RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals

In this chapter, we provide RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals pdf, free RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals

RD Sharma Class 9 Solution Chapter 14 Quadrilaterals Ex 14.1

Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels: [NCERT]
RD Sharma Class 9 Chapter 14 Quadrilaterals
Solution:
(i) ΔPCD and trapezium ABCD are on the same base CD and between the same parallels AB and DC.
(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.
(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.
(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.
(v) Parallelogram PQRS and trapezium SMNR on tire same base SR but they are not between the same parallels.
(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, and APSD are between the same parallels.

Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals Ex 14.2

Question 1.
In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
In ||gm ABCD,
Base AB = 16 cm
and altitude AE = 8 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
∴ Area = Base x Altitude
= AB x AE
= 16 x 8 = 128 cm2
Now area of ||gm ABCD = 128 cm2
Altitude CF = 10 cm
∴ Base AD = (frac { Area }{ Altitude }) = (frac { 128 }{ 10 }) = 12.8cm

Question 2.
In Q. No. 1, if AD = 6 cm, CF = 10 cm, AE = 8 cm, find AB.
Solution:
Area of ||gm ABCD,
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
= Base x Altitude
= AD x CF
= 6 x 10 = 60 cm2
Again area of ||gm ABCD = 60 cm2
Altitude AE = 8 cm
∴ Base AB =(frac { Area }{ Altitude }) = (frac { 60 }{ 8 }) = (frac { 15 }{ 2 }) cm = 7.5 cm

Question 3.
Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution:
Area of ||gm ABCD = 124 cm2
E and F are the mid points of sides AB and CD respectively. E, F are joined.
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Draw DL ⊥ AB
Now area of ||gm ABCD = Base x Altitude
= AB x DL = 124 cm2
∵ E and F are mid points of sides AB and CD
∴ AEFD is a ||gm
Now area of ||gm AEFD = AE x DL
= (frac { 1 }{ 2 })AB x DL [∵ E is mid point of AB]
= (frac { 1 }{ 2 }) x area of ||gm ABCD
= (frac { 1 }{ 2 }) x 124 = 62 cm2

Question 4.
If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = (frac { 1 }{ 2 })ar( ||gm ABCD).
Solution:
Given : In ||gm ABCD, BD and AC are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 - 4
To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = (frac { 1 }{ 2 })ar(||gm ABCD)
Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in area When BD is the diagonal, then
∴ ar(∆ABD) = ar(∆BCD) = (frac { 1 }{ 2 })ar(||gm ABCD) …(i)
Similarly, when AC is the diagonal, then
ar(∆ABC) = ar(∆ADC) = (frac { 1 }{ 2 })ar(||gm ABCD) …(ii)
From (i) and (ii),
ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = (frac { 1 }{ 2 }) ar(||gm ABCD)

RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals Ex 14.3

Question 1.
In the figure, compute the area of the quadrilateral.
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
Solution:
In the quadrilateral ABCD,
∠A = 90°, ∠CBD = 90°, AD = 9 cm, BC = 8 cm and CD = 17 cm
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
In right ∆BCD,
CD = BC2 + BD2 (Pythagoras Theorem)
⇒ (17)2 = (8)2 + BD2
⇒ 289 = 64 + BD2
⇒ BD2 = 289 – 64 = 225 = (15)2
∴ BD = 15 cm
Now in right ∆ABD,
BD2 = AB2 + AD2
⇒ (15)2 = AB2 + (9)2
⇒ 225 = AB2 + 81
⇒ AB2= 225 – 81 = 144 = (12)2
∴ AB = 12 cm
RD Sharma Class 9 Book Chapter 14 Quadrilaterals
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions

Question 2.
In the figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ∆OTS if PQ = 8 cm.
RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals
Solution:
In square PQRS, T and U are the mid-points of the sides PS and QR
TU, QS and US are joined
PQ = 8 cm
∴ T and U are mid-points of the opposites sides PS and QR
∴ TU || PQ TO || PQ
In RQS,
T is mid-point of PS and TO || PQ
∴ O is the mid point of SQ 1 1
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals

Question 3.
Compute the area of trapezium PQRS in the figure.
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals
Solution:
In ∆TQR, ∠RTQ = 90°
∴ QR2 = TQ2 + RT2
⇒ (17)2 = (8)2 + RT2
⇒ 289 = 64 + RT2
⇒ RT2 = 289 – 64 = 225 = (15)2
∴ RT = 15 cm
and PQ = 8 + 8 = 16 cm
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals

Question 4.
In the figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ∆AOB.
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
Solution:
In ∆AOB, ∠AOB = 90°
C is a point on AB such that AC = BC Join OC
Maths RD Sharma Class 9 Chapter 14 Quadrilaterals
Since C is the mid-point of hypotenuse of right ∆AOB
∴ AC = CB = OC = 6.5 cm
∴ AB = 6.5 + 6.5 = 13 cm
Now in right ∆AOB
⇒ AB2 = AO2 + OB(Pythagoras Theorem)
⇒ (13)2 = (12)2 + OB2
⇒ 169 = 144 + OB2
⇒ OB2 = 169 – 144 = 25 = (5)2
∴ OB = 5 cm
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 5.
In the figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
In the trapezium ABCD,
AB = 7 cm
AL = BM = 4 cm
AD = BC = 5 cm
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals

Question 6.
In the figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.If OE = 2 (sqrt { 5 } ) , find the area of the rectangle.
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Solution:
Radius of the quadrant of circle = 2(sqrt { 5 } ) units
∴ OD diagonal of rectangle = 10 units (∵ OD = OB = OA = 10 cm)
DE = 2 (sqrt { 5 } ) cm
∴ In right ∆OED,
OD2 = OE2 + DEv
(10)2 = OE2 + (2(sqrt { 5 } ))2
100 = OE2 + 20
OE2 = 100 – 20 = 80
⇒ OE2 = (4(sqrt { 5 } ))2
∴ OE = 4(sqrt { 5 } ) cm
∴ Area of rectangle = lxb
= DE x OE
= 2(sqrt { 5 } ) x 4(sqrt { 5 } )
= 8 x 5 = 40 cm2

Question 7.
In the figure, ABCD is a trapezium in which AB || DC. Prove that ar( ∆AOD = ar(∆BOC).
Quadrilaterals Class 9 RD Sharma Solutions
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O
∴ ∆ADB and ∆ACB are on the same base AB and between the same parallels
∴ ar(∆ADB = ar(∆ACD)
Subtracting, ar(AAOB) from both sides,
ar(∆ADB) – ar(∆AOB) = ar(∆ACD) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)

Question 8.
In the figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(∆ADE) = ar(∆BCF) [NCERT]
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
Solution:
Given : In the figure, ABCD, ABEF and CDEF are ||gms
To prove : ar(∆ADE) = ar(∆BCF)
Proof: ∴ ABCD is a ||gm
∴ AD = BC
Similarly, in ||gm ABEF
AE = BF
and in ||gm CDEF,
DE = CF
Now, in ∆ADE and ∆BCF
AD = BC (proved)
DE = CF (proved)
AE = BF (proved)
∴ ∆ADE ≅ ∆BCF
∴ ar(∆ADE) = ar(∆BCF) (∵ Congruent triangles are equal in area)

Question 9.
In the figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(∆ABC) = ar(∆ABC)
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
Solution:
Given : In the figure, ∆ABC and ∆ABD are on the same base AB and line CD is bisected by AB at O i.e. CO = OD
To prove : ar(∆ABC) = ar(∆ABD)
Construction : Draw CL ⊥ AB and DM ⊥ AB
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals

Question 10.
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid point of median AD, prove that ar(∆BGC) = 2ar(∆AGC).
Solution:
Given : In ∆ABC, AD is its median. G is mid point of AD. BG and CF are joined
To prove :
(i) ar(∆ADB) = ar(∆ADC)
(ii) ar(∆BGC) = 2ar(∆AGC)
Construction : Draw AL ⊥ BC
RD Sharma Class 9 Book Chapter 14 Quadrilaterals
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
A point D is taken on the side BC of a AABC such that BD = 2DC. Prove that ar(∆ABD) = 2ar(∆ADC)
Solution:
Given : In ∆ABC, D is a point on BC such that
BD = 2DC
AD is joined
To prove : ar(∆ABD) = 2ar(∆ADC)
Construction : Draw AL ⊥ BC
RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals

Question 12.
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that
(i) ar(∆ADO) = or (∆CDO)
(ii) ar(∆ABP) = ar(∆CBP).
Solution:
Given : In ||gm ABCD, Diagonals AC and BD intersect each other at O
P is any point on BO
AP and CP are joined
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals
To prove :
(i) ar(∆ADO) = ar(∆CDO)
(ii) ar(∆ABP) = ar(∆CBP)
Proof:
(i) In ∆ADC,
O is the mid point of AC
∴ ar(∆ADO) = ar(∆CDO)
(ii) Since O is the mid point of AC
∴ PO is the median of ∆APC
∴ af(∆APO) = or(∆CPO) …(i)
Similarly, BO is the median of ∆ABC
∴ ar(∆ABO) = ar(∆BCO) …(ii)
Subtracting (i) from (ii),
ar(∆ABO) – ar(∆APO) = ar(∆BCO) – ar( ∆CPO)
⇒ ar(∆ABP) = ar(∆CBP)
Hence proved.

