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 9 Maths are prepared by experts and are 100% accurate.
Page No 387:
Question 1:
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:
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.
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(âABD) + ar(ââDCB)
= 2 ar(ââABD) [âµ arâ(âABD) = ar(ââDCB)]
∴ ar(ââABD) = ar(quad. ABCD) ...(i)
Again, ar(quad. ABCD) = ar(âABC) + ar(ââCDA)
= 2 ar(ââ ABC) [âµ arâ(âABC) = ar(ââCDA)]
∴ ar(ââABC) = ar(quad. ABCD) ...(ii)
From (i) and (ii), we have:
ar(ââABD) = ar(ââABC) = AB ⨯ BD = AB ⨯ CL
⇒ CL = BD
⇒ DC |ââ| AB
Similarly, AD |ââ| BC.
Hence, ABCD is a paralleogram.
∴ ar(â|â| gm ABCD) = base â⨯ height = 5 â⨯ 7 = 35 cm2
Page No 387:
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(âABD) + ar(ââDCB)
= 2 ar(ââABD) [âµ arâ(âABD) = ar(ââDCB)]
∴ ar(ââABD) = ar(quad. ABCD) ...(i)
Again, ar(quad. ABCD) = ar(âABC) + ar(ââCDA)
= 2 ar(ââ ABC) [âµ arâ(âABC) = ar(ââCDA)]
∴ ar(ââABC) = ar(quad. ABCD) ...(ii)
From (i) and (ii), we have:
ar(ââABD) = ar(ââABC) = AB ⨯ BD = AB ⨯ CL
⇒ CL = BD
⇒ DC |ââ| AB
Similarly, AD |ââ| BC.
Hence, ABCD is a paralleogram.
∴ ar(â|â| gm ABCD) = base â⨯ height = 5 â⨯ 7 = 35 cm2
Answer:
ar(parallelogram ABCD) = base â⨯ height
⇒ AB â⨯DL = AD â⨯ BM
⇒ 10 ââ⨯ 6 = AD â⨯ BM
⇒ AD â⨯ 8 = 60 cm2
⇒ AD = 7.5 cm
∴ AD = 7.5 cm
Page No 388:
Question 4:
ar(parallelogram ABCD) = base â⨯ height
⇒ AB â⨯DL = AD â⨯ BM
⇒ 10 ââ⨯ 6 = AD â⨯ BM
⇒ AD â⨯ 8 = 60 cm2
⇒ AD = 7.5 cm
∴ AD = 7.5 cm
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 â ABC, we have:
PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Page No 388:
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 â ABC, we have:
PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Answer:
ar(trapezium) = ⨯ (sum of parallel sides) ⨯ (distance between them)
= ⨯ (9 + 6) ⨯ 8
= 60 cm2
Hence, the area of the trapezium is 60 cm2.
Page No 388:
Question 6:
ar(trapezium) = ⨯ (sum of parallel sides) ⨯ (distance between them)
= ⨯ (9 + 6) ⨯ 8
= 60 cm2
Hence, the area of the trapezium is 60 cm2.
Answer:
(i) In BCD,
Ar(BCD) =
In BAD,
Ar(DAB) =
Area of quad. ABCD = Ar(DAB) + Ar(BCD) = 54 + 60 = 114 cm2 .
(ii) Area of trap(PQRS) =
Page No 388:
Question 7:
(i) In BCD,
Ar(BCD) =
In BAD,
Ar(DAB) =
Area of quad. ABCD = Ar(DAB) + Ar(BCD) = 54 + 60 = 114 cm2 .
(ii) Area of trap(PQRS) =
Answer:
âADL is a right angle triangle.
So, DL =
Similarly, in âBMC, we have:
MC =
∴ DC = DL + LM + MC = 3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = ⨯ (sum of parallel sides) ⨯ (distance between them)
= ⨯ (7 + 13) ⨯ 4
= 40 cm2
âHence, DC = 13 cm and area of trapezium = 40 cm2
Page No 388:
Question 8:
âADL is a right angle triangle.
So, DL =
Similarly, in âBMC, we have:
MC =
∴ DC = DL + LM + MC = 3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = ⨯ (sum of parallel sides) ⨯ (distance between them)
= ⨯ (7 + 13) ⨯ 4
= 40 cm2
âHence, DC = 13 cm and area of trapezium = 40 cm2
Answer:
ar(quad. ABCD) = ar(ââABD) + ar (âDBC)
ar(âABD) = ⨯ base ⨯ height = ⨯ BD ⨯ AL ...(i)
ar(âDBC) = ⨯ BD ⨯ CL ...(ii)
From (i) and (ii), we get:
âar(quad ABCD) = ⨯BD ⨯â AL + ⨯ BD ⨯â CL
âar(quad ABCD) = ⨯ BD ⨯ â(AL + CL)
Hence, proved.
