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

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Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 11 Areas Of Parallelograms And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelograms And Triangles are extremely popular among Class 9 students for Maths Areas Of Parallelograms And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 387:

Answer:

(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:

Answer:

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
Similarly, AD || BC.
Hence, ABCD is a paralleogram.
∴ ar(|| gm ABCD) = base ⨯ height = 5 ⨯ 7 = 35 cm²

Page No 387:

Answer:

ar(parallelogram ABCD) = base ⨯ height
⇒ AB ⨯DL = AD ⨯ BM
⇒ 10 ⨯ 6 = AD ⨯ BM
⇒ AD ⨯ 8 = 60 cm²
⇒ AD = 7.5 cm
∴ AD = 7.5 cm

Page No 388:

Answer:

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

Page No 388:

Answer:

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:

Answer:

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

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

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

Page No 388:

Answer:

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

Page No 388:

Answer:

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
$\Rightarrow$ ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ (AL + CL)
Hence, proved.

Page No 388:

Answer:

Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
$ar (△ ABC) = ar (△ ACD) = \frac{1}{2} ar (ABCD)$
M is the midpoint of AB. So, CM is the median.
CM divides $△$ ABC in two triangles with equal area.
$ar (△ AMC) = ar (△ BMC) = \frac{1}{2} ar (△ ABC)$
ar(AMCD) = ar( $△$ ACD) + ar( $△$ AMC) = ar( $△$ ABC) + ar( $△$ AMC) = ar( $△$ ABC) + $\frac{1}{2}$ ar( $△$ ABC)
$\Rightarrow 24 = \frac{3}{2} ar (△ ABC) \Rightarrow ar (△ ABC) = 16 {cm}^{2}$

Page No 388:

Answer:

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:

Answer:

We know
ar(∆APB) = $\frac{1}{2} ar (ABCD)$ .....(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) = $\frac{1}{2} ar (ABCD)$ .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved

Page No 389:

Answer:

(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:

Answer:

∆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:

Answer:

∆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(∆DEC) + ar(∆ADE) = ar(∆DEB) + ar(∆ADE)
⇒ 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:

Answer:

Let AD is a median of ∆ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ∆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)

Page No 389:

Answer:

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:

Answer:

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)

Page No 389:

Answer:

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:

Answer:

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,
$ar (△ DPO) = ar (△ BPO) . . . . . (1) ar (△ ADO) = ar (△ ABO) . . . . . (2) Case 1 : (2) - (1) \Rightarrow ar (△ ADO) - ar (△ DPO) = ar (△ ABO) - ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

Case II:

$ar (△ ADO) + ar (△ DPO) = ar (△ ABO) + ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

Page No 389:

Answer:

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 ∆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)
∴ ar(∆ADC ) = ar(∆ABC)

Page No 389:

Answer:

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$
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:

Answer:

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

Page No 390:

Answer:

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

Page No 390:

Answer:

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:

Answer:

Given: ∆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 ∆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 = 90^o
∠AOL = ∠DOM (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
∴ OA = OD
Hence, BC bisects AD.

Page No 390:

Answer:

In $△$ MDA and $△$ MCP,
$∠$ DMA = $∠$ CMP (Vertically opposite angles)
$∠$ MDA = $∠$ MCP (Alternate interior angles)
AD = CP (Since AD = CB and CB = CP)
So, $△$ MDA $≅$ $△$ 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 ${cm}^{2}$
Area of parallelogram ABCD = 2 $\times$ ar( $△$ DBC) = 2 $\times$ 14 = 28 ${cm}^{2}$

Page No 390:

Answer:

Join BM and AC.
ar(∆ADC) = $\frac{1}{2} b h$ = $\frac{1}{2} \times DC \times h$
ar(∆ABM) = $\frac{1}{2} \times AB \times h$
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

Page No 390:

Answer:

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:

Answer:

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)

Page No 391:

Answer:

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 × $(\frac{1}{2} \times B D \times A L)$
⇒ ar(∆ABC) = 3 × ar(∆ABD)
∴ ar(∆ABD) = $\frac{1}{3}$ ar(∆ABC)

Page No 391:

Answer:

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) = $\frac{1}{2} ar (△ ABC)$ .....(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:

Answer:

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.
∴ 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)

Page No 391:

Answer:

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

Page No 391:

Answer:

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:
$\frac{ar (△ A B D)}{ar (∆ A D C} = \frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L} = \frac{B D}{D C} = \frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = m : n

Page No 391:

Answer:

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

Page No 391:

Answer:

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

Page No 391:

Answer:

In $△$ PAC,
PA || DE and E is the midpoint of AC
So, D is the midpoint of PC by converse of midpoint theorem.
Also, $DE = \frac{1}{2} PA$ .....(1)
Similarly, $DE = \frac{1}{2} AQ$ .....(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:

Answer:

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

Page No 395:

Answer:

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

Page No 396:

Answer:

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:

Answer:

(a) triangles of equal areas

Page No 396:

Answer:

(c)114 cm²

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

Page No 396:

Answer:

(c)124 cm²

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

Page No 396:

Answer:

Page No 396:

Answer:

(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:

Answer:

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

Page No 397:

Answer:

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).

