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#### Page No 387:

(i) No, it doesnt lie on the same base and between the same parallels.
(ii) No, it doesnt lie on the same base and between the same parallels.
(iii) Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
(iv) No, it doesnt lie on the same base and between the same parallels.
(v) Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
(vi) Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.

#### Page No 387:

Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB   (extend DC and AB). Join AC, the other diagonal of ABCD.

Proof: ar(quad. ABCD) = ar(∆ABD) + ar(​∆DCB)
= 2 ar(​∆ABD)                    [∵ ar​(∆ABD) = ar(​∆DCB)]
∴ ar(​∆ABD) = $\frac{1}{2}$ar(quad. ABCD)                 ...(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(​∆CDA)
= 2 ar(​∆ ABC)                    [∵ ar​(∆ABC) = ar(​∆CDA)]
∴ ar(​∆ABC) = $\frac{1}{2}$ar(quad. ABCD)                ...(ii)
From (i) and (ii), we have:
ar(​∆ABD) = ar(​∆ABC) = $\frac{1}{2}$ AB ⨯ BD = $\frac{1}{2}$ AB ⨯ CL
⇒ CL = BD
⇒ DC |​​| AB
Hence, ABCD is a paralleogram.
∴ ar(​|​| gm ABCD) = base ​⨯ height = 5 ​⨯ 7 = 35 cm2

#### Page No 387:

ar(parallelogram ABCD) = base ​⨯ height
AB ​⨯DL = AD ​⨯ BM
⇒ 10 ​​⨯ 6 = AD ​⨯ BM
⇒ AD ​⨯ 8 = 60 cm2

#### Page No 388:

Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]

Again, in ∆DACthe points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]

#### Page No 388:

ar(trapezium) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
= $\frac{1}{2}$ ⨯ (9 + 6) ⨯ 8
= 60 cm2
Hence, the area of the trapezium is 60 cm2.

#### Page No 388:

(i) In $△$BCD,

Ar($△$BCD) =
In $△$BAD,

Ar($△$DAB) =

Area of quad. ABCD = Ar($△$DAB) + Ar($△$BCD) = 54 + 60 = 114 cm.

(ii) Area of trap(PQRS) =

#### Page No 388:

ADL is a right angle triangle.
So, DL =
Similarly, in ∆BMC, we have:
MC
DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = $\frac{1}{2}$⨯ (sum of parallel sides) ⨯ (distance between them)
=$\frac{1}{2}$ ⨯ (7 + 13) ⨯ 4
= 40 cm2
​Hence, DC = 13 cm and area of trapezium = 40 cm2

#### Page No 388:

ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)
ar(∆ABD) = $\frac{1}{2}$⨯ base ⨯ height = $\frac{1}{2}$BD ⨯ AL             ...(i)
ar(∆DBC) = $\frac{1}{2}$BD ⨯ CL                  ...(ii)
From (i) and (ii), we get:
​ar(quad ABCD) = $\frac{1}{2}$ ⨯BD ⨯​ AL + $\frac{1}{2}$ ⨯ BD ⨯​ CL
$⇒$​ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ ​(AL + CL)
Hence, proved.

#### Page No 388:

Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$
M is the midpoint of AB. So, CM is the median.
CM divides $△$ABC in two triangles with equal area.
$\mathrm{ar}\left(△\mathrm{AMC}\right)=\mathrm{ar}\left(△\mathrm{BMC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
ar(AMCD) = ar($△$ACD) + ar($△$AMC) = ar($△$ABC) + ar($△$AMC) = ​ar($△$ABC) + $\frac{1}{2}$​ar($△$ABC)

#### Page No 388:

​ar(quad ABCD) = ar($△$ABD) + ar($△$BDC)
= $\frac{1}{2}$ ⨯BD ⨯​ AL  +$\frac{1}{2}$ ⨯BD ⨯​ CM
$\frac{1}{2}$ ⨯BD ⨯​ ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = ​$\frac{1}{2}$ ⨯ 14 ⨯ ( 8 + 6)
= 7 ​⨯14
= 98 cm2

#### Page No 388:

We know
ar(∆APB) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(1)                     [If a triangle 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]
Similarly,
ar(∆BQC) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved

#### Page No 389:

(i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = are(ABPQ)                   (Same base PQ and MB || PQ)                      .....(1)

(ii) If a parallelogram and a triangle 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.
So, ar(∆ATQ) = $\frac{1}{2}$ar(ABPQ)                 (Same base AQ and AQ || BP)                       .....(2)
From (1) and (2)
ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ)

#### Page No 389:

CDA and ​∆CBD lies on the same base and between the same parallel lines.
So, ar(​∆CDA) = ar(CDB)            ...(i)
Subtracting ar(​∆OCD) from both sides of equation (i), we get:
ar(​∆CDA) $-$ ar(​∆OCD) = ar(​​∆CDB) $-$ ar (​​∆OCD)
⇒ ar(​​∆AOD) = ar(​​∆BOC)

#### Page No 389:

DEC and ​∆DEB lies on the same base and between the same parallel lines.
So, ar(​∆DEC) = ar(∆DEB)                      ...(1)

(i) On adding​ ar(∆ADE)​ in both sides of equation (1), we get:
⇒ ar(​​∆ACD) = ar(​​∆ABE

(ii) On subtracting​ ar(ODE)​ from both sides of equation (1), we get:​
ar(​∆DEC) $-$ ar(∆ODE)​ = ar(∆DEB) $-$ ar(∆ODE)​ ​      ​
⇒ ar(​​∆OCE) = ar(​∆OBD)

#### Page No 389:

Let AD is a median of
ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ADC.
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with $\frac{1}{2}$AL on both sides, we get:
$\frac{1}{2}$ × BD × AL$\frac{1}{2}$ × DC × AL

#### Page No 389:

Let ABCD be a parallelogram and BD be its diagonal.

To prove: ar(∆ABD) = ar(∆CDB)

Proof:
In ∆ABD and ∆CDB, we have:
AB = CD                    [Opposite sides of a parallelogram]
AD = CB                   [Opposite sides of a parallelogram]​

BD  = DB                  [Common]
i.e., ∆ABD  CDB           [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)

#### Page No 389:

Line segment CD is bisected by AB at O                   (Given)
CO = OD                                .....(1)
In ΔCAO,
AO is the median.                (From (1))
So, arΔCAO = arΔDAO          .....(2)
Similarly,
In ΔCBD,
BO is the median                 (From (1))
So, arΔCBO = arΔDBO          .....(3)
From (2) and (3) we have
arΔCAO + arΔCBO = arΔDBO + arΔDAO
$⇒$ar(ΔABC) = ar(ΔABD)

#### Page No 389:

ar(∆BCD) = ar(∆BCE)                     (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC.

#### Page No 389:

Join BD.
Let BD and AC intersect at point O.
O is thus the midpoint of DB and AC.
PO is the median of $△$DPB,
So,

Case II:

$\mathrm{ar}\left(△\mathrm{ADO}\right)+\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)+\mathrm{ar}\left(△\mathrm{BPO}\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 389:

Given:  BO = OD
Proof
Since BO = OD, O is the mid point of BD.
We know that a median of a triangle divides it into two triangles of equal areas.
CO is a median of ∆BCD.
i.e., ar(∆COD) = ar (∆COB)            ...(i)

AO is a median of ∆ABD.
i.e., ar(∆AOD) = ar(∆AOB)              ...(ii)

From (i) and (ii), we have:
ar(∆COD) + ar(∆AOD) ar(∆COB) + ar(∆AOB)

#### Page No 389:

Given:  D is the midpoint of BC and E is the midpoint of AD.
To prove: $\mathrm{ar}\left(∆BEC\right)=\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = $\frac{1}{2}$ar(∆ABD)                 ...(i)
Also, ar(∆CDE ) =$\frac{1}{2}$ ar(∆ADC)             ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ $\frac{1}{2}$ ⨯​ ar(∆ABD)​ + $\frac{1}{2}$ ⨯​ ar(∆ADC)
⇒ ar(∆BEC )​ = $\frac{1}{2}$⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ​ar(∆BEC )​ =​ $\frac{1}{2}$ ⨯​ ar(∆ABC)

