NCERT Solutions for Class 9 Maths Areas of Parallelograms and Triangles [Old Book]

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Class : 9th
Subject : Maths
Chapter : 9
Chapter Name : Areas Of Parallelograms And Triangles

Exercise 9.1

Q1 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.

Answer. (i)

Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are
lying between the same parallel lines AB and CD.
(ii)

No. It can be observed that parallelogram PQRS and trapezium MNRS have a

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

common base RS. However, their vertices, (i.e., opposite to the common base) P, Q
of parallelogram and M, N of trapezium, are not lying on the same line.
(iii)

Yes. It can be observed that parallelogram PQRS and triangle TQR have a common
base QR and they are lying between the same parallel lines PS and QR.
(iv)

No. It can be observed that parallelogram ABCD and triangle PQR are lying between
same parallel lines AD and ac. However, these do not have any common base.
(v)

Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a
common base AD and these are lying between the same parallel lines AD and 3Q.
(vi)

No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However,
these do not lie between the same parallel lines.

Page : 155 , Block Name : Exercise 9.1

Exercise 9.2

Q1 In Figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF
= 10 cm, find AD.

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Answer. In parallelogram ABCD, CD = AB = 16 cm
[Opposite sides of a parallelogram are equal]
We know that
Area of a parallelogram = Base x Corresponding altitude
Area of parallelogram ABCD = CD x AE = AD x CF
Area of parallelogram = Base x Corresponding altitude
Area of parallelogram ABCD = CD x AE = AD x CF
16cm × 8cm = AD × 10cm
16×8
AD = cm = 12.8cm
10

Thus, the length of AD is 12.8 cm.

Page : 159 , Block Name : Exercise 9.2

Q2 If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
I
ar(EFGH) = ar(ABCD).
2

Answer.

Let us join HF.
In parallelogram ABCD,
AD = BC and AD∥BC ( Opposite sides of a parallelogram are equal and parallel)
AB = CD ( Opposite sides of a parallelogram are equal)
1 1
⇒ AD = BC and
2 2

AH∥BF

Therefore, ABFH is a parallelogram.
Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB
and HF,
Area (ABFH) … (1) Similarly, it can be proved that
1
∴ Area (ΔHEF) =
2

Area (ΔH GF ) = Area (HDCF) … (2)
1

2

On adding equations (1) and (2), we obtain
1 1
Area (ΔHEF) + Area (ΔHGF) = Area (ABFH) + 2 2
Area (HDCF)
1
= [ Area (ABFH) + Area (HDCF)]
2
1
⇒ Area (EFGH) = Area (ABCD)
2

Page : 159 , Block Name Exercise 9.2

Q3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

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Show that ar (APB) = ar (BQC).

Answer.

It can be observed that △BQC and paralleogram ABCD lie on the same base BC and these are between
the same parallel lines AD and BC.
1
∴ Area (ΔBQC) = Area (ABCD) … (1)
2

Similarly, △APB and parallelogram ABCD lie on the same base AB and between the same parallel
lines AB and DC.
Area (ABCD) … (2)
1
∴ Area (ΔAPB) =
2

From equation (1) and (2), we obtain
Area (ΔBQC) = Area (ΔAP B)

Page : 159 , Block Name Exercise 9.2

Q4 In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = ar(ABCD)
1

2

(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]

Answer.

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD,
AB∥EF(By construction ) … (1)

ABCD is a parallelogram.

∴ AD ∥ BC (Opposite sides of a parallelogram)

⇒ AE∥BF … (2)

From equations (1) and (2), we obtain
AB∥EF and AE∥BF

Therefore, quadrilateral ABFE is a parallelogram
It can be observed that △APB and parallelogram ABFE are lying on the same base AB and between the
same parallel lines AB and EF.
Area (ABFE) … (3)
1
∴ Area (ΔAPB) =
2

Similarly, for ΔPCD and parallelogram EFCD,

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Area (△PCD) = 1

2 Area (EF CD)…(4)

Adding equation (3) and (4), we obtain
1
(ΔAPB) + Area (ΔPCD) = [ Area (ABFE) + Area (EFCD)]
2

…(5)
1
(ΔAPB) + Area (ΔPCD) = Arca (ABCD)
2

(ii)

