RS Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 11 - Areas Of Parallelograms And Triangles

Question 1:

Question 2:

(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.

Question 3:

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 â¨¯ BD = $\frac{1}{2}$ AB â¨¯ CL
 ⇒ CL = BD
 ⇒ DC |​​| AB
Similarly, AD |​​| BC.
Hence, ABCD is a paralleogram.
∴ ar(​|​| gm ABCD) = base â€‹â¨¯ height = 5 ​⨯ 7 = 35 cm²

Question 4:

ar(parallelogram ABCD) = base â€‹â¨¯ height
⇒ AB â€‹â¨¯DL = AD â€‹â¨¯ BM
⇒ 10 â€‹â€‹â¨¯ 6 = AD â€‹â¨¯ BM
⇒ AD â€‹â¨¯ 8 = 60 cm²
⇒ AD =  7.5 cm
∴ AD = 7.5 cm

Question 5:

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 âˆ† ABC, we have:
PQ âˆ£âˆ£ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]
$\mathrm{PQ}=\frac{1}{2}\times 16=8 \mathrm{cm}$
Again, in âˆ†DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR âˆ£âˆ£ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]
$\mathrm{SR}=\frac{1}{2}\times 12=6 \mathrm{cm}$
$\mathrm{Area} \mathrm{of} \mathrm{PQRS}=\mathrm{length}\times \mathrm{breadth}=6\times 8=48 {\mathrm{cm}}^2$

Question 6:

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

Question 7:

(i) In $△$BCD, 
$$\begin{gathered} {\mathrm{DB}}^2+{\mathrm{BC}}^2={\mathrm{DC}}^2 \\ \Rightarrow {\mathrm{DB}}^2={17}^2-8^2=225 \\ \Rightarrow \mathrm{DB}=15 \mathrm{cm} \end{gathered}$$
Ar($△$BCD) = $\frac{1}{2}\times b\times h=\frac{1}{2}\times 8\times 15=60 {\mathrm{cm}}^2$
In $△$BAD,
$$\begin{gathered} {\mathrm{DA}}^2+{\mathrm{AB}}^2={\mathrm{DB}}^2 \\ \Rightarrow {\mathrm{AB}}^2={15}^2-9^2=144 \\ \Rightarrow \mathrm{AB}=12 \mathrm{cm} \end{gathered}$$
Ar($△$DAB) = $\frac{1}{2}\times b\times h=\frac{1}{2}\times 9\times 12=54 {\mathrm{cm}}^2$

Area of quad. ABCD = Ar($△$DAB) + Ar($△$BCD) = 54 + 60 = 114 cm² .

(ii) Area of trap(PQRS) = $\frac{1}{2}\left(8+16\right)\times 8=96 {\mathrm{cm}}^2$

Question 8:

∆ADL is a right angle triangle.
So, DL = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
Similarly, in ∆BMC, we have:
MC = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
∴ 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

Question 9:

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

Question 10:

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) = â€‹ar($△$ABC) + $\frac{1}{2}$​ar($△$ABC)
$$\begin{gathered} \Rightarrow 24=\frac{3}{2}\mathrm{ar}\left(△\mathrm{ABC}\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{ABC}\right)=16 {\mathrm{cm}}^2 \end{gathered}$$

Question 11:

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

Question 12:

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

Question 13:

(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)


 

Question 14:

∆CDA and â€‹âˆ†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)

Question 15:

∆DEC and â€‹âˆ†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(​∆DEC) + ar(∆ADE)​ = ar(∆DEB) + ar(∆ADE)​ â€‹                 
⇒ 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)​ â€‹      â€‹
⇒ ar(​​∆OCE) = ar(​∆OBD)

Question 16:

Let AD is a median of âˆ†ABC and D is the midpoint of BC. AD divides âˆ†ABC in two triangles: ∆ABD and âˆ†ADC.
To prove: ar(∆ABD) = ar(∆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 
⇒ ar(∆ABD) = ar(∆ADC)

Question 17:

Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(∆ABD) = ar(∆CDB)

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

Question 18:

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
$\Rightarrow$ar(ΔABC) = ar(ΔABD)



 

Question 19:

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

Question 20:

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, 
$$\begin{gathered} \mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{BPO}\right) .....\left(1\right) \\ \mathrm{ar}\left(△\mathrm{ADO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right) .....\left(2\right) \\ \mathrm{Case} 1: \\ \left(2\right)-\left(1\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{ADO}\right)-\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)-\mathrm{ar}\left(△\mathrm{BPO}\right) \end{gathered}$$
Thus, ar(∆ADP) = ar(∆ABP)

Case II: 

$\mathrm{ar}\left(△\mathrm{ADO}\right)+\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)+\mathrm{ar}\left(△\mathrm{BPO}\right)$
Thus, ar(∆ADP) = ar(∆ABP)

Question 21:

Given:  BO = OD
To prove: ar(∆ABC) = ar(∆ADC)
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 âˆ†BCD.
i.e., ar(∆COD) = ar (∆COB)            ...(i)

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

From (i) and (ii), we have:
ar(∆COD) + ar(∆AOD) = ar(∆COB) + ar(∆AOB)
∴ ar(∆ADC )​ = ar(∆ABC)

Question 22:

Given:  D is the midpoint of BC and E is the midpoint of AD.
To prove: $\mathrm{ar}(∆BEC)=\frac{1}{2}\mathrm{ar}(∆ABC)$
Proof: 
Since E is the midpoint of AD, BE is the median of âˆ†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)​ = $\frac{1}{2}$ â¨¯â€‹ ar(∆ABD)​ + $\frac{1}{2}$ ⨯​ ar(∆ADC)   
⇒ ar(∆BEC )​ = $\frac{1}{2}$⨯ [ar(∆ABD) + ar(∆ADC)] 
⇒ ​ar(∆BEC )​ =​ $\frac{1}{2}$ ⨯​ ar(∆ABC)

Question 23:

D is the midpoint of side BC of ∆ABC. 
$\Rightarrow$AD is the median of âˆ†ABC. 
$\Rightarrow$$\mathrm{ar}\left(△\mathrm{ABD}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
E is the midpoint of BD of âˆ†ABD, 
$\Rightarrow$AE is the median of âˆ†ABD
$\Rightarrow$$\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, 
$\Rightarrow$BO is the median of âˆ†ABE, 
$\Rightarrow$$\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)

 

Question 24:

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)
$\Rightarrow$ar($△$MQC) = ar($△$MPB)
$\Rightarrow$ar($△$MQC) + ar(APMCD) = ar($△$MPB) + ar(APMCD)
$\Rightarrow$ar(APQD) = ar(ABCD)

Question 25:

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.

Question 26:

Given: âˆ†ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof: 
Since ∆ABC and âˆ†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 âˆ†ALO and âˆ†DMO, we have:
AL = DM
∠ALO = ∠DMO =  90^o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ âˆ† DMO
​​∴​ OA = OD
Hence, BC bisects AD.

Question 27:

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

Question 28:

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

Question 29:

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 âˆ†ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC              [ By midpoint theorem] 
Again, in âˆ†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, âˆ†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)

Question 30:

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

Question 31:

Given: D is a point on BC of âˆ†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 ×​ $\left(\frac{1}{2}\times BD\times AL\right)$
⇒ ar(∆ABC)​ = 3 × ar(∆ABD)
∴ â€‹ar(∆ABD) = â€‹$\frac{1}{3}$​ar(∆ABC)

Question 32:

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)
 

Question 33:

Given:  ABCDE is a pentagon.  EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) =  ar( âˆ†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.  
 ∴ ar(∆ADE) = ar(∆ADG)                       ...(iv)

From (iii) and (iv), we have:
ar(∆DBC) + ar(∆ADE) = ar(∆DBF) + ar(∆ADG)
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) 

Question 34:

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

Question 35:

Given: D is a point on BC of âˆ† 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:
$$\begin{gathered} \frac{\mathrm{ar}(△ABD)}{\mathrm{ar}(∆ADC}=\frac{{\displaystyle \frac{1}{2}}\times BD\times AL}{{\displaystyle \frac{1}{2}}\times DC\times AL} \\ =\frac{BD}{DC} \\ =\frac{m}{n} \end{gathered}$$

∴ ar(∆ABD) : ar(∆ADC) = m : n

Question 36:

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. 
$$\begin{gathered} \mathrm{ar}\left(\mathrm{DCNM}\right)=\frac{1}{2}\times \mathrm{DP}\left(\mathrm{DC}+\mathrm{MN}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{MNBA}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{AB}+\mathrm{MN}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(DCNM) : ar(MNBA) = (a + 3b) : (3a + b)
 

Question 37:

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. 
$$\begin{gathered} \mathrm{ar}\left(\mathrm{ABFE}\right)=\frac{1}{2}\times \mathrm{AP}\left(\mathrm{AB}+\mathrm{EF}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{EFCD}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{CD}+\mathrm{EF}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{\mathrm{ar}\left(\mathrm{ABEF}\right)}{\mathrm{ar}\left(\mathrm{EFCD}\right)}=\frac{24+3\times 16}{16+3\times 24}=\frac{9}{11}$

Question 38:

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) 

Question 1:

In âˆ†RSC and âˆ†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, âˆ†RSC $\cong$ ∆PQB                   (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)
 

Question 2:

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

Question 3:

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

Question 4:

(a)  triangles of equal areas

Question 5:

(c)114 cm²

ar (quad. ABCD) =  ar (∆ ABC)  +  ar (∆ ACD)
In right angle triangle ACD, we have:
AC  = $\sqrt{\left({17}^2 - 8^2\right)} = \sqrt{225} = 15 \mathrm{cm}$
In right angle triangle ABC, we have:
BC = $\sqrt{\left({15}^2 - 9^2\right)} = \sqrt{144} = 12 \mathrm{cm}$
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