Page No 388:
Question 9:
ar(quad. ABCD) = ar(ââABD) + ar (âDBC)
ar(âABD) = ⨯ base ⨯ height = ⨯ BD ⨯ AL ...(i)
ar(âDBC) = ⨯ BD ⨯ CL ...(ii)
From (i) and (ii), we get:
âar(quad ABCD) = ⨯BD ⨯â AL + ⨯ BD ⨯â CL
âar(quad ABCD) = ⨯ BD ⨯ â(AL + CL)
Hence, proved.
Answer:
Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
M is the midpoint of AB. So, CM is the median.
CM divides ABC in two triangles with equal area.
ar(AMCD) = ar(ACD) + ar(AMC) = ar(ABC) + ar(AMC) = âar(ABC) + âar(ABC)
Page No 388:
Question 10:
Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.
M is the midpoint of AB. So, CM is the median.
CM divides ABC in two triangles with equal area.
ar(AMCD) = ar(ACD) + ar(AMC) = ar(ABC) + ar(AMC) = âar(ABC) + âar(ABC)
Answer:
âar(quad ABCD) = ar(ABD) + ar(BDC)
= ⨯BD ⨯â AL + ⨯BD ⨯â CM
= ⨯BD ⨯â ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = â ⨯ 14 ⨯ ( 8 + 6)
= 7 â⨯14
= 98 cm2
Page No 388:
Question 11:
âar(quad ABCD) = ar(ABD) + ar(BDC)
= ⨯BD ⨯â AL + ⨯BD ⨯â CM
= ⨯BD ⨯â ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = â ⨯ 14 ⨯ ( 8 + 6)
= 7 â⨯14
= 98 cm2
Answer:
We know
ar(âAPB) = .....(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(âBQC) = .....(2)
From (1) and (2)
ar(âAPB) = ar(âBQC)
Hence Proved
Page No 389:
Question 12:
We know
ar(âAPB) = .....(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(âBQC) = .....(2)
From (1) and (2)
ar(âAPB) = ar(âBQC)
Hence Proved
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(âATQ) = ar(ABPQ) (Same base AQ and AQ || BP) .....(2)
From (1) and (2)
ar(âATQ) = ar(MNPQ)
Page No 389:
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(âATQ) = ar(ABPQ) (Same base AQ and AQ || BP) .....(2)
From (1) and (2)
ar(âATQ) = ar(MNPQ)
Answer:
âCDA and ââCBD lies on the same base and between the same parallel lines.
So, ar(ââCDA) = ar(CDB) ...(i)
Subtracting ar(ââOCD) from both sides of equation (i), we get:
ar(ââCDA) ar(ââOCD) = ar(âââCDB) ar (âââOCD)
⇒ ar(âââAOD) = ar(âââBOC)
Page No 389:
Question 14:
âCDA and ââCBD lies on the same base and between the same parallel lines.
So, ar(ââCDA) = ar(CDB) ...(i)
Subtracting ar(ââOCD) from both sides of equation (i), we get:
ar(ââCDA) ar(ââOCD) = ar(âââCDB) ar (âââOCD)
⇒ ar(âââAOD) = ar(âââBOC)
Answer:
âDEC and ââDEB lies on the same base and between the same parallel lines.
So, ar(ââDEC) = ar(âDEB) ...(1)
(i) On addingâ ar(âADE)â in both sides of equation (1), we get:
ar(ââDEC) + ar(âADE)â = ar(âDEB) + ar(âADE)â â
⇒ ar(âââACD) = ar(âââABE)
(ii) On subtractingâ ar(ODE)â from both sides of equation (1), we get:â
ar(ââDEC) ar(âODE)â = ar(âDEB) ar(âODE)â â â
⇒ ar(âââOCE) = ar(ââOBD)
Page No 389:
Question 15:
âDEC and ââDEB lies on the same base and between the same parallel lines.
So, ar(ââDEC) = ar(âDEB) ...(1)
(i) On addingâ ar(âADE)â in both sides of equation (1), we get:
ar(ââDEC) + ar(âADE)â = ar(âDEB) + ar(âADE)â â
⇒ ar(âââACD) = ar(âââABE)
(ii) On subtractingâ ar(ODE)â from both sides of equation (1), we get:â
ar(ââDEC) ar(âODE)â = ar(âDEB) ar(âODE)â â â
⇒ ar(âââOCE) = ar(ââOBD)
Answer:
Let AD is a median of âABC and D is the midpoint of BC. AD divides âABC in two triangles: âABD and âADC.
