇒ar(△PTS)=ar(△PTQ) …(i)
In △SOQ, OT is the median of side QS
⇒ar(△STO)=(△QTO) …(ii)
Adding (i) and (ii), we have
ar(△PTS)+ar(△STO)=ar(△PTQ)+(△QTO)
⇒ar(△PSO)=ar(△PQO)
Hence proved.

Question:7

ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig.). AE intersects CD at F.
If ar (DFB) = 3cm2, find the area of the parallelogram ABCD.

Answer: [ar(ABCD)=12cm2]
Solution.
Given: ABCD is a parallelogram.
CE = BC
The lines BC and AF when produced, meet at E.
ar△DFB=3cm²
Now,
BC = EC & AB∥FC (given)
∴FC=12AB (By mid-point theorem)
Also, AB = DC (opposite sides of a parallelogram are equal)
∴FC=12DC
⇒ FC=DF=12DC
Now, let the perpendicular distance between parallel lines AB and CD be h
ar(△DFB)=12(Base)(Height)
=12(DF)(h)
=12(12CD)(h)
=14(CD)(h) …(i)
Area of ∥ABCD = (Base) (corresponding altitude)
= (CD) (h) …(ii)
From (i) and (ii)
Area of ∥ABCD = 4(ar(△DFB))
Area of ∥ABCD = 4(3cm²)=12cm²

Question:8
In trapezium ABCD, AB∥DC and L is the midpoint of BC. Through L, a line PQ∥AD has been drawn which meets AB in P and DC produced in Q (Fig.). Prove that ar (ABCD) = ar (APQD)

Answer:

Solution.
Given: Trapezium ABCD with AB∥CD and BL = CL
PQ∥AD
To prove: ar(ABCD) = ar(APQD)
Proof: DC is produced at Q and AB∥DC
So, DQ∥AP
Also, PQ∥AD(given)
Then APQD is a parallelogram
In △CLQ and △BLP
CL = BL [L is mid-point of BC]
∠LCQ=∠LBP [alternate interior angles]
∠CQL=∠LPB [alternate interior angle as PQ is a transverse]
∴△CLQ≅△BLP[AAS congruence rule]
ar(△CLQ)=ar(△BLP) [Congruent triangles have equal area] …(i)


Now,
ar(ABCD)=ar(APLCD)+ar(△BLP)…(ii)
ar(APQD)=ar(APLCD)+ar(△CLQ)…(iii)
From eq (i), (ii) and (iii)
⇒ ar(ABCD) = ar(APQD)

Question:9

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

Answer:

Solution.
Given that if the mid-points of the sides of a quadrilateral are joined in order, a parallelogram is formed. We have to find the area of this parallelogram.
To prove: ar (parallelogram PFRS) =12 ar(quadrilateral ABCD)

Let ABCD is a quadrilateral and P, F, R and S are the mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.
Construction:- Join BD and BR.
We know that the median of a triangle divides it into two triangles of equal area.
So, BR divides △BDA into two triangles of equal area.
∴ar(△BRA)=12ar(△BDA)…(i)
Similarly, median RS divides △BRA into two triangles of equal area.
∴ar(△ASR)=12ar(△BRA)…(ii)
From eq. (i) and (ii)
∴ar(△ASR)=14ar(△BDA) …(iii)
Similarly,
ar(△CFP)=14ar(△BCD) ….(iv)
On adding equations (iii) and (iv), we get
ar(△ASR)+ar(△CFP)=14ar(△BDA+△BCD)
⇒ar(△ASR)+ar(△CFP)=14ar (Quadrilateral BCDA) …(v)
Similarly,
ar(△DRF)+ar(△BSP)=14ar(BCDA)….(vi)
On adding (v) and (vi), we get

ar(△ASR)+ar(△CFP)+ar(△BSP)+ar(△BSP)
=12ar (quadrilateral BCDA) …(vii)
But
ar(△ASR)+ar(△CFP)+ar(△DRF)+ar(△BSP)=ar(∥gmPFRS)
=12ar (quadrilateral BCDA) …(viii)
from subtracting eq. (vii) from eq. (viii) we get
ar(parallelogram PFRS) =12ar(quadrilateral BCDA)
Hence proved.

NCERT Exemplar Class 9 Maths Solutions Chapter 9: Exercise 9.4
Page: 94-96, Total Questions: 10

Question: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)

Answer:

Solution.
According to the question, ABCD is a parallelogram.
E is a point on BC.
AE and DC are produced to meet at F.
Using the above information, we can make the figure as follows:

We know that, triangles on the same base and between the same parallels are equal in area.
Here, we have,
△ABC and △ABF are on same base AB and between parallels, AB and DF
Area (△ABC) = area (△ABF) …(1)
We also know that,
Diagonal of a parallelogram divides it into two triangles of equal area.
So, Area (△ABC) = area (△ACD) …(2)
Now,
Area (△ADF) = area (△ACD) + area (△ACF)
Area (△ADF) = area (△ABC) + area(△ACF) [ Q From equation (2)]
⇒ Area (△ADF) = area (△ABF) + area (△ACF) [ Q From equation (1)]
⇒ Area (△ADF) = area (ABFC)

Hence proved.