#### Page No 390:

D is the midpoint of side BC of ∆ABC.
$⇒$AD is the median of ∆ABC.
$⇒$$\mathrm{ar}\left(△\mathrm{ABD}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
E is the midpoint of BD of ∆ABD,
$⇒$AE is the median of ∆ABD
$⇒$$\mathrm{ar}\left(△\mathrm{ABE}\right)=\mathrm{ar}\left(△\mathrm{AED}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Also, O is the midpoint of AE,
$⇒$BO is the median of ∆ABE,
$⇒$$\mathrm{ar}\left(△\mathrm{ABO}\right)=\mathrm{ar}\left(△\mathrm{BOE}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABE}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{8}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Thus, ar(∆BOE) = $\frac{1}{8}$ar(∆ABC)

#### Page No 390:

In $△$MQC and $△$MPB,
MC = MB                            (M is the midpoint of BC)
$\angle$CMQ = $\angle$BMP                (Vertically opposite angles)
$\angle$MCQ = $\angle$MBP                (Alternate interior angles on the parallel lines AB and DQ)
Thus, $△$MQC $\cong$ $△$MPB   (ASA congruency)
$⇒$ar($△$MQC) = ar($△$MPB)
$⇒$ar($△$MQC) + ar(APMCD) = ar($△$MPB) + ar(APMCD)
$⇒$ar(APQD) = ar(ABCD)

#### Page No 390:

We have:
​ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar(∆ACP)​​ + ar(∆ABC)

ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ ACP)​
By adding ar(∆ABC) on both sides, we get:
ar(∆ACD) ar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)
⇒​ ar (quad. ABCD) = ar(∆ABP)
Hence, proved.

#### Page No 390:

GivenABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMO, we have:
AL = DM
ALO = ∠DMO =  90o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
​​
∴​ OA = OD

#### Page No 390:

In $△$MDA and $△$MCP,
$\angle$DMA = $\angle$CMP                  (Vertically opposite angles)
$\angle$MDA = $\angle$MCP                  (Alternate interior angles)
So, $△$MDA $\cong$ $△$MCP         (ASA congruency)
$⇒$DM = MC                         (CPCT)
$⇒$M is the midpoint of DC
$⇒$BM is the median of $△$BDC
$⇒$ar($△$BMC) = ar($△$DMB) = 7 cm2
ar($△$BMC) + ar($△$DMB) = ar($△$DBC) = 7 + 7 = 14 ${\mathrm{cm}}^{2}$
Area of parallelogram ABCD = $×$ ar($△$DBC) = 2 $×$ 14 = 28 ${\mathrm{cm}}^{2}$

#### Page No 390:

Join BM and AC.
ar(∆ADC) = $\frac{1}{2}bh$ = $\frac{1}{2}×\mathrm{DC}×h$
ar(∆ABM) = $\frac{1}{2}×\mathrm{AB}×h$
AB = DC                               (Since ABCD is a parallelogram)
$⇒$ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
$⇒$ar(∆ADM) = ar(ABMC)
Hence Proved

#### Page No 390:

Given:  ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof:
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC              [ By midpoint theorem]
Again, in
DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = $\frac{1}{2}$ × AC                                   [ By midpoint theorem]
Now, PQ
|| AC and SR || AC
​PQ || SR
Also, PQ = SR =
$\frac{1}{2}$ × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram​ PQRS) = ar(∆PSQ) + ar(∆SRQ)                       ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD)            ...(ii)

PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) =$\frac{1}{2}$ × ar(parallelogram ABQS)                         ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ × ar(parallelogram SQCD)                        ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABQS)​ + $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABCD)

#### Page No 390:

Figure
CF is median of $△$ABC.
$⇒$ar($△$BCF) = $\frac{1}{2}$($△$ABC)                     .....(1)
Similarly, BE is the median of $△$ABC,
$⇒$ar($△$ABE) = $\frac{1}{2}$($△$ABC)                     .....(2)
From (1) and (2) we have
ar($△$BCF) = ar($△$ABE)
$⇒$ar($△$BCF) $-$ ar($△$BFG) = ar($△$ABE) $-$ ar($△$BFG)
$⇒$ar(∆BCG) = ar(AFGE)