To prove: ar(âABD) = ar(âADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with AL on both sides, we get:
× BD × AL = × DC × AL
⇒ ar(âABD) = ar(âADC)
Page No 389:
Question 16:
Let AD is a median of âABC and D is the midpoint of BC. AD divides âABC in two triangles: âABD and âADC.
To prove: ar(âABD) = ar(âADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with AL on both sides, we get:
× BD × AL = × DC × AL
⇒ ar(âABD) = ar(âADC)
Answer:
Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(âABD) = ar(âCDB)
Proof:
In âABD and âCDB, we have:
AB = CD [Opposite sides of a parallelogram]
AD = CB [Opposite sides of a parallelogram]â
∴ ar(âABD) = ar(âCDB)
Page No 389:
Question 17:
Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(âABD) = ar(âCDB)
Proof:
In âABD and âCDB, we have:
AB = CD [Opposite sides of a parallelogram]
AD = CB [Opposite sides of a parallelogram]â
∴ ar(âABD) = ar(âCDB)
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
ar(ΔABC) = ar(ΔABD)
Page No 389:
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
ar(ΔABC) = ar(ΔABD)
Answer:
ar(âBCD) = ar(âBCE) (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC.
Page No 389:
Question 19:
ar(âBCD) = ar(âBCE) (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC.
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,
Thus, ar(âADP) = ar(âABP)
Case II:
Thus, ar(âADP) = ar(âABP)
Page No 389:
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,
Thus, ar(âADP) = ar(âABP)
Case II:
Thus, ar(âADP) = ar(âABP)
Answer:
Given: BO = OD
To prove: ar(âABC) = ar(âADC)
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 âBCD.
i.e., ar(âCOD) = ar (âCOB) ...(i)
AO is a median of âABD.
i.e., ar(âAOD) = ar(âAOB) ...(ii)
From (i) and (ii), we have:
ar(âCOD) + ar(âAOD) = ar(âCOB) + ar(âAOB)
∴ ar(âADC )â = ar(âABC)
Page No 389:
Question 21:
Given: BO = OD
To prove: ar(âABC) = ar(âADC)
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 âBCD.
i.e., ar(âCOD) = ar (âCOB) ...(i)
AO is a median of âABD.
i.e., ar(âAOD) = ar(âAOB) ...(ii)
From (i) and (ii), we have:
ar(âCOD) + ar(âAOD) = ar(âCOB) + ar(âAOB)
∴ ar(âADC )â = ar(âABC)
Answer:
Given: D is the midpoint of BC and E is the midpoint of AD.
To prove:
Proof:
Since E is the midpoint of AD, BE is the median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âBED ) = ar(âABD) ...(i)
Also, ar(âCDE ) = ar(âADC) ...(ii)
From (i) and (ii), we have:
ar(âBED) + ar(âCDE)â = ⨯â ar(âABD)â + ⨯â ar(âADC)
⇒ ar(âBEC )â = ⨯ [ar(âABD) + ar(âADC)]
⇒ âar(âBEC )â =â ⨯â ar(âABC)
Page No 390:
Question 22:
Given: D is the midpoint of BC and E is the midpoint of AD.
To prove:
Proof:
Since E is the midpoint of AD, BE is the median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âBED ) = ar(âABD) ...(i)
Also, ar(âCDE ) = ar(âADC) ...(ii)
From (i) and (ii), we have:
ar(âBED) + ar(âCDE)â = ⨯â ar(âABD)â + ⨯â ar(âADC)
⇒ ar(âBEC )â = ⨯ [ar(âABD) + ar(âADC)]
⇒ âar(âBEC )â =â ⨯â ar(âABC)
Answer:
D is the midpoint of side BC of âABC.
AD is the median of âABC.
E is the midpoint of BD of âABD,
AE is the median of âABD
Also, O is the midpoint of AE,
BO is the median of âABE,
Thus, ar(âBOE) = ar(âABC)
Page No 390:
Question 23:
D is the midpoint of side BC of âABC.
AD is the median of âABC.
E is the midpoint of BD of âABD,
AE is the median of âABD
Also, O is the midpoint of AE,
BO is the median of âABE,
Thus, ar(âBOE) = ar(âABC)
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)
ar(MQC) = ar(MPB)
ar(MQC) + ar(APMCD) = ar(MPB) + ar(APMCD)
ar(APQD) = ar(ABCD)
Page No 390:
Question 24:
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)
ar(MQC) = ar(MPB)
ar(MQC) + ar(APMCD) = ar(MPB) + ar(APMCD)
ar(APQD) = ar(ABCD)
Answer:
We have:
âar(quad. ABCD) = ar(âACD) + ar(âABC)
ar(âABP) = ar(âACP)ââ + ar(âABC)
âACD and âACP are on the same base and between the same parallels AC and DP.
∴ ar(âACD) = ar(â ACP)â
By adding ar(âABC) on both sides, we get:
ar(âACD) + ar(âABC) = ar(âACP)ââ + ar(âABC)
⇒â ar (quad. ABCD) = ar(âABP)
Hence, proved.
Page No 390:
Question 25:
We have:
âar(quad. ABCD) = ar(âACD) + ar(âABC)
ar(âABP) = ar(âACP)ââ + ar(âABC)
âACD and âACP are on the same base and between the same parallels AC and DP.
∴ ar(âACD) = ar(â ACP)â
By adding ar(âABC) on both sides, we get:
ar(âACD) + ar(âABC) = ar(âACP)ââ + ar(âABC)
⇒â ar (quad. ABCD) = ar(âABP)
Hence, proved.
Answer:
Given: âABC and âDBC are on the same base BC.
ar(âABC) = ar(âDBC)â
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since âABC and âDBC 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 âALO and âDMO, we have:
AL = DM
∠ALO = ∠DMO = 90o
∠âAOL = ∠DOâM (Vertically opposite angles)
i.e., â ALO ≅ â DMO
ââ∴â OA = OD
Hence, BC bisects AD.
Page No 390:
Question 26:
Given: âABC and âDBC are on the same base BC.
ar(âABC) = ar(âDBC)â
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since âABC and âDBC 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 âALO and âDMO, we have:
AL = DM
∠ALO = ∠DMO = 90o
∠âAOL = ∠DOâM (Vertically opposite angles)
i.e., â ALO ≅ â DMO
ââ∴â OA = OD
Hence, BC bisects AD.
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)
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
Area of parallelogram ABCD = 2 ar(DBC) = 2 14 = 28
Page No 390:
Question 27:
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)
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
Area of parallelogram ABCD = 2 ar(DBC) = 2 14 = 28
Answer:
Join BM and AC.
ar(âADC) = =
ar(âABM) =
AB = DC (Since ABCD is a parallelogram)
So, ar(âADC) = ar(âABM)
ar(âADC) + ar(âAMC) = ar(âABM) + ar(âAMC)
ar(âADM) = ar(ABMC)
Hence Proved
Page No 390:
Question 28:
Join BM and AC.
ar(âADC) = =
ar(âABM) =
AB = DC (Since ABCD is a parallelogram)
So, ar(âADC) = ar(âABM)
ar(âADC) + ar(âAMC) = ar(âABM) + ar(âAMC)
ar(âADM) = ar(ABMC)
Hence Proved
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 ) = × ar(parallelogram ABCD )
Proof:
In âABC, PQ || AC and PQ = × AC [ By midpoint theorem]
Again, in âDAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = × AC [ By midpoint theorem]
Now, PQ || AC and SR || AC
⇒ âPQ || SR
Also, PQ = SR = × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.
Now, ar(parallelogramâ PQRS) = ar(âPSQ) + ar(âSRQ) ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD) ...(ii)
âPSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(âPSQ ) = × ar(parallelogram ABQS) ...(iii)
Similarly, âSRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(âSRQ ) = × ar(parallelogram SQCD) ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
ar(parallelogramâ PQRS) = × ar(parallelogram ABQS)â + × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogramâ PQRS) = × ar(parallelogram ABCD)
Page No 390:
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 ) = × ar(parallelogram ABCD )
Proof:
In âABC, PQ || AC and PQ = × AC [ By midpoint theorem]
Again, in âDAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = × AC [ By midpoint theorem]
Now, PQ || AC and SR || AC
⇒ âPQ || SR
Also, PQ = SR = × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.
Now, ar(parallelogramâ PQRS) = ar(âPSQ) + ar(âSRQ) ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD) ...(ii)
âPSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(âPSQ ) = × ar(parallelogram ABQS) ...(iii)
Similarly, âSRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(âSRQ ) = × ar(parallelogram SQCD) ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
ar(parallelogramâ PQRS) = × ar(parallelogram ABQS)â + × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogramâ PQRS) = × ar(parallelogram ABCD)
Answer:
Figure
CF is median of ABC.
ar(BCF) = (ABC) .....(1)
Similarly, BE is the median of ABC,
ar(ABE) = (ABC) .....(2)
From (1) and (2) we have
ar(BCF) = ar(ABE)
ar(BCF) ar(BFG) = ar(ABE) ar(BFG)
ar(âBCG) = ar(AFGE)
Page No 391:
Question 30:
Figure
CF is median of ABC.
ar(BCF) = (ABC) .....(1)
Similarly, BE is the median of ABC,
ar(ABE) = (ABC) .....(2)
From (1) and (2) we have
ar(BCF) = ar(ABE)
ar(BCF) ar(BFG) = ar(ABE) ar(BFG)
ar(âBCG) = ar(AFGE)
Answer:
Given: D is a point on BC of âABC, such that BD = DC
To prove: ar(âABD) = ar(âABC)
Construction: Draw AL ⊥ BC.
Proof:
In âABC, we have:
BC = BD + DC
⇒â BDâ + 2 BD = 3 × BD
Now, we have:
ar(âABD)â = â ×â BD ×â AL
ar(âABC)â = â ×â BC ×â AL
⇒ ar(âABC) = ×â 3BD ×â AL = 3 ×â
⇒ ar(âABC)â = 3 × ar(âABD)
∴ âar(âABD) = ââar(âABC)
Page No 391:
Question 31:
Given: D is a point on BC of âABC, such that BD = DC
To prove: ar(âABD) = ar(âABC)
Construction: Draw AL ⊥ BC.
Proof:
In âABC, we have:
BC = BD + DC
⇒â BDâ + 2 BD = 3 × BD
Now, we have:
ar(âABD)â = â ×â BD ×â AL
ar(âABC)â = â ×â BC ×â AL
⇒ ar(âABC) = ×â 3BD ×â AL = 3 ×â
⇒ ar(âABC)â = 3 × ar(âABD)
∴ âar(âABD) = ââar(âABC)
Answer:
E is the midpoint of CA.
So, AE = EC .....(1)
Also, BD = 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) = .....(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(âABC) = 2ar(âDBC)
Page No 391:
Question 32:
E is the midpoint of CA.
So, AE = EC .....(1)
Also, BD = 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) = .....(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(âABC) = 2ar(âDBC)
Answer:
Given: ABCDE is a pentagon. EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) = ar( âDGF)
Proof:
ar(pentagon ABCDE )â = ar(âDBC) + ar(âADE ) + ar(âABD) ...(i)
Also, ar(âDGF) = ar(âDBF) + ar(âADG) + ar(âABD ) ...(ii)
Now, âDBC and âDBF lie on the same base and between the same parallel lines.
∴ ar(âDBC) = ar(âDBF) ...(iii)
Similarly, âADE and âADG lie on same base and between the same parallel lines.
∴ ar(âADE) = ar(âADG) ...(iv)
From (iii) and (iv), we have:
ar(âDBC) + ar(âADE) = ar(âDBF) + ar(âADG)
Adding ar(âABD) on both sides, we get:
ar(âDBC) + ar(âADE) + ar(âABD) = ar (âDBF) + ar(âADG) + ar(âABD)
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) = ar(âDGF)
Page No 391:
Question 33:
Given: ABCDE is a pentagon. EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) = ar( âDGF)
Proof:
ar(pentagon ABCDE )â = ar(âDBC) + ar(âADE ) + ar(âABD) ...(i)
Also, ar(âDGF) = ar(âDBF) + ar(âADG) + ar(âABD ) ...(ii)
Now, âDBC and âDBF lie on the same base and between the same parallel lines.
∴ ar(âDBC) = ar(âDBF) ...(iii)
Similarly, âADE and âADG lie on same base and between the same parallel lines.