Question 6:

(c)124 cm²

In the right angle triangle BEC, we have:
EC  = $\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{cm}$
ar(trapez. ABCD) = $\frac{1}{2}\times (\mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides})\times \mathrm{distance} \mathrm{between} \mathrm{them}=\frac{1}{2}\times 31\times 8=124$ cm²

Question 7:

(c) 34 cm²

ar(parallelogram ABCD) = base × height = 5 â€‹× 6.8 =  34 cm²

Question 8:

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


​

Question 9:

(a) 40 cm²
ar(||gm ABCD) = base × height =  10 â€‹× 4 =  40 cm²

Question 10:

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

Question 11:

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). 

Question 12:

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 âˆ†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).    
 

Question 13:

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
$$\begin{gathered} \mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right) \\ =2\mathrm{ar}\left(△\mathrm{AFE}\right) \\ =2\times \frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right) \end{gathered}$$
Hence, the correct answer is option (a). 

Question 14:

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

Question 15:

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

Question 16:

(b) 40 cm²

In right angled triangle MBC, we have:
MC = $\sqrt{5^2 - 4^2} = \sqrt{9} = 3 \mathrm{cm}$
In right angled triangle ADL, we have:
DL = $\sqrt{5^2 - 4^2} = \sqrt{9} = 3 \mathrm{cm}$

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) â€‹× 4
= 40 cm²

Question 17:

(b)128 cm²

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

∴ ar(quad ABCD) =​ 72 + 56 = 128 cm²

Question 18:

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

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC 
⇒ ∠BDC = ∠DBC = x^o   (Angles opposite to equal sides are equal)
Also, ∠BCD = 60^o
∴ x^o + x^o + 60^o = 180^o 
⇒​ 2x^o = 120^o
⇒​ x^o = 60^o
i.e., ∠BCD = ∠BDC = ∠DBC =  60^o
So, â€‹âˆ†BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
 $A{B}^2= O{A}^2 + O{B}^2$ 
$$\begin{gathered} \Rightarrow O{A}^2=A{B}^2-O{B}^2=a^2-{\left({\displaystyle \frac{a}{2}}\right)}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4} \\ \Rightarrow O{A}^2=\frac{3a^2}{4} \\ \Rightarrow OA=\sqrt{\frac{3a^2}{4}} \\ \Rightarrow OA=\frac{\sqrt{3}a}{2} \\ \mathrm{Now}, AC=2\times OA=2\times \frac{\sqrt{3}a}{2}=\sqrt{3}a\therefore AC:BD=\sqrt{3}a:a =\sqrt{3} :1 \end{gathered}$$

Question 19:

(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

Question 20:

(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 =  $\frac{a}{2}$
∴ $\frac{\mathrm{ar}(∆BDE)}{\mathrm{ar}(∆ABC)}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}A{B}^2}{{\displaystyle \frac{\sqrt{3}}{4}}B{E}^2} =\frac{{\left({\displaystyle \frac{a}{2}}\right)}^2}{a^2}= \frac{1}{4}$
So, required ratio = 1 : 4

Question 21:

(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

Question 22:

(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               

Question 23:

(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) = $\frac{1}{2}$⨯ $\frac{1}{2}$​ â¨¯â€‹ ar(∆ABC)
∴​ ar(∆BED)​ = $\frac{1}{4}$⨯ ar(∆ABC)  
 ar(∆ABC)

Question 24:

(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 ) = $\frac{1}{2}\left(∆ABD+ADC\right)=\frac{1}{2}∆ABC$    

Question 25:

(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 âˆ†ABC.
E is the midpoint of BC, so AE is the median of âˆ†ABD. O is the midpoint of AE, so BO is median of âˆ†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 ) =$\frac{1}{2}$​ â¨¯â€‹ 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 ) = $\frac{1}{2}$ â¨¯â€‹ $\frac{1}{2}$​ ⨯​ $\frac{1}{2}$​ ⨯​ ar(∆ABC)​
∴​​  ar(∆BOE )​ = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)                    

Question 26:

(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

Question 27:

(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.
 $$\begin{gathered} \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \mathrm{ar}(\mathrm{trap} ABEF)=\frac{1}{2}\left(a+\frac{a+b}{2}\right)d =\frac{\left(3a+b\right)d}{4} \\ \mathrm{ar}(\mathrm{trap} EFCD)=\frac{1}{2}\left(b+\frac{a+b}{2}\right)d=\frac{\left(a+3b\right)d}{4} \\ \therefore \mathrm{Required} \mathrm{ratio}= \frac{\left(3a+b\right)d}{4} : \frac{\left(a+3b\right)d}{4} = (3a+b) : ( a+3b) \end{gathered}$$ 

Question 28:

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

Question 29:

(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:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF