give me the solutions of the exercise 8.1
Asked by paddipawar
| 2nd Dec, 2014,
09:22: PM
Expert Answer:
12.
Extend AB. Draw a line through C, which is parallel to AD, intersecting AE at point E.
Now, AECD is a parallelogram.
(i) AD = CE (opposite sides of parallelogram AECD)
But AD = BC (given)
So, BC = CE
CEB =
CBE (angle opposite to equal sides are also equal)
Now consider parallel lines AD and CE. AE is transversal line for them
A +
CEB = 180 (angles on the same side of transversal)
A+
CBE = 180 (using the relation
CEB =
CBE) ... (1)
But
B +
CBE = 180 (linear pair angles) ... (2)
From equations (1) and (2), we have
A =
B
(ii) AB || CD
A +
D = 180 (angles on the same side of transversal)
Also
C +
B = 180 (angles on the same side of transversal)
A +
D =
C +
B
But
A =
B [using the result obtained proved in (i)]
C =
D
(iii) In ABC and BAD
AB = BA (common side)
BC = AD (given)
B =
A (proved before)
ABC
BAD (SAS congruence rule)
(iv)
ABC
BAD
AC = BD (by CPCT)
11.
(i) Here AB = DE and AB || DE.
Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be
a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.
(ii) Again BC = EF and BC || EF.
Therefore, quadrilateral BEFC is a parallelogram.
(iii) Here ABED and BEFC are parallelograms.
AD = BE, and AD || BE
(Opposite sides of parallelogram are equal and parallel)
And BE = CF, and BE || CF
(Opposite sides of parallelogram are equal and parallel)
AD = CF, and AD || CF
(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and
parallel to each other,
so, it is a parallelogram.
(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each
other.
AC || DF and AC = DF
(vi)
ABC and
DEF.
AB = DE (given)
BC = EF (given)
AC = DF (ACFD is a parallelogram)

ABC
DEF (by SSS congruence rule)
10.
(i) In
APB and
CQD
APB =
CQD (each 90o)
AB = CD (opposite sides of parallelogram ABCD)
ABP =
CDQ (alternate interior angles for AB || CD)
APB
CQD (by AAS congruency)
(ii) By using the result obtained as above
APB
CQD, we have
AP = CQ (by CPCT)
9.
(i) In
APD and
CQB
ADP =
CBQ (alternate interior angles for BC || AD)
AD = CB (opposite sides of parallelogram ABCD)
DP = BQ (given)

APD
CQB (using SAS congruence rule)
(ii) As we had observed that
APD
CQB
AP = CQ (CPCT)
(iii) In
AQB and
CPD
ABQ =
CDP (alternate interior angles for AB || CD)
AB = CD (opposite sides of parallelogram ABCD)
BQ = DP (given)

AQB CPD (using SAS congruence rule)
(iv) As we had observed that
AQB
CPD
AQ = CP (CPCT)
(v) From the result obtained in (ii) and (iv), we have
AQ = CP and AP = CQ
Since opposite sides in quadrilateral APCQ are equal to each other. So, APCQ is a
parallelogram.
You have not mentioned the book for which you want to get the solutions
We have provided solutions of 9,10,11,12 of Ex.8.1 of NCERT.
If you need all the solutions, please enroll for textbook solutions from topper's site.
Extend AB. Draw a line through C, which is parallel to AD, intersecting AE at point E.
Now, AECD is a parallelogram.
Now, AECD is a parallelogram.
(i) AD = CE (opposite sides of parallelogram AECD)
But AD = BC (given)
So, BC = CE
CEB =
CBE (angle opposite to equal sides are also equal)
Now consider parallel lines AD and CE. AE is transversal line for them
A +
CEB = 180 (angles on the same side of transversal)
A+
CBE = 180 (using the relation
CEB =
CBE) ... (1)
But
B +
CBE = 180 (linear pair angles) ... (2)
From equations (1) and (2), we have
A =
B
(ii) AB || CD
A +
D = 180 (angles on the same side of transversal)
Also
C +
B = 180 (angles on the same side of transversal)
A +
D =
C +
B
But AD = BC (given)
So, BC = CE


Now consider parallel lines AD and CE. AE is transversal line for them






But


From equations (1) and (2), we have


(ii) AB || CD


Also






But




(iii) In ABC and BAD
AB = BA (common side)
BC = AD (given)




(iv)


11.
(i) Here AB = DE and AB || DE.
Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be
Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be
a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.
(ii) Again BC = EF and BC || EF.
Therefore, quadrilateral BEFC is a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.
(ii) Again BC = EF and BC || EF.
Therefore, quadrilateral BEFC is a parallelogram.
(iii) Here ABED and BEFC are parallelograms.
AD = BE, and AD || BE
(Opposite sides of parallelogram are equal and parallel)
And BE = CF, and BE || CF
(Opposite sides of parallelogram are equal and parallel)
AD = CF, and AD || CF
(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and
AD = BE, and AD || BE
(Opposite sides of parallelogram are equal and parallel)
And BE = CF, and BE || CF
(Opposite sides of parallelogram are equal and parallel)
(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and
parallel to each other,
so, it is a parallelogram.
(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each
so, it is a parallelogram.
(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each
other.
AC || DF and AC = DF
(vi)
ABC and
DEF.
AB = DE (given)
BC = EF (given)
AC = DF (ACFD is a parallelogram)

ABC
DEF (by SSS congruence rule)
(vi)


AB = DE (given)
BC = EF (given)
AC = DF (ACFD is a parallelogram)


10.
(i) In
APB and
CQD
APB =
CQD (each 90o)
AB = CD (opposite sides of parallelogram ABCD)
ABP =
CDQ (alternate interior angles for AB || CD)
APB
CQD (by AAS congruency)
(ii) By using the result obtained as above
APB
CQD, we have
AP = CQ (by CPCT)
9.




AB = CD (opposite sides of parallelogram ABCD)



(ii) By using the result obtained as above


AP = CQ (by CPCT)
9.
(i) In
APD and
CQB
ADP =
CBQ (alternate interior angles for BC || AD)
AD = CB (opposite sides of parallelogram ABCD)
DP = BQ (given)

APD
CQB (using SAS congruence rule)
(ii) As we had observed that
APD
CQB
AP = CQ (CPCT)




AD = CB (opposite sides of parallelogram ABCD)
DP = BQ (given)


(ii) As we had observed that


(iii) In
AQB and
CPD
ABQ =
CDP (alternate interior angles for AB || CD)
AB = CD (opposite sides of parallelogram ABCD)
BQ = DP (given)

AQB CPD (using SAS congruence rule)
(iv) As we had observed that
AQB
CPD
AQ = CP (CPCT)




AB = CD (opposite sides of parallelogram ABCD)
BQ = DP (given)

(iv) As we had observed that


(v) From the result obtained in (ii) and (iv), we have
AQ = CP and AP = CQ
Since opposite sides in quadrilateral APCQ are equal to each other. So, APCQ is a
parallelogram.
You have not mentioned the book for which you want to get the solutions
We have provided solutions of 9,10,11,12 of Ex.8.1 of NCERT.
If you need all the solutions, please enroll for textbook solutions from topper's site.
Answered by Vimala Ramamurthy
| 3rd Dec, 2014,
08:52: AM
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