∴ ar(âADE) = ar(âADG) ...(iv)
From (iii) and (iv), we have:
ar(âDBC) + ar(âADE) = ar(âDBF) + ar(âADG)
Adding ar(âABD) on both sides, we get:
ar(âDBC) + ar(âADE) + ar(âABD) = ar (âDBF) + ar(âADG) + ar(âABD)
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) = ar(âDGF)
Answer:
(Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
Hence Proved
Page No 391:
Question 34:
(Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
Hence Proved
Answer:
Given: D is a point on BC of â ABC, such that BD : DC = m : n
To prove: ar(âABD) : ar(âADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(âABD)â = ×â BD ×â AL ...(i)
ar(âADC)â = â ×â DC ×â AL ...(ii)
Dividing (i) by (ii), we get:
∴ ar(âABD) : ar(âADC) = m : n
Page No 391:
Question 35:
Given: D is a point on BC of â ABC, such that BD : DC = m : n
To prove: ar(âABD) : ar(âADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(âABD)â = ×â BD ×â AL ...(i)
ar(âADC)â = â ×â DC ×â AL ...(ii)
Dividing (i) by (ii), we get:
∴ ar(âABD) : ar(âADC) = m : n
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 =
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) : ar(MNBA) = (a + 3b) : (3a + b)
Page No 391:
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 =
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) : ar(MNBA) = (a + 3b) : (3a + b)
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 =
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(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So,
Page No 391:
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 =
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(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So,
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, .....(1)
Similarly, .....(2)
From (1) and (2) we have
PA = AQ
âABQ and âACP are on same base PQ and between same parallels PQ and BC
ar(âABQ) = ar(âACP)
Page No 392:
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, .....(1)
Similarly, .....(2)
From (1) and (2) we have
PA = AQ
âABQ and âACP are on same base PQ and between same parallels PQ and BC
ar(âABQ) = ar(âACP)
Answer:
In âRSC and âPQB
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, âRSC âPQB (AAS congruency)
Thus, ar(âRSC) = ar(âPQB)
Page No 395:
Question 1:
In âRSC and âPQB
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, âRSC âPQB (AAS congruency)
Thus, ar(âRSC) = ar(âPQB)
Answer:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Page No 396:
Question 2:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Answer:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) âââParallelogram ABPQ
Page No 396:
Question 3:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) âââParallelogram ABPQ
Answer:
(a) triangles of equal areas
Page No 396:
Question 4:
(a) triangles of equal areas
Answer:
(c)114 cm2
ar (quad. ABCD) = ar (â ABC) + ar (â ACD)
In right angle triangle ACD, we have:
AC =
In right angle triangle ABC, we have:
BC =
Now, we have the following:
ar(âABC) = × 12 × 9 = 54 cm2
ar(âADC) = × 15 × 8 = 60 cm2
ar(quad. ABCD) =â 54 + 60 = 114 cm2
Page No 396:
Question 5:
(c)114 cm2
ar (quad. ABCD) = ar (â ABC) + ar (â ACD)
In right angle triangle ACD, we have:
AC =
In right angle triangle ABC, we have:
BC =
Now, we have the following:
ar(âABC) = × 12 × 9 = 54 cm2
ar(âADC) = × 15 × 8 = 60 cm2
ar(quad. ABCD) =â 54 + 60 = 114 cm2
Answer:
(c)124 cm2
In the right angle triangle BEC, we have:
EC =
ar(trapez. ABCD) = cm2
Page No 396:
Question 6:
(c)124 cm2
In the right angle triangle BEC, we have:
EC =
ar(trapez. ABCD) = cm2
Answer:
(c) 34 cm2
ar(parallelogram ABCD) = base × height = 5 â× 6.8 = 34 cm2
Page No 396:
Question 7:
(c) 34 cm2
ar(parallelogram ABCD) = base × height = 5 â× 6.8 = 34 cm2
Answer:
(d) 13 cm2
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of âOAB = ⨯ ar(||gm ABCD)
⇒ ar(âOAB) = ââ ⨯â 52 = 13 cm2
â
Page No 396:
Question 8:
(d) 13 cm2
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of âOAB = ⨯ ar(||gm ABCD)
⇒ ar(âOAB) = ââ ⨯â 52 = 13 cm2
â
Answer:
(a) 40 cm2
ar(||gm ABCD) = base × height = 10 â× 4 = 40 cm2
Page No 397:
Question 9:
(a) 40 cm2
ar(||gm ABCD) = base × height = 10 â× 4 = 40 cm2
Answer:
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:
Question 10:
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).
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:
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).
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 âBMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(âBMP) = ar(||gm ABPQ) .....(2)
From (1) and (2) we have
ar(âBMP) = ar(||gm ABCD)
Thus, the given statement is true.
Hence, the correct answer is option (a).
Page No 397:
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 âBMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(âBMP) = ar(||gm ABPQ) .....(2)
From (1) and (2) we have
ar(âBMP) = ar(||gm ABCD)
Thus, the given statement is true.
Hence, the correct answer is option (a).
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
Hence, the correct answer is option (a).
Page No 397:
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
Hence, the correct answer is option (a).