Let us draw a line segment MN, passing through point P and parallel to line segment AD.
In parallelogram ABCD,
MN || AD (By construction)...(6)
ABCD is a parallelogram.
∴ AB ∥ DC (Opposite sides of a parallelogram)

⇒ AM∥DN … (7)

From equations (6) (7), we obtain
MN∥AD and AM∥DN

Therefore, quadrilateral AMND is a parallelogram.
It can be observed that △AP D and parallelogram AMND are lying on the same base
AD and between the same parallel lines AD and MN.
Area (AMND) … (8)
1
∴ Area (ΔAP D) =
2

Similarly, for △PCB and parallelogram MNCB,
Area (ΔPCB) = Area (MNCB) … (9)
1

2

Adding equations (8) and (9), we obtain
1
(ΔAPD) + Area (ΔPCB) = [ Area (AMND) + Area (MNCB)]
2

= Arca(ABCD)
1
(ΔAPD) + Area (ΔPCB) … (10)
2

On comparing equations (5) and (10), we obtain
(ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)

Page : 159 , Block Name : Exercise 9.2

Q5 In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar(PQRS) = ar(ABRS)
(ii) ar(AXS) = ar (PQRS)
1

2

Answer. (i) It can be observed that paralleogram PQRS and ABRS lie on the same base SR and also,
these lie in between the same parallel lines SR and PB.
∴ Area (PQRS) = Area (ABRS) … (1)

(ii) Consider △AXS and paralleogram ABRS.
As these lie on the same base and are between the same parallel lines AS and BR,
Area (ABRS) … (2)
1
∴ Area (ΔA × S) =
2

From equations (1) and (2), we obtain
1
Area (ΔAXS) = = Area(P QRS)2

Page : 159 , Block Name : Exercise 9.2
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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Q6 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?

Answer.

From the figure, it can be that point A divides the field into three parts.
These parts are triangular in shape - △PSA, ΔPAQ, and △QRA
Are of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of ∥gm pQRS … (1)
1
∴ Area (ΔPAQ) = Area (PQRS) … (2)
2

From equation (1) and (2), we obtain
Area (ΔPSA) + Area (ΔQRA) = Area(P QRS). . . (3)
1

2

Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ and pulses in other
two triangular parts PSA and QRA or wheat in triangular parts PSA and QRA and pulses in triangular
parts PAQ.

Page : 160 , Block Name : Exercise 9.2

Exercise 9.3

Q1 In Figure, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).

Answer. AD is the median of △ABC. Therefore, it will divide △ABC in to two triangles of equal
areas.
∴ Area (ΔABD) = Area (ΔACD) … (1)

ED is the median of △EBC
∴ Area (ΔEBD) = Area (ΔECD) … (2)

On subtracting equation (2) from equations (1), we obtain
Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD)
Area (ΔABE) = Area (ΔACE)

Page : 162 , Block Name : Exercise 9.3

Q2 In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar(ABC).

Answer.

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

ar(BED) = (1/2) × BD × DE

As E is the mid-point of AD

Thus, AE = DE

As AD is the median on side BC of triangle ABC,

Thus, BD = DC

Therefore,

DE = (½)AD ….(i)
BD = (½) BC …(ii)
From (i) and (ii)
ar(BED) = (1/2) × (1/2)BC × (1/2)AD

⇒ ar(BED) = (1/2) × (1/2) ar(ABC)

⇒ ar(BED) = 1/4 ar(ABC)

Page : 162 , Block Name : Exercise 9.3

Q3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer.

We know that diagonals of parallelogram bisect each other.
Therefore, O is the mid-point of AC and BD.
BO is the median in △ABC. Therefore, it will divide it into two triangles of equal areas.
∴ Area (ΔAOB) = Area (ΔBOC) … (1)

In △BCD, CO is the nedian.
∴ Area (ΔBOC) = Area (ΔCOD) … (2)

Similarly, Area (ΔCOD) = Area (ΔAOD) … (3)
From equations (1), (2), and (3), we obtain
Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)
Therefore, it is evident that the diagonals of a paralleogram divide it into four triangles of equal area.

Page : 162 , Block Name : Exercise 9.3

Q4 In Figure, 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).

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Answer. Consider △ACD
Line-segment CD is bisected by AB at O. Therefore, AO is the median of △ACD
∴ Area (ΔACO) = Area (ΔADO) … (1)

Considering ΔBCD, BO is the median.
∴ Area (ΔBCO) = Area (ΔBDO) … (2)

Adding equations (1) and (2), we obtain
Area (ΔACO) + Area (ΔBCO) = Area (△ADO) + Area (ΔBDO)
⇒ Area (ΔABC) = Area (ΔABD)

Page : 162 , Block Name : Exercise 9.3

Q5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ ABC. Show that
(i) BDEF isaparallelogram.
(ii) ar(DEF) = ar(ABC)
1

4
1
(iii) ar(BDEF) = 2
ar(ABC)

Answer.

(i) In ∆ ABC
EF ∥BC and EF = 1/2BC (by mid point theorem)

also,

BD = 1/2BC(D is the mid point)

So , BD = EF

also,

BF and DE will also parallel and equal to each other.

Thus, the pair opposite sides are equal in length and parallel

to each other.

∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i)
BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔBFD) = ar(ΔDEF)( For parallelogram BDEF ) …. (i)

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Also, ar(ΔAFE) = ar(ΔDEF)( For parallelogram DCEF) …. (ii)
ar(ΔCDE) = ar(ΔDEF)( For parallelogram AFDE) …. (iii)

From (i), (ii) and (iii)

ar(ΔBF D) = ar(ΔAF E) = ar(ΔCDE) = ar(ΔDEF )

⇒ ar(ΔBF D) + ar(ΔAF E) + ar(ΔCDE) + ar(ΔDEF ) =

arar(ΔABC)

⇒ 4 ar(ΔDEF ) = ar(ΔABC)

⇒ ar(DEF ) = 1/4 ar(ABC)

(iii) Area (parallelogram BDEF) = ar (∆ DEF) + ar(∆ BDE)
= ar (parallelogram BDEF) = ar (∆ DEF) + ar(∆ DEF)
= ar (parallelogram BDEF) = 2 x ar (∆ DEF)
ar(parallelogram BDEF )= 2 × 1/4 ar( (ΔABC) ⇒
=
ar(parallelogram BDEF) = 1/2 ar(ΔABC)

Page : 163 , Block Name : Exercise 9.3

Q6 In Figure, 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.]

Answer.

Let us draw DN ⊥ AC and BN ⊥ AC.
(i) In △DON and △BOM
∠DNO = ∠BMO (By construction)

∠DON = ∠BOM (Vertically opposite angles)

OD = OB(Given)

By AAS congruence rule,
△DON ≅ △BOM

DN = BM …(1)
We know that congruent triangles have equal areas.
Area (ΔDON) = Area (ΔBOM) …(2)
In △DNC and △BMA,
∠DN C = ∠BM A(By construction )

CD = AB(given)

DN = BM[Using Equation (1)]

∴ △DNC ≅ △BMA(RHS congruence rule)

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∴ Area (ΔDNC) = Area (ΔBMA) ..(3)
On adding Equations (2) and (3), we obtain
Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)
Therefore, Area (ΔDOC) = Area (△AOB)
(ii) We obtained,
Area (ΔDOC) = Area (ΔAOB)
∴ Area (ΔDOC) + Area (ΔOCB) = (ΔAOB) + Area (ΔOCB)

(Adding Area (△OCB)to both sides)
∴ Area (ΔDCB) = Ar ca(ΔACB)

(iii) We obtained,
Area (ΔDCB) = Area (△ACB)
If two triangles have the same base and equal areas, then these will lie between the same parallels.
∴ DA∥CB …(4)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD ) and the other part of opposite
sides is parallel (DA∥CB)

Page : 163 , Block Name : Exercise 9.3

Q7 D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC).
Prove that DE || BC.

Answer.

Since △BCE and △BCD are lying on a common base BC and also have equal areas,
△BCE and ΔBCD will lie between the same parallel lines.

∴ DE∥BC

Page : 163 , Block Name : Exercise 9.3

Q8 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)

Answer.

It is given that
XY ∥BC = EY ∥BC

BE∥AC = BE∥CY

Therefore,EBYC is a parallelogram.
It is given that
XY ∥BC = XF ∥BC

F C∥AB = F C∥XB

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Therefore, BCFX is a parallelogram.
Parallelograms EBYC and BCFX are on the same base BC and between the same parallels BC and EF.
1
∴ Area (EBCY ) = Area (BCF X) … (1)
2

Consider parallelogram EBYC and △AEB
These lie on the same base BE and are between the same parallels BE and AC.
∴ Area (ΔABE) =
1

2
..(2)
Also, parallelogram ΔCFX and △ACF are on the same base CF and between the same parallels CF
and AB.
… (3)
1
∴ Area (ΔACF) = Area (BCFX)
2

From equations (!), (2), and (3), we obtain
Area (△ABE) = Area (ΔACF )

Page : 163 , Block Name : Exercise 9.3

Q9 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 Figure). Show that ar
(ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Answer.

Let us join AC and PQ.
△ACQ and △AQP are the same base AQ and between the same parallels AQ and CP.

Area (ΔACQ) = Area (ΔAP Q)
(ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)

(ΔABC) = Area (ΔQBP) …(1)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively.
Area (ΔABC) = Area (ABCD) … (2)
1

2
1
Area (ΔQBP ) = Area (PBQR) … (3)
2

From equations(!),(2) and (3), we obtain
1 1
Area (ABCD) = Area(P BQR)
2 2

Area(ABCD) = Area(PBQR)

Page : 163 , Block Name : Exercise 9.3

Q10 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar
(AOD) = ar (BOC).

Answer.

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It can be observed that △DAC and ΔDBC lie on the same base DC and between the same parallels
AB and CD.
Area (ΔDAC) = Arca(ΔDBC)
Area (ΔDAC) − Area(ΔDOC) = Arca(ΔDBC) − Arca(ΔDOC)
(ΔAOD) = Arca (ΔBOC)

Page : 163 , Block Name : Exercise 9.3

Q11 In Figure, 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)

Answer. (i) △ACB and △ACF lie on the same base AC and are between
The same parallels AC and BF.
(ii) It can be observed that
Area (ΔACB) = Area (ΔACF)
Area (△ACB) + Area (ACDE) = Area (ΔACF) + Area (ACDE)
Area (ABCDE) = Area (AEDF)

Page : 163 , Block Name : Exercise 9.3

Q12 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.

Answer.

Let quadrilateral ABCD be the original shape of the field.
The proposal may be implemented as follows.
Join diagonal BD and draw a line parallel to BD through point A. Let it meet the extended side CD of
ABCD at point E. Join BE and AD. Let them intersect each other at O. Then, portion △AOB can be cut
from the original field so that the new shape of the field will b ΔBCE.(se figure).

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We have to prove that the area of △AOB (portion that was cut so as to construct Health Centre) is equal
to the area of △DEO (portion added to the field so as to make the area of the new field so formed equal
to the area of the original field).

It can be observed that △DEB and △DAB lie on the same base BD and are between the same
parallels BD and AE.
Area (ΔDEB) = Area (ΔDAB)
Area (ΔDEB) − Area (ΔDOB) = (ΔDAB) − Area (ΔDOB)
Area (ΔDEO) = Area(ΔAOB)

Page : 164 , Block Name : Exercise 9.3

Q13 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.]

Answer.

It can be observed that △ADX and △ACX lie on the same base AX and are between the same
parallels AB and DC.
Area (ΔADX) = Arca (ΔACX) …(1)
△ACY and △ACX lie on the same base AC and are between the same parallels AC and XY.

Area (ΔACY ) = Area (ACX) …(2)
From Equations (1) and (2), we obtain
Area (ΔADX) = Area(ΔACY )

Page : 164 , Block Name : Exercise 9.3

Q14 In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Answer. Since △ABQ and △P BQ lie on the same base BQ and are between the same parallels AP
and BQ.
∴ Area (ΔABQ) = Arca (ΔPBQ) …(1)

Again, △BCQ and △BRQ lie on the same BQ and are between the same parallels BQ and CR.
∴ Area (ΔBCQ) = Area (\DeltaBRQ) … ….(2)

On adding Equations(1) and (2), we obtain
Area (ΔABQ) + Arca(ΔBCQ) = (ΔP BQ) + Area (ΔBRQ)

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∴ Area (ΔAQC) = Area (ΔPBR)

Page : 164 , Block Name : Exercise 9.3

Q15 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.

Answer.
It is given that
Area (△AOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = (ΔBOC) + Area (ΔAOB)
Area (ΔADB) = Area (△ACB)
We know that triangles on the same base having area equal to each other lie between the same parallels.
Therefore, these triangles, △ADB and △ACB, are lying between the same parallels.
i.e., AB || CD
Therefore, ABCD is a trapezium.

Page : 164 , Block Name : Exercise 9.3

Q16 In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD
and DCPR are trapeziums.

It is given that
Area (ΔDRC) = Area (ΔDP C)
As △DRC and ΔDP C lie on the same base DC and have equal area, therefore, they must lie
between the same parallel lines.

∴ DC∥ RP

Therefore, DCPR is a trapezium
It is also given that
Area (ΔBDP ) = Area (△ARC)
Area (ΔBDP ) − Area (ΔDP C) = (ΔARC) − Area (ΔDRC)
∴ Area (ΔBDC) = Area(ΔADC)

Since △BDC and △ADC are on the same base CD and have equal areas, they must lie between the
same parallel lines.
∴ AB∥CD

Therefore, ABCD is a trapezium.

Page : 164 , Block Name : Exercise 9.3

Exercise 9.4

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Q1 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.

Answer. As the parallelogram and the rectangle have the same base and equal area, therefore, these will
also lie between the same parallels.
Consider the parallelogram ABCD and rectangle ABEF as follows.

Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels
AB and CF.
We know that opposite sides of parallelogram or a rectangle are of equal lengths.
Therefore,
AB = EF (For rectangle)
AB = CD (For parallelogram)
∴ CD = EF

∴ AB + CD = AB + EF ...(1)
Of all the lines segments that can be drawn to a given line from a point not lying on it, the perpendicular
line segment is the shortest.
∴ AF < AD

And similarly, BE < BC
∴ AF + BE < AD + BC . . . (2)

FRom equations (1) and (2), we obtain
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD.

Page : 164 , Block Name : Exercise 9.4

Q2 In Figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE)
= ar (AEC).

Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the
field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD,
ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points
of division so obtained to the opposite vertex of BC, you can divide ∆ABC into n triangles of equal
areas.]

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Answer. Let us draw a line segment AL ⊥ BC

We know that,
Area of a triangle = 1

2
× Base × Altitude

Area (△ADE) = 1

2
× DE × AL

Area (ΔABD) = 1

2
× BD × AL

Area (△AEC) = × EC × AL
1

2

It is given that DE = BD =EC
1 1 1
× DE × AL = × BD × AL = × EC × AL
2 2 2

Area (△ADE) = Area (ΔABD) = Area (△AEC)
It can be observed that Bhudhia has divided her field into 3 equal parts.

Page : 165 , Block Name : Exercise 9.4

Q3 In Figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Answer. It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are
equal.
∴ AD = BC …(1)

Similarly, for parallelograms DCEF and ABFE, it can be proved that
DE = CF…(2)
And, EA = FB….(3)
In △ADE and ΔBCF,
AD = BC [ Using equation (1) ]
DE = CF [ Using equation (2) ]
EA = FB [ Using Equation (3)]
∴ △ADE ≅ △BCF(SSS congruence rule)

∴ (ΔADE) = Arca (ΔBCF )

Page : 165 , Block Name : Exercise 9.4

Q4 In Figure, 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.]

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Answer. It is given that ABCD is a parallelogram.
AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other )
Join point A to point C.

Consider △APC and △BPC
△APC and ΔBPC are lying on the same base PC and between the same parallels PC and AB.