Question 13.
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(∆ADF) = ar(∆ECF)
(ii) If the area of ∆DFB = 3 cm2, find the area of ||gm ABCD.
Solution:
Given : In ||gm ABCD, BC is produced to E such that CE = BC
AE intersects CD at F
To prove :
(i) ar(∆ADF) = ar(∆ECF)
(ii) If ar(∆DFB) = 3 cm2, find the area of (||gm ABCD)
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals

Question 14.
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(∆POA) = ar(∆QOC).
Solution:
Given : In ||gm ABCD, diagonals AC and BD intersect at O
A line through O intersects AB at P and CD at Q
Maths RD Sharma Class 9 Chapter 14 Quadrilaterals
To prove : ar(∆POA) = ar(∆QOC)
Proof : In ∆POA and ∆QOC,
OA = OC (O is mid-point of AC)
∠AOD = ∠COQ (Vertically opposite angles)
∠APO = ∠CQO (Alternate angles)
∴ ar(∆POA) ≅ ar(∆QOC) (AAS criterian)
∴ ar(∆POA) = ar(∆QOC)

Question 15.
In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar(∆ABD) = ar(∆ADE) = ar(∆AEC). [NCERT]
RD Sharma Class 9 Chapter 14 Quadrilaterals
Solution:
Given : D and E are two points on BC such that BD = DE = EC
AD and AE are joined
To prove : ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Construction : From A, draw AL ⊥ BC and XAY || BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Proof: ∵ BD = DE = EC
and ∆ABD, ∆ADE and ∆AEC have equal bases and from the common vertex A
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)

Question 16.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that: ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)
Solution:
Given : In quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar(∆APB) x ar(∆CPD) = ar(APD) x ar(∆BPC)
Construction : Draw AL and CN perpendiculars on BD
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals

Question 17.
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
Given : In ||gm ABCD, P is any point in the ||gm
AP and BP are joined
To prove : ar(∆APB) < (frac { 1 }{ 2 }) ar(||gm ABCD)
Construction : Draw DN ⊥AB and PM ⊥ AM
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Quadrilaterals Class 9 RD Sharma Solutions

Question 18.
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is a point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Given : In ||gm ABCD, E is a point on AB such that BE = 2EA and F is a point on CD such that DF = 2FC. AE and CE are joined
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals

Question 19.
In a ∆ABC, P and Q are respectively, the mid-points of AB and BC and R is the mid-point of AP. Prove that
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
Solution:
Given : In ∆ABC,
P and Q are mid-pionts of AB and BC R is mid-point of AP, PQ, RC, RQ are joined
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
RD Sharma Class 9 Book Chapter 14 Quadrilaterals
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions
RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals

Question 20.
ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals
(v) Find what portion of the area of parallelogram is the area of AEFG.
Solution:
Given : ABCD is a parallelogram and AG = 2GB, CE = 2DE and BF = 2FC
To prove :
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals
(v) Find what portion of the area of parallelogram is the area of AFEG.
Construction : Draw EP ⊥ AB and EQ ⊥ BC
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
RD Sharma Class 9 Chapter 14 Quadrilaterals
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Quadrilaterals Class 9 RD Sharma Solutions

Question 21.
In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ACBX.
(ii) Prove that or(∆ZDE) = ar(∆CZA).
(iii) Prove that ar(∆CZY) = ar(∆EDZ).
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
Solution:
Given : In the figure,
CP || AE and CY || BA
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
To prove :
(i) Name a triangle equal in area of ∆CBX
(ii) Prove that ar(∆ZDE) = ar(∆CZA)
(iii) ar(BCZY) = ar(∆EDZ)
Proof:
(i) ∆CBX and ∆CYX are on the same base BY and between same parallels.
∴ ar(∆CBX) = ar(∆CYX)
(ii) ∆ADE and ∆ACE are on the same base AE
and between the same parallels (AE || CD)
∴ ar(∆ADE) = ar(∆ACE)
Subtracting ar(∆AZE) from both sides
⇒ ar(∆ADE) – ar(∆AZE) = ar(∆ACE) – ar(∆AZE)
⇒ ar(∆ZDE) = ar(∆ACZ)
⇒ ar∆ZDE = ar∆CZA
(iii) ∵ As ACY and BCY are on the same base CY and between the same parallels
∴ ar(∆ACY) = ar(∆BCY)
Now ar(∆ACZ) = ar(∆ZDE) (Proved)
⇒ ar(∆ACY) + ar(∆CYZ) = ar(∆EDZ)
⇒ ar(∆BCY) + ar(∆CYZ) = ar(∆EDZ)
∴ ar quad. (BCZY) = ar(EDZ)
Hence proved.

Question 22.
In the figure, PSD A is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar(∆PQE) = ar(∆CFD).
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
Solution:
Given : In the figure, PSDA is a ||gm in
which PQ = QR = RS and
AP || BQ || CR || DS
To prove : ar(∆PQE) = ar(∆CFD)
Construction : Join PD
Proof : ∵ PA || BQ || CR || DS
and PQ – QR = RS (Given)
∴ AB = BC = CD
∴ PQ = CD
Now in ABED, F is mid point of ED
∴ EF = FD
Similarly, EF = PE
⇒ PE = FD
In ∆PQE and ∆CFD,
∴ ∠EPQ = ∠FDC (Alternate angles)
PQ = CD
PE = FD (Proved)
∴ APQE ≅ ACFD (SAS cirterion)
∴ ar(∆PQE) = ar(∆CFD)

Question 23.
In the figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = (frac { 9 }{ 11 }) ar(trap. XYBA)
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
Solution:
Given : In the figure, ABCD is a trapezium in which AB || DC
DC = 40 cm, AB = 60 cm
X and Y are the mid-points of AD and BC respectively
To prove :
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = (frac { 9 }{ 11 }) m(trap. XYBA)
Construction : Join DY and produce it to meet AB produced at P
RD Sharma Class 9 Book Chapter 14 Quadrilaterals
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions
RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals

Question 24.
D is the mid-point of side BC of ∆ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(∆BOE) = (frac { 1 }{ 8 }) ar(∆ABC).
Solution:
Given : In ∆ABC, D is mid point of BC, E is mid point BE and O is the mid point of AE. BO, AE, AD are joined.
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals

Question 25.
In the figure, 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).
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals
Solution:
Given : In ∆ABC, X and Y are the mid pionts of AC and AB respectively. Through A, a line parallel to BC is drawn. Join BX and CY and produce them to meet the parallel line through A, at P and Q respectively and intersect each other at O.
To prove : ar(∆ABP) = ar(∆ACQ)
Construction : Join XY and produce it to both sides
Proof : ∵ X and Y are mid points of sides AC and AB
∴ XY || BC
Similarly, XY || PQ
∆BXY and ∆CXY are on the same base XY and between the same parallels
∴ ar(∆BXY) = ar(∆CXY) …(i)
Now, trap. XYAP and trap. XYAQ are on the same base XY and between the same parallels
∴ ar(XYAP) = ar(XYAQ) …(ii)
Adding (i) and (ii),
∴ ar(∆BXY) + ar(∆YAP)
= ar(CXY) + ar(XYAQ)
⇒ ar(∆ABP) = ar(∆ACQ)

Question 26.
In the figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD).
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
Solution:
Given : Two ||gm ABCD and ||gm AEFD are on the same base AD. EF is produced to meet CD at Q. Join AF and PD also
To prove :
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD)
Maths RD Sharma Class 9 Chapter 14 Quadrilaterals
Proof:
(i) In ∆AEP and DFQ,
AE = DF (Opposite sides of a ||gm)
∠AEP = ∠DFQ (Corresponding angles)
∠APE = ∠DQF (Corresponding angles)
∴ ∆AEP ≅ ∆DFQ (AAS axiom)
∴ PE = QF (c.p.c.t.)
(ii) and ar(∆AEP) = ar(∆DFQ) …(i)
(iii) ∵ ∆PFA and ∆PFD are on the same base PF and between the same parallels
∴ ar(∆PFA) = ar(∆PFD) …(ii)
From (i) and (ii),
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 27.
In the figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar(||gm DLOP) = ar(||gm BMOQ).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC. PQ || AB and LM || BC
To prove : ar(||gm DLOP) = ar(||gm BMOQ)
Proof : ∵ Since, a diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(∆ADC) = or(∆ABC)
⇒ ar(∆APO) + or(||gm DLOP) + ar(∆OLC)
= ar(∆AOM) + ar(||gm BMOQ) + ar( ∆OQC) …(i)
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively,
∴ ar(∆APO) = ar(∆AMO) …(ii)
And, ar(∆OLC) = ar(∆OQC) …(Hi)
Subtracting (ii) and (iii) from (i), we get ar(||gm DLOP) = ar(||gm BMOQ)