Page No 397:

Answer:

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:

Answer:

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:

Answer:

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

Page No 397:

Answer:

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

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Answer:

(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:

Answer:

(b) 40 cm²

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

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) × 4
= 40 cm²

Page No 397:

Answer:

(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:

Answer:

(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, ∆BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
$A B^{2} = O A^{2} + O B^{2}$
$\Rightarrow O A^{2} = A B^{2} - O B^{2} = a^{2} - {(\frac{a}{2})}^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3 a^{2}}{4} \Rightarrow O A^{2} = \frac{3 a^{2}}{4} \Rightarrow O A = \sqrt{\frac{3 a^{2}}{4}} \Rightarrow O A = \frac{\sqrt{3} a}{2} Now, A C = 2 \times O A = 2 \times \frac{\sqrt{3} a}{2} = \sqrt{3} a ∴ A C : B D = \sqrt{3} a : a = \sqrt{3} : 1$

Page No 398:

Answer:

(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

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Answer:

(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 = $\frac{a}{2}$
∴ $\frac{ar (∆ B D E)}{ar (∆ A B C)} = \frac{\frac{\sqrt{3}}{4} A B^{2}}{\frac{\sqrt{3}}{4} B E^{2}} = \frac{{(\frac{a}{2})}^{2}}{a^{2}} = \frac{1}{4}$
So, required ratio = 1 : 4

Page No 398:

Answer:

(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:

Answer:

(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:

Answer:

(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:

Answer:

(a) $\frac{1}{2} ar (∆ A B C)$

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 ) = $\frac{1}{2} (∆ A B D + A D C) = \frac{1}{2} ∆ A B C$

Page No 399:

Answer:

(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:

Answer:

(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:

Answer:

(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.
$Now, we have : ar (trap A B E F) = \frac{1}{2} (a + \frac{a + b}{2}) d = \frac{(3 a + b) d}{4} ar (trap E F C D) = \frac{1}{2} (b + \frac{a + b}{2}) d = \frac{(a + 3 b) d}{4} ∴ Required ratio = \frac{(3 a + b) d}{4} : \frac{(a + 3 b) d}{4} = (3 a + b) : (a + 3 b)$

Page No 399:

Answer:

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

Page No 399:

Answer:

(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

Page No 399:

Answer:

(b) 40 cm²

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

$Now, A B = \sqrt{A C^{2} - B C^{2}} = \sqrt{10^{2} - {(2 \sqrt{5})}^{2}} = \sqrt{80} = 4 \sqrt{5} cm ∴ Area of the rectangle = A B \times A D = 2 \sqrt{5} \times 4 \sqrt{5} = 40 {cm}^{2}$

Page No 400:

Answer:

(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) $\neq$ ar(∆ADC)
Subtracting ar(∆AOD) from both sides, we get:
ar(∆ADB) $-$ ar(∆AOD) $\neq$ ar(∆ADC) $-$ ar(∆AOD)
Or ar(∆ AOB) $\neq$ ar(∆ COD)

Page No 400:

Answer:

(b) $Area of a | | gm = \frac{1}{2} \times base \times corresponding height$

Area of a parallelogram = base × corresponding height

Page No 400:

Answer:

(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
$\Rightarrow$ AB × DE = AD × BF
$\Rightarrow$ 10 × 6 = 8 × AD
$\Rightarrow$ AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.

Page No 401:

Answer:

(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:

Answer:

(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:
$O A^{2} = A B^{2} - O B^{2} = a^{2} - {(\frac{a}{2})}^{2} = \frac{3 a^{2}}{4} \Rightarrow O A = \frac{\sqrt{3} a}{2} \Rightarrow A C = 2 \times \frac{\sqrt{3} a}{2} = \sqrt{3} a$
$∴ A C : B D = \sqrt{3} a : a = \sqrt{3} : 1$
Thus, assertion (A) is also true, but reason (R) does not give (A).
Hence, the correct answer is (b).

Page No 401:

Answer:

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

Page No 401:

Answer:

(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 ) = $\frac{\sqrt{3}}{4} \times (side)^{2} = (\frac{\sqrt{3}}{4} \times 8 \times 8) = 16 \sqrt{3} {cm}^{2}$
Thus, reason (R) is true.

Assertion (A):
Area of trapezium = $\frac{1}{2} \times (25 + 15) \times 6 = 120 {cm}^{2}$
Thus, assertion (A) is true, but reason (R) does not give assertion (A).

Page No 401:

Answer:

(d) Assertion is false and Reason is true.

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

So, the assertion is false.

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