Question: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.

Answer:

Solution.
Given in parallelogram ABCD, diagonals intersect at O and draw a line PQ, which intersects AD at P and BC at Q.
To prove: PQ divides the parallelogram ABCD into two part of equals area.
ar (ABQP) = ar(CDPQ)

We know that, diagonals of a parallelogram bisect each other.
OA = OC and OB = OD …(i)
In △AOB and △COD
OA = OC and OB = OD [From eq. (i)]
and <AOB = <COD [Vertically opposite angles]
∴△AOB≅△COD [SAS congruence rule]
ar(△AOB)=ar(△COD) …(ii) [Congruent figures have equal area]
In △AOP and △COQ
<PAO = <OCQ [alternate interior angles]
OA = OC [From eq. (i)]
<AOP = <COQ [Vertically opposite angles]
△AOP≅△COQ [ASA congruence rule]
ar(△AOP)=ar(△COQ) …(iii) [Congruent figures have equal area]
In △POD and △BOQ
<PDO = <OBQ [alternate interior angles]
OD = OB [From eq. (i)]
<DOP = <BOQ [Vertically opposite angles]
△DOP≅△BOQ [ASA congruence rule]
ar(CDPQ)=ar(△COQ)+ar(△COD)+ar(△POD) …(iv) [Congruent figures have equal area]
=ar(△AOP)+ar(△AOB)+ar(△BOQ) [From ii, iii, iv]
= ar(ABQP)
⇒ ar(ABQP) = ar(CDPQ)
Hence Proved.

Question:3
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of △GBC = area of the quadrilateral AFGE.

Answer:

Solution.
Given that the medians BE and CF of a triangle ABC intersect at G
We know that the median divides a triangle into two triangles of the same area.


So, ar(△BEC)=12ar(△ABC) …(1)
And, ar(△ACF)=12ar(△ABC) ….(2)
from (1) and (2)
ar(△ACF)=ar(△BEC)
∴ar(△BGC)=ar(GFAE)
This can be written as:
Area of △GBC = area of the quadrilateral AFGE.
Hence proved

Question:4

In Fig., CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

Answer:

Solution.
Given: CD∥AE and CY∥BA
To prove: ar(△CBX) = ar(△AXY)
Proof: We know that triangles on the same base and between the same parallels have equal area.
Consider CY∥BA
△ABY and △ABC both lie on the same base AB and between the same parallels CY and BA.
ar(△ABY)=ar(△ABC)…(1)
Now, ar(△ABY)=ar(ABX)+ar(AXY)
and, ar(△ABC)=ar(ABX)+ar(CBX)
Putting the above values in eq. (1)
⇒ ar(ABX) + ar(AXY) = ar(ABX) + ar(CBX)
⇒ ar(AXY) = ar(CBX)
Hence proved

Question:6

In △ABC, if L and M are the points on AB and AC, respectively, such that LM∥BC. Prove that ar (LOB) = ar (MOC)

Answer:

Solution.
Given: △ABC with L and M points on AB and AC respectively, such that LM∥BC.

To prove: ar(△LOB)=ar(△MOC)
Proof:
We know that triangles on the same base and between the same parallels are equal in area.
Hence, △LBC and △MBC lie on the same base BC and between the same parallels BC and LM.
So, ar(△LBC)=ar(△MBC) …(1)
ar(△LBC)=ar(△LOB)+ar(△BOC)
ar(△MBC)=ar(△MOC)+ar(△BOC)
Using these values in equation 1, we get:
⇒ ar(△LOB)+ar(△BOC)=ar(△MOC)+ar(△BOC)
On eliminating ar(△BOC) form both sides, we get
ar(△LOB)=ar(△MOC)
Hence proved.

Question:5

ABCD is a trapezium in which AB∥DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 79 ar (XYBA)

Answer:

Solution.
Given: Trapezium ABCD with AB∥DC,
DC = 30 cm and AB = 50 cm.
X and Y are the midpoints of AD and BC, respectively.
To prove: ar(DCYX)=79ar(XYBA)

Construction, Join DY and extend it to meet AB produced at P.
Proof:
In △DCY and △PBY
CY = BY [Given, Y is the mid-point of BC]
∠DCY=∠PBY [alternate interior angles]
∠2=∠3[Vertically opposite angles]
∠DCY≅∠PBY[ASA congruence rule]
then, DC = BP [CPCT]
DC = 30 cm (given)
DC = BP = 30 cm
Now,
AP=AB+BP=50+30=80cm
In △ADP;XY∥AP, X is the mid-point of AD and Y is the mid-point of DP.
XY=12AP=12×80=40cm [mid-point theorem ]
Let distance between AB, XY and XY, DC be h cm
Area of trapezium is given as 1/2 [Sum of parallel side × distance between them]
Now, area of trapezium DCYX=1/2 h(30+40)
= 1/2 h(70) = 35h cm²
Area of trapezium XYBA=12h(40+50)=12h×90=45hcm²
ar(DCYX)ar(XYBA)=35h45h=79
ar(DCYX)=79ar(XYBA)
Hence proved.

Question:7
In Fig., ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)

Answer:

Solution.
Given: a pentagon ABCDE, BP∥AC, and EQ∥AD.
To prove:ar(ABCDE)=ar(△APQ)
Proof: We know that triangles on the same base and between the same parallel lines are equal in area.
Now, △ADQ and △ADE lie on the same base AD and lie between the same parallels AD and EQ. So,
ar(△ADE)=ar(△ADQ)…(i)
Similarly,
△CAB and △CAP lie on the same base AC and lie between the same parallels AC and PB
ar(△CAB)=ar(△CAP) …(ii)
On adding equations (i) and (ii), we get
ar(△ADE)+ar(△CAB)=ar(△ADQ)+ar(△CAP)
Adding ar (DCDA) to both sides, we get
ar(△ADE)+ar(△CAB)+ar(△CDA)=ar(△ADQ)+ar(△CAP)+ar(△CDA)
⇒ar(ABCDE)=ar(△APQ)
Hence proved.

Question:8
If the medians of a △ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = 13 ar (ABC)

Answer:

Solution.
Given: △ABC with medians AM, BN & CL
ar(AGB)=ar(AGC)=ar(BGC)=13ar(ABC)


Proof: We know that a median divides the triangle into two triangles of same area.
Let the area of small triangles be denoted as 1, 2, 3, 4, 5, 6 as shown in the figure
AM is the median
ar(△AMC)=ar(△AMB)
ar(1)+ar(2)+ar(6)=ar(3)+ar(4)+ar(5) …(i)
CL is the median
ar(△CAL)=ar(△CBL)
ar(1)+ar(6)+ar(5)=ar(2)+ar(3)+ar(4)…(ii)
BN is the median
ar(△BNA)=ar(△BNC)
ar(4)+ar(5)+ar(6)=ar(1)+ar(2)+ar(3)
(i) – (ii)
ar(1)+ar(2)+ar(6)−ar(1)−ar(6)−ar(5)=ar(3)+ar(4)+ar(5)−ar(2)−ar(3)−ar(4)
So, ar(2)=ar(5)
(ii) – (iii)
ar(1)+ar(6)+ar(5)−ar(4)−ar(5)−ar(6)=ar(2)+ar(3)+ar(4)−ar(1)−ar(2)−ar(3)
ar(1)−ar(4)=ar(4)−ar(1)
So, ar(1)=ar(4)
Similarly we can prove that area of ar(1)=ar(2)=ar(3)=ar(4)=ar(5)=ar(6)
Hence the triangle is divided into 6 triangles of equal area.
△AGB,△AGCand △BGC consists of 2 triangles each
So they have equal area which is equal to twice the area of the smaller triangles.
ar(△AGB)=ar(△AGC)=ar(△BGC)=2 (area of 1 small triangle)
Hence, ar(△AGB)=ar(△AGC)=ar(△BGC)=13ar(ABC)
Hence proved.

Question:9
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).

Answer:

Solution.
Given: X and Y are the mid points of AC and AB respectively.
QP∥BC
CYQ, BXP are straight lines,
To prove: ar(△ABP)=ar(△ACQ)
Proof:
Since, X and Y are the mid-point of AC and AB.
So, XY∥BC
We know that triangles on the same base and between the same parallels are equal in area
Consider parallel lines BC and XY;
△BYC and △BXC lie on same base BC and between the same parallels BC and XY.
So, ar(△BYC)=ar(△BXC)
ar(△BOC)+ar(△BOY)=ar(△BOC)+ar(△XOC)
⇒ar(△BOY)=ar(△COX)
Adding ar(XOY) to LHS and RHS
⇒ ar(BOY)+ar(XOY)=ar(△COX)+ar(△XOY)
⇒ ar(△BYX)=ar(△CXY)…(i)
We observe that the quadrilateral XYAP and XYAQ have the same base XY and these are between the same parallel lines XY and PQ.
∵ar(XYAP)=ar(YXAQ) …(ii)
on adding eq. (i) and (ii), we get
ar(△BYX)+ar(XYAP)=ar(△CXY)+ar(YXAQ)
⇒ ar(△ABP)=ar(△ACQ)
Hence Proved.

Question:10

In Fig., ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD)

Answer:

Solution.
Given: Two parallelograms ABCD and AEFD.
To prove: ar(△PEA)=ar(△QFD)
Proof:
PQDA is a parallelogram, so AP∥DQ,PQ∥AD
Now, parallelograms PQDA and AEFD are on the same base AD and lie between the same parallel lines AD and EQ. Hence,
ar(∥gmPQDA)=ar(∥gmAEFD)
ar(PFDA)+ar(△QFD)=ar(△PEA)+ar(PFDA)
So, we get:
ar(△QFD)=ar(△PEA)
Hence proved.