Therefore,
Area (ΔAPC) = Area (ΔBPC) … (1)
In quadrilateral ACDQ, it is given that
AD = CQ
Since ABCD is a parallelogram,
AD || BC (Opposite sides of a parallelogram are parallel)
CQ is a line segment which is obtained when line segment BC is produced.
∴ AD || CQ

We have,
AC = DQ and AC || DQ
Hence, ACQD is a parallelogram.
Consider BDCQ and BACQ
These are on the same base CQ and between the same parallels CQ and AD.
Therefore,
Area (ΔDCQ) = A rca(ΔACQ)
∴ Area (ΔDCQ) − Arca (ΔP QC) = Area (ΔACQ) − Area (ΔP QC)

∴ Area (ΔDPQ) = Arca (ΔAPC) − (2)

From equations (1) and (2), we obtain
Area (ΔBP C) = Area (ΔDP Q)

Page : 165 , Block Name : Exercise 9.4

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Q5 In Figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE
intersects BC at F, show that

(i) ar(BDE) = 1

4
ar(ABC)

(ii) ar(BDE) = ar(BAE)
1

2

(iii) ar(ABC) = 2ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2ar(FED)
(vi) ar(FED) = ar(AFC)
1

8

[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

Answer. (i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side
BC and also its length will be half of the length of 3C (mid-point theorem).

1
∴ GH = BC and GH ∥BD
2

∴ GH = BD = DC and GH∥BD(D is the mid-point of BC)

Similarly,
GD = HC = HA
HD = AG BG
Therefore, clearly △ABC is divided into 4 equal equilateral triangles viz
△BGD, △AGH , ΔDH C and ΔGH D

In other words, ΔBGD = 1

4
ΔABC

In other words, ΔBGD = ΔABC 1

4

Now consider △BDG and ΔBDE
BD = BD(Common base)
As both triangles are equilateral triangle, we can say
BG = BE
DG = DE

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

Therefore, ΔBDG ≅ △BDE[By SSS congruency ]
Thus, area (ΔBDG) = area(ΔBDE)
1
ar(ΔBDE) = ar [ΔABC)
4

Hence proved

(ii) Area (ΔBDE) = Area (△AED) (Common base DE and DE || AB)
Area (ΔBDE) − Arca (ΔF ED) = Area(ΔAED) − Area(ΔF ED)
Area (ΔBEF ) = Area(ΔAF D) ..(1)
Now, Area (ΔABD) = Arca (ΔABF ) + Area (ΔAF D)
Area (ΔABD) = Area (ΔABF )+ Area (ΔBEF ) [From equation (1)]
Area (ΔABD) = Area (ΔABE) … (2)
AD is the median in △ABC
1
(ΔABD) = ar (ΔABC)
2
4
=
2
ar(ΔBDE) (As proved earlier)
ar (ΔABD) = 2 ar (ΔBDE) (3)
From (2) and (3), we obtain
2 ar ( (\Delta BDE)=\text { ar }(\Delta A B E) \)
ar (BDE) = ar (BAE)
1

2

(iii) ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar ( (\Delta A B F)+\text { ar }(\Delta B E F)= \) ar (△BEC)
Using equations (1), we obtain
ar(ΔABF) + ar(ΔAFD) = ar(ΔBEC)

ar(ΔABD) = ar(ΔBEC)
1
ar(ΔABC) = ar(ΔBEC)
2

ar(ΔABC) = 2 ar (ΔBEC)

(iv) It is seen that △BDE and ar △AED lie on the base (DE) and between the parallels DE and AB.
∴ ar(ΔBDE) = ar(ΔAED)

∴ ar(ΔBDE) − ar(ΔF ED) = ar(ΔAED) − ar(ΔF ED)

∴ ar(ΔBF E) = ar(ΔAF D)

(v) Let h be the height of vertex E, corresponding to the side BD in △BDE .
Let H be the height of vertex A, corresponding to the side BC in △ABC.
In (i), it was shown that ar(BDE) = ar(ABC) 1

4

1 1 1
∴ × BD × h = ( × BC × H )
2 4 2

1
⇒ BD × h = (2BD × H )
4
1
⇒ h = H
2

In (iv), it was shown that ar(ΔBF E) = ar(ΔAF D)
∴ ar ( △BF E) = ar (ΔAFD)

= 2 ar (ΔF ED)
Hence,
(vi) Area (AFC) = area(AFD) + area(ADC)
1
= ar(BFE) + ar(ABC) [ln(iv), ar(BFE) = ar(AFD); AD is median of ΔABC]
2

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

1 1
= ar(BFE) + × 4 ar(BDE) [In(i), ar(BDE) = ar(ABC)]
2 4

= ar(BFE) + 2 ar(BDE) …(5)
Now, by (v), ar (BFE) = 2 ar(FED) ….(6)
arc(BDE) = ar(BFE) + ar(FED) = 2 ar(FED) + ar(FED) = 3 ar(FED) ….(7)
Therefore, from equations (5), (6), and (7), we ger:
ar(AFC) = 2 ar(FED) + 2 × 3 ar(FED) = 8ar(FED)

∴ ar(AFC) = sar(FED)

Hence, ar(FED) = 1

8
ar(AFC)

Page : 165 , Block Name : Exercise 9.4

Q6 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.]

Answer. A quadrilateral ABCD, in which diagonals AC and BD intersect each other at point E.

To Prove : ar (△AED) × arc(ΔBEC)
= arc(△ABE) × arc(ΔCDE)

Construction From A, draw AM ⊥ BD and CN ⊥ BD
Proof : : ar(ΔABE) = × BE × AM …..(i)
1

2

ar(ΔAED) =
1

2
× DE × AM ….(ii)
Dividing eq.(ii) by (i), we get,
1
ar(ΔAED) ×DE×AM
2
= 1
ar(GABE)
×BE×AM
2

ar(ΔAED)
….(iii)
DE
⇒ =
ar(ΔABE) BE

ar(ΔCDE) DE
Similarly ar(ΔBEC)
−
BE
….(iv)
From eq.(iii) and (iv), we get
ar(ΔAED) ar(ΔCDE)
= =
ar(△ABE) ar(ΔBEC)

⇒ arc(ΔAED) × ar(ΔBEC) = arc(AABE) × arc(ΔCDE)

Hence proved.

Page : 166 , Block Name : Exercise 9.4

Q7 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
1
(i) ar(PRQ) = ar(ARC) 2

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

3
(ii) ar(RQC) = ar(ABC)
8

(iii) ar(PBQ) = ar(ARC)

Answer. Take a point S on AC such that S is the mid-point of AC.
Extend PQ to T such that PQ = QT.
Join TC, QS, PS, and AQ.

In △ABC, P and Q are the mis-points of AB and BC respectively. Hence, by using mid-point theorem,
we obatin
1
PQ ∥
∥AC and PQ = AC
2

∴ PQ || AC and PQ = AS (As S is the mid-point of AC)
∴ PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of

triangles.
∴ ar (ΔPAS) = ar(ΔSQP) = ar(ΔPAQ) = ar(ΔSQA)

Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, AND PSQB are also parallelograms
and therefore,
ar(ΔPSQ) = ar(ΔCQS)( For parallelogram PSCQ)

ar(ΔQSC) = ar(ΔCTQ)( For parallelogram QSCT)

ar(ΔPSQ) = ar(ΔQBP)( For parallelogram PSQB)

Thus,
ar (ΔPAS) = ar(ΔSQP) = ar(ΔPAQ) = ar(ΔSQA) = ar(ΔQSC) = ar(ΔCTQ) = ar

(ΔQBP) … (1)

(ΔABC) = ar(ΔPBQ) + ar(ΔPAS) + ar(ΔPQS) + ar(ΔQSC)

ar(ΔABC) = ar(ΔPBQ) + ar(ΔPBQ) + ar(ΔPBQ) + ar(ΔPBQ)

= ar(ΔP BQ) + ar(ΔP BQ) + ar(ΔP BQ) + ar(ΔP BQ)

= 4 ar (ΔPBQ)
1
∴ ar ar(ΔPBQ) = 4
ar(ΔABC) … (2)

(i) Join point P to C.

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

In ΔPAQ, QR is the median.
1 1 1 1
∴ ar(ΔPRQ) = ar(ΔPAQ) = × ar(ΔABC) = ar(ΔABC) ……(3)
2 2 4 8

In △ABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem,
we obtain
PQ = 1/2 AC
AC = 2PQ ⇒ AC = PT

Also, PQ∥AC ⇒ PT∥AC
Hence, PACT is a parallelogram.
ar(PACT) = ar(PACQ) + ar(ΔQTC)

= ar(PACQ) + ar(ΔPBQ[U sing equation (1)]

∴ ar(PACT) = ar(ΔABC) … (4)
1
ar(ΔARC) = ar(ΔPAC) (CR is the median of ΔPAC)
2
1 1
= × ar(PACT)(PC is the diagonal of parallelogram PACT)
2 2
1 1
= ar(ΔPACT) = ar(ΔABC)
4 4
1 1
⇒ ar(ΔARC) = ar(ΔABC)
2 8

⇒
1

2
ar(ΔARC) = ar(ΔPRQ)[ Using equation (3)] … (5)
(ii)

ar(PACT) = ar(ΔPRQ) + ar(ΔARC) + ar(ΔQTC) + ar(ΔRQC)

(1), (2), (3), (4), and (5), we obtain
1 1 1
ar(ΔABC) = ar(ΔABC) + ar(ΔABC) + ar(ΔABC) + ar(ΔRQC)
8 4 4
5
ar(ΔABC) = ar(ΔABC) + ar(ΔRQC)
8

5
ar(ΔRQC) = (1 − ) ar(ΔABC)
8

3
ar(ΔRQC) = ar(ΔABC)
8

(iii) In parallelogram PACT,

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

1
ar(ΔARC) = ar(ΔPAC) (CR is the median of ΔPAC)
2
1 1
= × ar(PACT)(PC is the diagonal of parallelogram PACT)
2 2
1
= ar(ΔPACT)
4

1
= ar(ΔABC)
4

= ar(ΔPBQ)

Page : 166 , Block Name : Exercise 9.4

Q8 In Figure, 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:

(i) ΔMBC ≅ △ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ △ACE
(v) ar(CYXE) = 2ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCFD) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem
in Class X.

Answer. (i) We know that each angle of a square is 90 .
∘

Hence, ∠ABM = ∠DBC = 90 ∘

∴ ∠ABM + ∠ABC = ∠DBC + ∠ABC

∴ ∠M BC = ∠ABD

In △M BC and △ABD
∠M BC = ∠ABD( Proved above)

MB = AB( Sides of square ABMN)

BC = BD( Sides of square BCED)

∴ △MBC ≅ △ABD (SAS congruence rule)
(ii) We have
△M BC ≅ △ABD

∴ (ΔM BC) = ar (ΔABD) … (1)

It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
∴ BD∥AX (Two lines perpendicular to same line are parallel to each other)

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Book : Mathematics Ncert Solutions | Chapter-9 Maths

△ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
Area (ΔY XD) = 2 Area (ΔM BC) [Using equation (1)]...(2)
(iii) △MBC and parallelogram ABMN are lying on the same base MB and between same parallels MB
and NC.
2 ar (ΔM BC) = ar(ABM N )
ar (ΔY XD) = ar (ABM N ) [ Using equation (2)]...(3)
(iv) We know that each angle of a square is 90
∘

∘
∴ ∠F CA = ∠BCE = 90

∴ ∠F CA + ∠ACB = ∠BCE + ∠ACB

∴ ∠F CB = ∠ACE

In ΔF CB and △ACE
∠F CB = ∠ACE

FC = AC( Sides of square ACFG)

CB = CE( Sides of square BCED)

ΔFCB ≅ △ACE (SAS congruence rule)
(v) It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)
Hence, CE || AX (Two lines perpendicular to th same line are parallel to each other)
Consider BACE and parallelogram CYXE
BACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.
∴ ar (ΔY XE) = 2 ar (ΔACE) … (4)

We had proved that
△ΔFCB ≅ △ACE

ar(ΔF CB) ≅ ar(ΔACE) − (5)

On comparing equations (4) and (5), we obtain
(CYXE) = 2 ar (ΔFCB) ..(6)

(vi) Consider BFCB and parallelogram ACFG
BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and
BG.
∴ ar (ACFG) = 2ar(ΔFCB)

∴ ar (ACFG) = ar(CYXE) [ Using equation(6)]...(7)
(vii) From the figure, it is evident that
ar (ΔCED) = ar(ΔY XD) + ar(CY XE)
∴ ar (ΔCED) = ar(ABM N ) + ar(ACF G) [Using equations (3) and (7)].

Page : 166 , Block Name : Exercise 9.4

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