#### Page No 391:

Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$DC
To prove:  ar(∆ABD) = $\frac{1}{3}$ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof:
In ∆ABC, we have:
BC = BD + DC

⇒​ BD​ + 2 BD = 3 × BD
Now, we have:
ar(∆ABD)​ = $\frac{1}{2}$​ ×​ BD ×​ AL
ar(∆ABC)​ = $\frac{1}{2}$​ ×​ BC ×​ AL
⇒  ar(∆ABC) = $\frac{1}{2}$ ×​ 3BD ×​ AL = 3 ×​ $\left(\frac{1}{2}×BD×AL\right)$
⇒ ar(∆ABC)​ = 3 × ar(∆ABD)
∴ ​ar(∆ABD) = ​$\frac{1}{3}$​ar(∆ABC)

#### Page No 391:

E is the midpoint of CA.
So, AE = EC                            .....(1)
Also, BD = $\frac{1}{2}$ CA                    (Given)
So, BD = AE                            .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of $△$ABC
so, ar($△$BCE) = ar($△$ABE) = $\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$                  .....(1)
ar($△$DBC) = ar($△$BCE)                  .....(2)                  (Triangles on the same base and between the same parallels are equal in area)
From (1) and (2)
ar(∆ABC) = 2ar(∆DBC)

#### Page No 391:

Given:  ABCDE is a pentagon.  EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) =  ar( DGF)
Proof:
ar(pentagon ABCDE )​ = ar(∆DBC) + ar(∆ADE ) + ar(∆ABD)               ...(i)
Also, ar(DGF) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD )                ...(ii)

Now, ∆DBC and ∆DBF lie on the same base and between the same parallel lines.
∴ ar(∆DBC) = ar(∆DBF)                         ...(iii)
Similarly, ∆ADE and ∆ADG lie on same base and between the same parallel lines.

From (iii) and (iv), we have:
Adding ar(∆ABD) on both sides, we get:
ar(∆DBC) + ar(∆ADE) + ar(∆ABD) = ar (∆DBF) + ar(∆ADG) + ar(∆ABD
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) =  ar(DGF)

#### Page No 391:

$\mathrm{ar}\left(△\mathrm{CFA}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)$                            (Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
$⇒\mathrm{ar}\left(△\mathrm{CFA}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(△\mathrm{AFG}\right)=\mathrm{ar}\left(△\mathrm{CBG}\right)$
Hence Proved

#### Page No 391:

Given: D is a point on BC of ∆ ABC, such that BD : DC =  m : n
To prove:  ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(∆ABD)​ = $\frac{1}{2}$ ×​ BD ×​ AL                     ...(i)
ar(∆ADC)​ = $\frac{1}{2}$​ ×​ DC ×​ AL                   ...(ii)
Dividing (i) by (ii), we get:
$\frac{\mathrm{ar}\left(△ABD\right)}{\mathrm{ar}\left(∆ADC}=\frac{\frac{1}{2}×BD×AL}{\frac{1}{2}×DC×AL}\phantom{\rule{0ex}{0ex}}=\frac{BD}{DC}\phantom{\rule{0ex}{0ex}}=\frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = mn

#### Page No 391:

Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.
Let length DQ = h
Given, M and N are the midpoints of AD and BC respectively.
So, MN || AB || DC and MN = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
M is the mid point of AD and MP || AB so by converse of mid point theorem,
MP || AQ and P will be the mid point of DQ.
$\mathrm{ar}\left(\mathrm{DCNM}\right)=\frac{1}{2}×\mathrm{DP}\left(\mathrm{DC}+\mathrm{MN}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right)\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\mathrm{MNBA}\right)=\frac{1}{2}×\mathrm{PQ}\left(\mathrm{AB}+\mathrm{MN}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right)\phantom{\rule{0ex}{0ex}}$
ar(DCNM) : ar(MNBA) = (a + 3b) : (3a + b)

#### Page No 391:

Construction: Draw a perpendicular from point D to the opposite side CD, meeting CD at Q and EF at P.
Let length AQ = h
Given, E and F are the midpoints of AD and BC respectively.
So, EF || AB || DC and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
E is the mid point of AD and EP || AB so by converse of mid point theorem,
EP || DQ and P will be the mid point of AQ.
$\mathrm{ar}\left(\mathrm{ABFE}\right)=\frac{1}{2}×\mathrm{AP}\left(\mathrm{AB}+\mathrm{EF}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right)\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\mathrm{EFCD}\right)=\frac{1}{2}×\mathrm{PQ}\left(\mathrm{CD}+\mathrm{EF}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right)\phantom{\rule{0ex}{0ex}}$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{\mathrm{ar}\left(\mathrm{ABEF}\right)}{\mathrm{ar}\left(\mathrm{EFCD}\right)}=\frac{24+3×16}{16+3×24}=\frac{9}{11}$

#### Page No 391:

In $△$PAC,
PA || DE and E is the midpoint of AC
So, D is the midpoint of PC by converse of midpoint theorem.
Also, $\mathrm{DE}=\frac{1}{2}\mathrm{PA}$                         .....(1)
Similarly, $\mathrm{DE}=\frac{1}{2}\mathrm{AQ}$                  .....(2)
From (1) and (2) we have
PA = AQ
∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC
ar(∆ABQ) = ar(∆ACP)

#### Page No 392:

In ∆RSC and ∆PQB
$\angle$CRS = $\angle$BPQ                       (CD || AB) so, corresponding angles are equal)
$\angle$CSR = $\angle$BQP                        ( SC || QB so, corresponding angles are equal)
SC = QB                                    (BQSC is a parallelogram)
So, ∆RSC $\cong$ ∆PQB                   (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)

#### Page No 395:

In this figure, both the triangles are on the same base (QR) but not on the same parallels.

#### Page No 396:

In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ​​​Parallelogram ABPQ

#### Page No 396:

(a)  triangles of equal areas

#### Page No 396:

(c)114 cm2

ar (quad. ABCD) =  ar (∆ ABC)  +  ar (∆ ACD)
In right angle triangle ACD, we have:
AC  =
In right angle triangle ABC, we have:
BC
Now, we have the following:
ar(∆ABC) = $\frac{1}{2}$ × 12 × 9 = 54 cm2
ar(∆ADC) = $\frac{1}{2}$ × 15 × 8 = 60 cm2
ar(quad. ABCD) =​ 54 + 60 = 114 cm2

#### Page No 396:

(c)124 cm2

In the right angle triangle BEC, we have:
EC  =
ar(trapez. ABCD) = cm2

#### Page No 396:

(c) 34 cm2

ar(parallelogram ABCD) = base × height = 5 ​× 6.8 =  34 cm2

#### Page No 396:

(d) 13 cm2
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of ∆OAB$\frac{1}{4}$ ⨯ ar(||gm ABCD)
⇒ ar(∆OAB) = ​​$\frac{1}{4}$ ⨯​ 52 = 13 cm2

#### Page No 396:

(a) 40 cm2
ar(||gm ABCD) = base × height =  10 ​× 4 =  40 cm2

#### Page No 397:

Area of a parallelogram is base into height.
Height = DL = NB
Base = AB = CD
So, area of parallelogram ABCD = DC $×$ DL
Hence, the correct answer is option (c).

#### Page No 397:

Parallelograms on the same base and between the same parallels are equal in area.
So, the ratio of their areas will be 1 : 1.
Hence, the correct answer is option (b).

#### Page No 397:

We know parallelogram on the same base and between the same parallels are equal in area.
Here, AB is the common base and AB || PD
Hence, ar(ABCD) = ar(ABPQ)                             .....(1)
Also, when a triangle and a parallelogram are on the same base and between the same parallels then the
area of triangle is half the area of the parallelogram.
Here, for the ∆BMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABPQ)                      .....(2)
From (1) and (2) we have
ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABCD)
Thus, the given statement is true.
Hence, the correct answer is option (a).

#### Page No 397:

D, E and F are the midpoints of sides BC, AC and AB respectively.
On joining FE, we divide $△$ABC into 4 triangles of equal area.
Also, median of a triangle divides it into two triangles with equal area
$\mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{ar}\left(△\mathrm{AFE}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Hence, the correct answer is option (a).