Question 28.
In a ∆ABC, if L and M are points on AB and AC respectively such that LM || BC.
Prove that:
(i) ar(∆LCM) = ar(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC).
Solution:
Given : In ∆ABC,
L and M are mid points on AB and AC
LM, LC and MB are joined
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
To prove :
(i) ar(∆LCM) = or(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC)
Proof: ∵ L and M are the mid points of AB and AC
∴ LM || BC
(i) Now ∆LBM and ∆LCM are on the same base LM and between the same parallels
∴ar(∆LBM) = ar(∆LCM) …(i)
⇒ ar(∆LCM) = ar(∆LBM)
(ii) ∵ ∆LBC and ∆MBC are on the same base
BC and between the same parallels
∴ ar(∆LBC) = ar(∆MBC) …(ii)
(iii) a(∆LMB) = ar(∆LMC) [From (i)]
⇒ ar(∆ALM) + ar(∆LMB)
= ar(∆ALM) + ar(∆LMC) [Adding or(∆ALM) to both sides]
⇒ ar(∆ABM) = ar(∆ACL)
(iv) ∵ ar(∆LBC) = ar(∆MBC) [From (ii)]
⇒ ar(∆LBC) – ar(∆BOC) = ar(∆MBC) – ar(∆BOC)
ar(∆LBO) = ar(∆MOC)

Question 29.
In the figure, ABC and BDC are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F.
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Solution:
Given : ABC and BDE are two equilateral triangles and D is mid point of BC. AE intersects BC in F
To prove :
Quadrilaterals Class 9 RD Sharma Solutions
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
RD Sharma Class 9 Book Chapter 14 Quadrilaterals

Question 30.
In the 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 [NCERT]
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(∆CFG)
(vii) ar(BCED) = ar(AMBN) + ar(ACFG)
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions
Solution:
Given : In ∆ABC, ∠A = 90°
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AX ⊥ DE meeting DE at X
To prove :
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = 2ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = or(AMBN) + ar(ACFG)
Construction : Join AD, AE, BF and CM
Proof:
(i) In ∆MBC and ∆ABD,
MB=AB (Sides of square)
BC = BD
∠MBC = ∠ABD (Each angle = 90° + ∠ABC)
∴ ∆MBC ≅ ∆ABD (SAS criterian)
∴ ar(∆MBC) = ar(∆ABD) …(i)
(ii) ∵ ∆ABD and rectangle BYXD are on the same base BD and between the same parallels
∴ ar(∆ABD) = (frac { 1 }{ 2 }) ar(rect. BYXD)
⇒ ar(rect. BYXD) = 2ar(∆ABD)
⇒ ar(rect. BYXD) = 2ar(∆MBC) …(ii)
(iii) Similarly, ∆MBC and square MBAN are on the same base MB and between the same parallels
∴ ar(∆MBC) = ar(sq. ABMN) …(iii)
From (ii) and (iii)
ar(sq. ∆BMN) = ar(rect. BYXD)
(iv) In AFCB and ∆ACE,
FC = AC
CB = CE (Sides of squares)
∠FCB = ∠ACE (Each = 90° + ∠ACB)
∴ ∆FCB = ∆ACE (SAS criterian)
(v) ∵ ∆FCB ≅ ∆ACE (Proved)
∴ ar(∆FCB) = ar(∆ACE)
∵∆ACE and rectangle CYXE are on the same base and between the same parallels
∴ 2ar(∆ACE) = ar(CYXC)
⇒ 2ar(∆FCB) = ar(CYXE) …(iv)
(vi) ∵ AFCB and rectangle FCAG are on the base FC and between the same parallels
∴ 2ar(∆FCB) = ar(FCAG) …(v)
From (iv) and (v)
ar(CMXE) = ar(ACFG)
(vii) In ∆ACB.
BC2 = AB2 + AC2 (By Pythagoras Theorem)
⇒ BC x BD = AB x MB + AC x FC
⇒ ar(BCED) = ar(ABMN) + ar(ACFG)
Hence proved.

RD Sharma Class 9 Book Chapter 14 Quadrilaterals VSAQS

Question 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(∆ABC) : ar(∆BDE).
Solution:
ABC and BDE are two equilateral triangles and D is the mid-point of BC
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals
Let each side of AABC = a
Then BD = (frac { a }{ 2 })
∴ Each side of triangle BDE will be (frac { a }{ 2 })
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals

Question 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals
Solution:
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
∴ Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm2
∵ DC || AB and AB is produced to F and DC is produced to G
∴ DG || AF
∵ Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
∴ ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm2

Question 3.
In the figure of Q. No. 2, find the area of ∆GEF.
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
Solution:
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
∵ ABCD is a rectangle
∴DC || AB,
DC is produced to E and AB is produced to G
∴DE || AG
∵ Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
∴ ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm2
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
∴ area ∆EFG = (frac { 1 }{ 2 }) ar(||gm ABEF)
= (frac { 1 }{ 2 }) x 50 = 25 cm2

Question 5.
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 find or(∆RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Radius of the circle = 13 cm
A is any point on PQ
AR and AS are joined, PS = 5 cm
In right ∆PRS,
PR2 = PS2 + SR2
⇒ (132 = (5)2+ SR2
⇒ 169 = 25 + SR2
⇒ SR2 = 169 – 25 = 144 = (12)2
∴ SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm2
∵ Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
∴ Area ARAS = (frac { 1 }{ 2 }) area rect. PQRS 1
= (frac { 1 }{ 2 }) x 60 = 30 cm2

Question 6.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB.
Solution:
In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ and BD are joined which intersect each other at O
Side of square AB = 8 cm
Quadrilaterals Class 9 RD Sharma Solutions
∴ Area of square ABCD = (side)2
∵ Diagonal BD bisects the square into two triangle equal in area
∴ Area ∆ABD = (frac { 1 }{ 2 }) x area of square ABCD
= (frac { 1 }{ 2 }) x 64 = 32 cm2
∵ P is mid point of AB of AABD, and PQ || AD
∴ O is the mid point of BD
∴ OP = (frac { 1 }{ 2 })AD = (frac { 1 }{ 2 }) x 8 = 4 cm
and PB = (frac { 1 }{ 2 }) AB = (frac { 1 }{ 2 }) x 8 = 4 cm
∴ Area ∆OPB = (frac { 1 }{ 2 })PB x OP
= (frac { 1 }{ 2 }) x4x4 = 8 cm2

Question 7.
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm2, find the area of ∆DEF.
Solution:
In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm2
FD is joined
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
∵ D is mid point of BC
∴ AD is the median and median divides the triangle into two triangles equal in area
area ∆ADC = (frac { 1 }{ 2 }) ar(∆ABC)
= (frac { 1 }{ 2 }) x 16 = 8 cm2
Similarly, E is mid point of DC
∴ area (∆ADE) = (frac { 1 }{ 2 }) ar(∆ADC)
= (frac { 1 }{ 2 }) x 8 = 4 cm2
∵ F is mid point of AE of ∆ADE
∴ ar(∆DEF) = (frac { 1 }{ 2 })area (∆ADE)
= (frac { 1 }{ 2 }) x 4 = 2 cm2

Question 8.
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ∆PBQ is 17 cm2, find the area of ∆ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of ∆PBQ = 17 cm2
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
∆ABQ and ∆ABR are on the same base AB and between the same parallels
∴ ar(∆ABQ) = ar(∆ABR) …(i)
Similarly, ∆ABP and ∆ABS are on the same base and between the same parallels
∴ ar(ABP) = ar(∆ABS) …(ii)
Adding (i) and (ii)
ar( ∆ABQ) + ar( ∆ABP)
= ar(∆ABR) + ar(∆ABS)
⇒ ar(∆PBQ) = ar(∆ASR)
Put ar(PBQ) = 17 cm2
∴ ar(∆ASR) = 17 cm2

Question 9.
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3 : 1. If ar(∆PBQ) = 10 cm2, find the area of parallelogram ABCD.
Solution:
In ||gm ABCD, P is mid point on AB,
PC and BD intersect each other at Q
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
CQ : QP = 3 : 1
ar(∆PBQ) = 10 cm2
In ||gm ABCD,
BD is its diagonal
∴ ar(∆ABD) = ar(∆BCD) = (frac { 1 }{ 2 }) ar ||gm ABCD
∴ ar(||gm ABCD) = 2ar(∆ABD) …(i)
In ∆PBC CQ : QP = 3 : 1
∵ ∆PBQ and ∆CQB have same vertice B
∴ 3 x area ∆PBQ = ar(∆CBQ)
⇒ area(∆CBQ) = 3 x 10 = 30 cm2
∴ ar(∆PBC) = 30 + 10 = 40 cm2
Now ∆ABD and ∆PBC are between the
same parallel but base PB = (frac { 1 }{ 2 }) AB
∴ ar(∆ABD) = 2ar(∆PBC)
= 2 x 40 = 80 cm2
But ar(||gm ABCD) = 2ar(∆ABD)
= 2 x 80 = 160 cm2

Question 10.
P is any point on base BC of ∆ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(∆ABC) = 12 cm2, then find area of ∆EPC.
Solution:
P is any point on base of ∆ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(∆ABC) = 12 cm2
AD and PE are joined
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
∵ D is mid point of BC
∴ AD is median
∴ ar(∆ABD) = ar(∆ACD)
= (frac { 1 }{ 2 }) (∆ABC) = (frac { 1 }{ 2 }) x 12 = 6 cm2 …(i)
∵ ∆PED and ∆ADE are on the same base DE and between the same parallels
∴ ar(∆PED) = ar(∆ADE)
Adding ar(∆DCE) to both sides,
ar(∆PED) + ar(∆DCE) = ar(∆ADE) + ar(∆DCE)
ar(∆EPC) = ar(∆ACD)
⇒ ar(∆EPC) = ar(∆ABD) = 6 cm2 [From (i)]
∴ ar(∆EPC) = 6 cm2