Answer:
(b) 96 cm2
Area of the rhombus = × product of diagonals = ×â 12 â× 16 = 96 cm2
Page No 397:
Question 14:
(b) 96 cm2
Area of the rhombus = × product of diagonals = ×â 12 â× 16 = 96 cm2
Answer:
(c) 65 cm2
Area of the trapezium = × (sum of parallel sides) × distance between them
= ×â ( 12 + 8) â× 6.5
= 65 cm2
Page No 397:
Question 15:
(c) 65 cm2
Area of the trapezium = × (sum of parallel sides) × distance between them
= ×â ( 12 + 8) â× 6.5
= 65 cm2
Answer:
(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 = × (sum of parallel sides) × distance between them
= ×â ( 13 + 7) â× 4
= 40 cm2
Page No 397:
Question 16:
(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 = × (sum of parallel sides) × distance between them
= ×â ( 13 + 7) â× 4
= 40 cm2
Answer:
(b)128 cm2
ar(quad ABCD) = ar (â ABD) + âar (â DBC)
â
We have the following:
ar(âABD) = × base â× height â= â× 16 â× 9 = 72 cm2
ar(âDBC) = × base â× height â= â × 16 â× 7 = 56 cm2
∴ ar(quad ABCD) =â 72 + 56 = 128 cm2
Page No 398:
Question 17:
(b)128 cm2
ar(quad ABCD) = ar (â ABD) + âar (â DBC)
â
We have the following:
ar(âABD) = × base â× height â= â× 16 â× 9 = 72 cm2
ar(âDBC) = × base â× height â= â × 16 â× 7 = 56 cm2
∴ ar(quad ABCD) =â 72 + 56 = 128 cm2
Answer:
(a)
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
⇒â 2xo = 120o
⇒â xo = 60o
i.e., ∠BCD = ∠BDC = ∠DBC = 60o
So, ââBCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in âOAB, we have:
Page No 398:
Question 18:
(a)
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
⇒â 2xo = 120o
⇒â xo = 60o
i.e., ∠BCD = ∠BDC = ∠DBC = 60o
So, ââBCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in âOAB, we have:
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 ABFEâ) = ar(||gm ABCDâ) = 25 cm2
⇒ ar(âBCF ) = ar(||gm ABFEâ)â ar(quad EABCâ)â = ( 25 17) = 8 cm2
Page No 398:
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 ABFEâ) = ar(||gm ABCDâ) = 25 cm2
⇒ ar(âBCF ) = ar(||gm ABFEâ)â ar(quad EABCâ)â = ( 25 17) = 8 cm2
Answer:
(b) 1:4
âABC and âBDE are two equilateral trianglesâ and D is the midpoint of BC.
Let AB = BC = AC = a
Then BD = BE = ED =
∴
So, required ratio = 1 : 4
Page No 398:
Question 20:
(b) 1:4
âABC and âBDE are two equilateral trianglesâ and D is the midpoint of BC.
Let AB = BC = AC = a
Then BD = BE = ED =
∴
So, required ratio = 1 : 4
Answer:
(a) 8 cm2
Let the distance between AB and CD be h cm.
Then ar(||gm APQDâ) = AP ×â h
= ×â AB ×âh (AP = AB )
= ×â ar(||gm ABCD) [ ar(|| gm ABCD) = AB ×âh )
∴ ar (||gm APQDâ) = ×â 16 = 8 cm2
Page No 398:
Question 21:
(a) 8 cm2
Let the distance between AB and CD be h cm.
Then ar(||gm APQDâ) = AP ×â h
= ×â AB ×âh (AP = AB )
= ×â ar(||gm ABCD) [ ar(|| gm ABCD) = AB ×âh )
∴ ar (||gm APQDâ) = ×â 16 = 8 cm2
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 PQRSâ) = ×â product of diagonals = ×â 8 ×â 6 = 24 cm2
Page No 398:
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 PQRSâ) = ×â product of diagonals = ×â 8 ×â 6 = 24 cm2
Answer:
(c) ar (â ABC )
Since D is the mid point of BC, AD is a median of âABC and BE is the median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âABD ) = ar(âABC) ...(i)
⇒ ar(âBED) = â ar(âABD) ...(ii)
From (i) and (ii), we have:
ar(âBED) = ⨯ â ⨯â ar(âABC)
∴â ar(âBED)â = ⨯ ar(âABC)
ar(âABC)
Page No 399:
Question 23:
(c) ar (â ABC )
Since D is the mid point of BC, AD is a median of âABC and BE is the median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âABD ) = ar(âABC) ...(i)
⇒ ar(âBED) = â ar(âABD) ...(ii)
From (i) and (ii), we have:
ar(âBED) = ⨯ â ⨯â ar(âABC)
∴â ar(âBED)â = ⨯ ar(âABC)
ar(âABC)
Answer:
(a)
Since E is the midpoint of AD, BE is a median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âBED) = ⨯ ar(âABD) ...(i)
Since E is the midpoint of AD, CE is a median of âADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âCED ) = ⨯
Adding (i) and (ii), we have:
ar(âBED ) + ar(âCED ) = ⨯
⇒ ar (â BEC ) =
Page No 399:
Question 24:
(a)
Since E is the midpoint of AD, BE is a median of âABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âBED) = ⨯ ar(âABD) ...(i)
Since E is the midpoint of AD, CE is a median of âADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âCED ) = ⨯
Adding (i) and (ii), we have:
ar(âBED ) + ar(âCED ) = ⨯
⇒ ar (â BEC ) =
Answer:
(d) ar (â 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 âABC.
E is the midpoint of BC, so AE is the median of âABD. O is the midpoint of AE, so BO is median of âABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âABD ) = ⨯ â
ar(âABE ) =â ⨯â ar(âABD) ...(ii)
ar(âBOE) = â⨯â ar(âABE) ...(iii)
From (i), (ii) and (iii), we have:
ar(âBOE ) =â ar(âABE)
ar(âBOE ) = ⨯â â ⨯â â ⨯â ar(âABC)â
∴ââ ar(âBOE )â = ar(âABC)
Page No 399:
Question 25:
(d) ar (â 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 âABC.
E is the midpoint of BC, so AE is the median of âABD. O is the midpoint of AE, so BO is median of âABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(âABD ) = ⨯ â
ar(âABE ) =â ⨯â ar(âABD) ...(ii)
ar(âBOE) = â⨯â ar(âABE) ...(iii)
From (i), (ii) and (iii), we have:
ar(âBOE ) =â ar(âABE)
ar(âBOE ) = ⨯â â ⨯â â ⨯â ar(âABC)â
∴ââ ar(âBOE )â = ar(âABC)
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 = × area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram = : 1 = 1 : 2
Page No 399:
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 = × area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram = : 1 = 1 : 2
Answer:
(c) (3a +b) : (a +3b)
Clearly, EF = (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:
Question 27:
(c) (3a +b) : (a +3b)
Clearly, EF = (a + b) [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.
Answer:
(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas.
Page No 399:
Question 28:
(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas.
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:
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
Answer:
(b) 40 cm2
Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm
Page No 400:
Question 30:
(b) 40 cm2
Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm
Answer:
(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(âAOB) = ar(âCOD).
Consider âADB and âADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(âADB) ar(âADC)
Subtracting ar(âAOD) from both sides, we get:
ar(âADB) ar(âAOD) ar(âADC) ar(âAOD)
Or ar(â AOB) ar(â COD)
Page No 400:
Question 31:
(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(âAOB) = ar(âCOD).
Consider âADB and âADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(âADB) ar(âADC)
Subtracting ar(âAOD) from both sides, we get:
ar(âADB) ar(âAOD) ar(âADC) ar(âAOD)
Or ar(â AOB) ar(â COD)
Answer:
(b)
Area of a parallelogram = âbase â× corresponding height
Page No 400:
Question 32:
(b)
Area of a parallelogram = âbase â× corresponding height
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
AB × DE = AD × BF
10 × 6 = 8 × AD
AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.
Page No 401:
Question 33:
(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.
Answer:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
In trapezium ABCD, âABC and âABD are on the same base and between the same parallel lines.
∴ ar(âABC) = ar(âABD)
⇒ ar(âABC) ar(âAOB) = ar(âABD) ar(âAOB)
⇒ âar(âBOC) = ar(âAOD)
∴ Assertion (A) is true and, clearly, reason (R) gives (A).
Page No 401:
Question 34:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
In trapezium ABCD, âABC and âABD are on the same base and between the same parallel lines.
∴ ar(âABC) = ar(âABD)
⇒ ar(âABC) ar(âAOB) = ar(âABD) ar(âAOB)
⇒ âar(âBOC) = ar(âAOD)
∴ Assertion (A) is true and, clearly, reason (R) gives (A).
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)â 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
⇒â2x = 120
⇒â x = 60
∴ ∠BCD = ∠BDC = ∠DBC = 60
So, ââBCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB =
Now, in â OAB, we have:
Thus, assertion (A)â is also true, but reason (R) does not give (A).
âHence, the correct answer is (b).
Page No 401:
Question 35:
(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)â 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
⇒â2x = 120
⇒â x = 60
∴ ∠BCD = ∠BDC = ∠DBC = 60
So, ââBCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB =
Now, in â OAB, we have:
Thus, assertion (A)â is also true, but reason (R) does not give (A).
âHence, the correct answer is (b).
Answer:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Page No 401:
Question 36:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
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(âABC ) =
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:
Question 37:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not a correct explanation of Assertion (A).
Explanation:
Reason (R):
∴ ar(âABC ) =
Thus, reason (R) is true.
Assertion (A):
Area of trapezium =
Thus, assertion (A) is true, but reason (R) does not give assertion (A).
Answer:
(d) Assertion is false and Reason is true.
Clearly, reason (R) is true.
Assertion: Area of a parallelogram = base × height
AB ×â DE = AD × BF
AD = (16 × 8) ÷ 10 = 12.8 cm
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