#### Page No 397:

(b) 96 cm2
Area of the rhombus = $\frac{1}{2}$ × product of diagonals = $\frac{1}{2}$ ×​ 12 ​× 16 = 96 cm2

#### Page No 397:

(c) 65 cm2
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
$\frac{1}{2}$ ×​ ( 12 + 8) ​× 6.5
= 65 cm2

#### Page No 397:

(b) 40 cm2

In right angled triangle MBC, we have:
MC
In right angled triangle ADL, we have:
DL

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
$\frac{1}{2}$ ×​ ( 13 + 7) ​× 4
= 40 cm2

#### Page No 397:

(b)128 cm2

ar(quad ABCD) = ar (∆ ABD) + ​ar (∆ DBC)

We have the following:
ar(∆ABD) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$× 16 ​× 9 = 72 cm2
ar(∆DBC) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$ × 16 ​× 7 = 56 cm2

∴ ar(quad ABCD) =​ 72 + 56 = 128 cm2

#### Page No 398:

(a)$\sqrt{3}:1$

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC
⇒ ∠BDC = ∠DBC = xo   (Angles opposite to equal sides are equal)
Also, ∠BCD = 60o
xo + xo + 60o = 180o
⇒​ 2xo = 120o
⇒​ xo = 60o
i.e., ∠BCD = BDC = ∠DBC 60o
So, ​∆BCD is an equilateral triangle.
BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:

#### Page No 398:

(d) 8 cm2

Since ||gm  ABCD and ||gm ABFE are on the same base and between the same parallel lines, we have:
ar(||gm ABFE​) = ar(||gm ABCD​) = 25 cm2
⇒ ar(∆BCF )ar(||gm ABFE​)​ $-$ ar(quad EABC​)​ = ( 25 $-$ 17) = 8 cm2

#### Page No 398:

(b) 1:4

ABC and ∆BDE are two equilateral triangles​ and D is the midpoint of BC.
Let AB = BC = AC =  a
Then BD = BE = ED =  $\frac{a}{2}$
∴
So, required ratio = 1 : 4

#### Page No 398:

(a) 8 cm2

Let the distance between AB and CD be h cm.
Then ar(||gm APQD​) = AP ×​ h
= $\frac{1}{2}$ ×​ AB ×​h               (AP = $\frac{1}{2}$AB )
$\frac{1}{2}$ ×​ ar(||gm ABCD)                [ ar(|| gm ABCD) = AB ×​h )
∴ ar (||gm APQD​) = $\frac{1}{2}$ ×​ 16 = 8 cm2

#### Page No 398:

(d) rhombus of 24 cm2

We know that the figure formed by joining the midpoints of the adjacent sides of a rectangle is a rhombus.
So, PQRS is a rhombus and SQ and PR are its diagonals.
i.e., SQ = 8 cm and PR = 6 cm
∴ ar(rhombus PQRS​) = $\frac{1}{2}$ ×​ product of diagonals = $\frac{1}{2}$ ×​ 8  ×​ 6 = 24 cm2

#### Page No 398:

(c) $\frac{1}{4}$ar (∆ ABC )

Since D is the mid point of BC, AD is a median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(ABD ) =
$\frac{1}{2}$ ar(ABC)                      ...(i)

⇒ ar(BED) =
$\frac{1}{2}$ ar(ABD)                      ...(ii)

From (i) and (ii), we have:
ar(BED)
$\frac{1}{2}$⨯ $\frac{1}{2}$​ ⨯​ ar(∆ABC)
∴​ ar(∆BED)​ =
$\frac{1}{4}$⨯ ar(∆ABC)
ar(ABC)

#### Page No 399:

(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$

Since E is the midpoint of AD, BE is a median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED) $\frac{1}{2}$ar(∆ABD)                ...(i)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆CED ) $\frac{1}{2}$ ar(ADC)               ...(ii)

Adding (i) and (ii), we have:
ar(∆BED ) + ar(∆CED ) = $\frac{1}{2}$ ar(∆ABD) + $\frac{1}{2}$ ar(∆ADC)
⇒ ar (∆ BEC ) = $\frac{1}{2}\left(∆ABD+ADC\right)=\frac{1}{2}∆ABC$

#### Page No 399:

(d) $\frac{1}{8}$ ar (∆ ABC)

Given: D is the midpoint of BC, E is the midpoint of BD and O is the mid point of AE.
Since D is the midpoint of BC, AD is the median of ∆ABC.
E is the midpoint of BC, so AE is the median of ∆ABD. O is the midpoint of AE, so BO is median of ABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD ) = $\frac{1}{2}$ ⨯ ​ar(∆ABC)                ...(i)
ar(∆ABE ) =$\frac{1}{2}$​ ⨯​ ar(∆ABD)                        ...(ii)
ar(∆BOE) = $\frac{1}{2}$⨯​ ar(∆ABE)                       ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE ) = $\frac{1}{2}$ar(∆ABE)
ar(∆BOE = $\frac{1}{2}$ ⨯​ $\frac{1}{2}$​ ⨯​ $\frac{1}{2}$​ ⨯​ ar(∆ABC)​
∴​​  ar(∆BOE )​ = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)

#### Page No 399:

(a) 1:2

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangles is half the area of the parallelogram.
i.e., area of triangle = $\frac{1}{2}$× area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram =$\frac{1}{2}$ : 1 = 1 : 2

#### Page No 399:

(c) (3a +b) : (a +3b)

Clearly, EF$\frac{1}{2}$ (a + b)                   [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.

#### Page No 399:

(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas.

#### Page No 399:

(c) perimeter of ABCD > perimeter of ABEF

Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
∴ Perimeter of ABCD > perimeter of ABEF

#### Page No 399:

(b) 40 cm2

Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm

#### Page No 400:

(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(∆AOB) = ar(∆COD).

Consider ∆ADB and ∆ADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(∆ADB)$\ne$ ar(∆ADC)
Subtracting ar(∆AOD) from both sides, we get:
ar(∆ADB) $-$ ar(∆AOD) $\ne$ ar(∆ADC) $-$ ar(∆AOD)
Or ar(∆ AOB)$\ne$ ar(∆ COD)

#### Page No 400:

(b)

Area of a parallelogram  = ​base ​× corresponding height

#### Page No 400:

(c) I and II

Statement I is true, because if a parallelogram and a rectangle lie on the same base and between the same parallel lines, then they have the same altitude and therefore equal areas.
Statement II is also true as area of a parallelogram = base × height
$⇒$AB × DE = AD × BF
$⇒$10 × 6 = 8 × AD
$⇒$ AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.

#### Page No 401:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

In trapezium ABCD, ∆ABC and ABD  are on the same base and between the same parallel lines.
∴ ar(∆ABC) = ar(∆ABD)
⇒ ar(∆ABC) $-$ ar(∆AOB) = ar(∆ABD) $-$ ar(∆AOB)
⇒ ​ar(BOC) = ar(AOD
∴ Assertion (A) is true and, clearly, reason (R) gives (A).

#### Page No 401:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Reason (R) is clearly true.
The explanation of assertion (A)​ is as follows:
ABCD is a rhombus. So, all of its sides are equal.
Now, BC = DC
⇒∠BDC = ∠DBC = x$°$
Also, ∠BCD = 60$°$
x$°$ + x$°$ + 60$°$ = 180$°$
⇒​2x$°$ = 120$°$
⇒​ x$°$ = 60$°$
∴ ∠BCD = ∠BDC = ∠DBC =  60$°$
So, ​∆BCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB = $\frac{a}{2}$
Now, in ∆ OAB, we have:

Thus,
assertion (A)​ is also true, but reason (R) does not give (A).
​Hence, the correct answer is (b).

#### Page No 401:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Page No 401:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not a correct explanation of Assertion (A).

Explanation:
Reason (R):
∴ ar(∆ABC ) =
Thus, reason (R) is true.

Assertion (A):
Area of trapezium =
Thus, assertion (A) is true, but reason (R) does not give assertion (A).

#### Page No 401:

(d) Assertion is false and Reason is true.

Clearly, reason (R) is true.
Assertion: Area of a parallelogram = base × height
$⇒$AB ×​ DE = AD × BF
$⇒$AD = (16 × 8) ÷ 10 = 12.8 cm

So, the assertion is ​false.

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