RD Sharma Class 9 Solution Chapter 14 Quadrilaterals MCQS

Question 1.
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Solution:
Two parallelograms which are on the same base and between the same parallels are equal in area
∴ Ratio in their areas =1 : 1 (c)

Question 2.
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Solution:
A triangle and a parallelogram which are on the same base and between the same parallels, then area of triangle is half the area of the parallelogram
∴ Their ratio =1:2 (c)

Question 3.
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ∆ABC. Then the area of ∆PQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units
Solution:
Area of ∆ABC = 24 sq. units
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 4.
The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Solution:
The median of a triangle divides it into two triangles equal in area (d)

Question 5.
In a ∆ABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If
ar(∆ABC) = 16 cm2, then ar(trapezium FBCE) =
(a) 4 cm²
(b) 8 cm²
(c) 12 cm²
(d) 10 cm²
Solution:
In ∆ABC, D, E and F are the mid points of sides BC, CA and AB respectively
ar(∆ABC) = 16 cm²
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 6.
ABCD is a parallelogram. P is any point on CD. If ar(∆DPA) = 15 cm² and ar(∆APC) = 20 cm², then ar(∆APB) =
(a) 15 cm²
(b) 20 cm²
(c) 35 cm²
(d) 30 cm²
Solution:
In ||gm ABCD, P is any point on CD
AP, AC and PB are joined
ar(∆DPA) =15 cm²
ar(∆APC) = 20 cm²
Adding, ar(∆ADC) = 15 + 20 = 35 cm²
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
∵ AC divides it into two triangles equal in area
∴ ar(∆ACB) = ar(∆ADC) = 35 cm²
∵ ∆APB and ∆ACB are on the same base
AB and between the same parallels
∴ ar(∆APB) = ar(∆ACB) = 35 cm²(c)

Question 7.
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
(a) 28 cm²
(b) 48 cm²
(c) 96 cm²
(d) 24 cm²
Solution:
In rhombus ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively and are joined in order to get a quad. PQRS
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals

Question 8.
A, B, C, D are mid points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm²,then ar(ABCD) =
(a) 24 cm²
(b) 18 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
A, B, C and D are the mid points of a ||gm PQRS
Area of PQRS = 36 cm²
Quadrilaterals Class 9 RD Sharma Solutions
The area of ||gm formed by joining AB, BC, CD and DA
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals

Question 9.
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 rhombus of area 24 cm²
(b) a rectangle of area 24 cm²
(c) a square of area 26 cm²
(d) a trapezium of area 14 cm²
Solution:
Let P, Q, R, S be the mid points of sides of a rectangle ABCD. Whose sides 8 cm and 6 cm
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
Their PQRS is a rhombus
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions

Question 10.
If AD is median of ∆ABC and P is a point on AC such that ar(∆ADP) : ar(∆ABD) = 2:3, then ar(∆PDC) : ar(∆ABC) is
(a) 1 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Solution:
AD is the median of ∆ABC,
P is a point on AC such that
ar(∆ADP) : ar(∆ABD) = 2:3
Let area of ∆ADP = 2×2
Then area of ∆ABD = 3×2
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
But area of AABD = (frac { 1 }{ 2 }) area AABC
∴ Area ∆ABC = 2 x area of ∆ABD
= 2 x 3x² = 6x²
and area of ∆PDC = area ∆ADC – (ar∆ADP) = area ∆ABD – ar ∆ADP
= 3x² – 2x² = x²
∴ Ratio = x² : 6x²
= 1 : 6 (c)

Question 11.
Medians of AABC, intersect at G. If ar(∆ABC) = 27 cm2, then ar(∆BGC) =
(a) 6 cm2
(b) 9 cm2
(c) 12 cm2
(d) 18 cm2
Solution:
In ∆ABC, AD, BE and CF are the medians which intersect each other at G
RD Sharma Class 9 Book Chapter 14 Quadrilaterals

Question 12.
In a ∆ABC if D and E are mid-points of BC and AD respectively such that ar(∆AEC) = 4 cm², then ar(∆BEC) =
(a) 4 cm²
(b) 6 cm²
(c) 8 cm²
(d) 12 cm²
Solution:
In ∆ABC, D and E are the mid points of BC and AD
Join BE and CE ar(∆AEC) = 4 cm²
Quadrilaterals With Solutions PDF RD Sharma Class 9 Solutions
In ∆ABC,
∵ AD is the median of BC
∴ ar(∆ABD) = ar(∆ACD)
Similarly in ∆EBC,
ED is the median
∴ ar(∆EBD) = ar(∆ECD)
and in ∆ADC, CE is the median
∴ ar(∆FDC) = ar(∆AEC)
= 4 cm
∴ar∆EBC = 2 x ar(∆EDC)
= 2 x 4 = 8 cm (c)

Question 13.
In the figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
(a) 3 cm
(b) 6 cm
(c) 8 cm
(d) 10.5 cm
RD Sharma Class 9 Maths Book Questions Chapter 14 Quadrilaterals
Solution:
In ||gm ABCD, AB = 12 cm AE = 7.5 cm
∴ Area of ||gm ABCD = base x height = AB x AE = 12 x 7.5 cm² = 90 cm²
Now area ||gm ABCD = 90 cm²
and altitude CF = 15 cm
∴ Base AD = (frac { Area }{ Altitude }) = (frac { 90 }{ 15 }) = 6 cm (b)

Question 14.
In the figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||gm XQRY) : ar(∆QSR) =
(a) 1 : 4
(b) 2 : 1
(c) 1 : 2
(d) 1 : 1
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
In ||gm PQRS, X and Y are the mid points of PQ and SR respectively XY and SQ are joined.
∵ XY bisects PQ and SR
∴ PXYS and XQRY are also ||gms and ar(∆PXYS) = nr(∆XQRY)
∵ ||gm PQRS and AQSR are on the same base and between the same parallel lines
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals

Question 15.
Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ∆AOD is
(a) ∆AOB
(b) ∆BOC
(c) ∆DOC
(d) ∆ADC
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O. AB || DC
∆ABC and ∆ABD are on the same base and between the same parallels
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals
∴ ar(∆ABC) = or(∆ABD)
Subtracting ar(∆AOB)
ar(∆ABC) – ar(∆AOB) = ar(∆ADB) – ar(∆AOB)
⇒ ar(∆BOC) = ar(∆AOD)
ar(∆AOD) = ar(∆BOC) (c)

Question 16.
ABCD is a trapezium in which AB || DC. If ar(∆ABD) = 24 cm² and AB = 8 cm, then height of ∆ABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Solution:
In trapezium ABCD, AB || DC
AC and BD are joined
ar(∆ABD) = 24 cm2
AB = 8 cm,
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals

Question 17.
ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)
Solution:
In quadrilateral ABCD, E and F are the mid points of AD and BC
AB = a, CD = b
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
Let h be the height of trapezium ABCD then height of each quadrilateral
ABFE = altitude of quadrilateral EFCD = (frac { h }{ 2 })
Now area of trap. ABFE = (frac { 1 }{ 2 }) (sum of parallel sides) x altitude
Maths RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 18.
ABCD is a rectangle with O as any point in its interior. If or(∆AOD) = 3 cm2 and ar(∆BOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2
Solution:
In rectangle ABCD, O is any point
ar(∆AOD) = 3 cm2
and ar(∆BOC) = 6 cm2
Join OA, OB, OC and OD
RD Sharma Class 9 Chapter 14 Quadrilaterals
We know that if O is any point in ABCD Then ar(AOB) + ar(COD) = ar(AOB) + ar(BOC)
= 3 + 6 = 9 cm
∴ ar(rect. ABCD) = 2 x 9 = 18 cm (d)

Question 19.
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
P,Q and R the mid points of the sides of a ∆ABC then area of any parallelogram formed by the mid points and one vertex of the given triangle has area = (frac { 1 }{ 2 }) area ∆ABC (b)

Question 20.
In the figure, ABCD and FECG are parallelograms equal in area. If ar(∆AQE) = 12 cm2, then ar(||gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Solution:
In the figure, ABCD and EFCG are parallelograms equal in area and ar(∆AQE) = 12 cm2
In ||gm AQED, AE is the diagonal
∴ ar(∆AQE) = (frac { 1 }{ 2 }) ar(||gm AQED)
⇒ 12 cm2 = (frac { 1 }{ 2 }) ar(||gm AQED)
∴ ar(||gm AQED) = 24 cm2
∵ ar ||gm ABCD = ar ||gm FECG
⇒ ar(||gm ∆QED) + ar(|| gm QBCE)
= ar(||gm QBCE) + ar(||gm FGBQ)
⇒ ar(||gm ∆QED) = ar(||gm FGBQ)
= 24 cm2 (c)

All Chapter RD Sharma Solutions For Class 9 Maths

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