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Question Bank In Mathematics Class IX (Term II)
8
QUADRILATERALS
A. SUMMATIVE ASSESSMENT
(ii) opposite angles are equal or
8.1 PROPERTIES OF PARALLELOGRAM
1. Sum of the angles of a quadrilateral is
360°.
2. A diagonal of a parallelogram divides it
into the two congruent triangles.
3. In a parallelogram,
(i) opposite sides are equal
(ii) opposite angles are equal
(iii) diagonals bisect each other
4. A quadrilateral is a parallelogram, if
(i) opposite sides are equal or
(iii) diagonals bisect each other or
(iv) a pair of opposite sides is equal and
parallel
5. Diagonals of a rectangle bisect each other
and are equal and vice-versa.
6. Diagonals of a rhombus bisect each other
at right angles and vice-versa.
7. Diagonals of a square bisect each other at
right angles and are equal, and vice-versa.
ER
Proof : In ABC and ABD
AB = AB [Common]
BC = AD
[Opposite sides of a parallelogram]
AC = BD
[Given]
 ABC BAD [SSS congruence]
ABC = BAD …(i) [CPCT]
Since, ABCD is a parallelogram, thus,
ABC + BAD = 180° …(ii)
[Consecutive interior angles]
ABC + ABC = 180°
 2ABC
= 180° [From (i) and (ii)]
 ABC
= BAD = 90°
This shows that ABCD is a parallelogram
one of whose angle is 90°.
Hence, ABCD is a rectangle. Proved.
Q.3. Show that if the diagonals of a
quadrilateral bisect each other at right angles,
then it is a rhombus.
[2011 (T-II)]
Sol. Given : A quadrilateral ABCD, in which
diagonals AC and BD bisect each other at right
angles.
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Q.1. The angles of a quadrilateral are in the
ratio 3 : 5 : 9 : 13. Find all the angles of the
quadrilateral.
[2011 (T-II)]
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TEXTBOOK’S EXERCISE 8.1
Sol. Suppose the measures of four angles are 3x,
5x, 9x and 13x.
 3x + 5x + 9x + 13x = 360°
[Angle sum property of a quadrilateral]
30x= 360° 
x=
3x= 3 × 12° = 36°,
9x= 9 × 12° = 108°,
360
= 12°
30
5x= 5 × 12° = 60°,
13x= 13 × 12° = 156°
 the angles of the quadrilateral are 36°, 60°, 108°
and 156°
Q.2. If the diagonals of a parallelogram are
equal, then show that it is a rectangle. [2010]
Sol. Given : ABCD is a parallelogram in which
AC = BD.
To Prove : ABCD is a rectangle.
1
OAB = OCD [Alternate angles]

AOB COD [AAS congruence]
AO = OC
[CPCT]
Similarly by taking AOD and BOC, we can
show that OB = OD.
In ABC, BAC + BCA = 90° [ B = 90°]
 2BAC = 90° [BAC = BCA, as BC = AD]
 BCA = 45° or BCO = 45°
Similarly, CBO = 45°
In BCO, BCO + CBO + BOC = 180°
90° + BOC = 180° BOC = 90°
BO OC BO AC
Hence, AC = BD, AC  BD, AO = OC and
OB = OD. Proved.
To Prove : ABCD is a rhombus.
Proof : In AOB and BOC
AO = OC
[Diagonals AC and BD bisect each other]
AOB = COB
[Each = 90°]
BO = BO
[Common]
AOB BOC [SAS congruence]
AB = BC
…(i)
[CPCT]
Since, ABCD is a quadrilateral in which
AB = BC
[From (i)]
Hence, ABCD is a rhombus.
[ if the diagonals of a quadrilateral bisect each
other, then it is a parallelogram and opposite sides
of a parallelogram are equal] Proved.
Q.5. Show that if the diagonals
of a
quadrilateral are equal and bisect each other at
right angles, then it is a square.
Sol. Given : A quadrilateral ABCD, in which
diagonals AC and BD are equal and bisect each
other at right angles,
To Prove : ABCD is a square.
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Sol. Given : ABCD is a square in which AC and
BD are diagonals.
To Prove : AC = BD and AC bisects BD at right
angles, i.e. AC  BD. AO = OC, OB = OD
Proof : In ABC and BAD,
AB = AB
[Common]
BC = AD
[Sides of a square]
ABC = BAD = 90°
[Angles of a square]

ABC BAD
[SAS congruence]

AC = BD
[CPCT]
ER
[2011 (T-II)]
S
Q.4. Show that the diagonals of a square are
equal and bisect each other at right angles.
Proof : Since ABCD is a quadrilateral whose
diagonals bisect each other, so it is a
parallelogram. Also, its diagonals bisect each
other at right angles, therefore, ABCD is a
rhombus.
 AB = BC = CD = DA [Sides of a rhombus]
In ABC and BAD, we have
AB = AB
[Common]
BC = AD
[Sides of a rhombus]
AC = BD
[Given]
ABC BAD
[SSS congruence]
 ABC = BAD
[CPCT]
But, ABC + BAD = 180°
[Consecutive interior angles]
Now in AOB and COD,
AB = DC
[Sides of a square]
AOB = COD
[Vertically opposite angles]
2
ABC = BAD = 90°
A = B = C = D = 90°
[Opposite angles of a || gm]
 ABCD is a rhombus whose angles are of 90°
each.
Hence, ABCD is a square. Proved.
Q.6. Diagonal AC of a parallelogram ABCD
bisects A (see Fig.). Show that
(i) it bisects C also,
(ii) ABCD is a rhombus.
[HOTS]
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Similarly, ADB = CDB and ABD = CBD.
Hence, diagonal AC bisects A as well as C and
diagonal BD bisects B as well as D. Proved.
Q.8. ABCD is a rectangle in which diagonal AC
bisects  A as well as C. Show that :
(i) ABCD is a square (ii) diagonal BD
bisects B as well as D.
[HOTS]
Sol. Given : ABCD is a
rectangle in which diagonal
AC bisects A as well as C.
To Prove : (i) ABCD is a square.
(ii) Diagonal BD bisects B as well as D.
Proof : (i) In ABC and ADC, we have
BAC = DAC [Given]
BCA = DCA [Given]
AC = AC
ABC ADC [ASA congruence]
 AB = AD and CB = CD
[CPCT]
But, in a rectangle opposite sides are equal,
i.e., AB = DC and BC = AD
 AB = BC = CD = DA
Hence, ABCD is a square. Proved.
(ii) In ABD and CDB, we have
AD = CD 
[Sides of a square]
 [CPCT]
AB = BC 
BD = BD
[Common]
ABD CBD
[SSS congruence]
So,ABD = CBD  [CPCT]
[CPCT]

ADB = CDB 
Hence, diagonal BD bisects B as well as D.
Proved.
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Given : A parallelogram ABCD, in which
diagonal AC bisects A, i.e., DAC = BAC.
To Prove : (i) Diagonal AC bisects, C
i.e., DCA = BCA
(ii) ABCD is a rhombus.
Proof : (i) DAC = BCA [Alternate angles]
BAC = DCA
[Alternate angles]
But,
DAC = BAC
[Given]

BCA = DCA
Hence, AC bisects DCB
Or, AC bisects C Proved.
(ii) In ABC and CDA
AC = AC
[Common]
BAC = DAC
[Given]
and BCA = DCA
[Proved above]
 ABC ADC
[ASA congruence]
 BC DC
[CPCT]
But AB = DC
[Given]
 AB BC = DC = AD
Hence, ABCD is a rhombus Proved.
Proof : In ABC and CDA, we have
AB = AD
[Sides of a rhombus]
AC = AC
[Common]
BC = CD
[Sides of a rhombus]
ABC ADC [SSS congruence]
So, DAC = BAC 
 [CPCT][CPCT]
BCA = DCA 
Q.7. ABCD is a rhombus. Show that diagonal
AC bisects  A as well as C and diagonal BD
bisects B as well as D.
[2011 (T-II)]
Sol. Given : ABCD is a rhombus, i.e., AB = BC
= CD = DA.
To Prove : DAC = BAC, BCA = DCA,
ADB = CDB, ABD = CBD
3
Sol. Given : ABCD is a parallelogram and AP
and CQ are perpendiculars from vertices A and C
on BD.
To Prove : (i) APB CQD (ii) AP = CQ
Proof : (i) In APB and CQD, we have
ABP = CDQ
[Alternate angles]
AB = CD [Opposite sides of a parallelogram]
APB = CQD
[Each = 90°]
APB CQD
[AAS congruence]
(ii) So, AP = CQ
[CPCT] Proved.
Q.9. In parallelogram ABCD, two points P and
Q are taken on diagonal BD such that DP = BQ
(see Fig.). Show that :
(i)  APD  CQB
[2010]
(ii) AP = CQ
[2010]
(iii)  AQB  CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
ER
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(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi)  ABC   DEF
Sol. Given : In ABC and DEF, AB = DE,
AB || DE, BC = EF and BC || EF. Vertices A, B
and C are joined to vertices D, E and F.
To Prove : (i) ABED is a parallelogram
(ii) BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) ACFD is a parallelogram
(v) AC = DF
(vi) ABC DEF
Proof : (i) In quadrilateral ABED, we have
AB = DE and AB || DE.
[Given]
ABED is a parallelogram.
[One pair of opposite sides is
parallel and equal]
(ii) In quadrilateral BEFC, we have
BC = EF and BC || EF
[Given]
BEFC is a parallelogram.
[One pair of opposite sides is
parallel and equal]
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Sol. Given : ABCD is a parallelogram and P
and Q are points on diagonal BD such that
DP = BQ.
To Prove : (i) APD CQB
(ii) AP = CQ
(iii) AQB CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Proof : (i) In APD and CQB, we have
AD = BC
[Opposite sides of a ||gm]
DP = BQ
[Given]
ADP = CBQ
[Alternate angles]
APD CQB
[SAS congruence]
(ii)  AP = CQ
[CPCT]
(iii) In AQB and CPD, we have
AB = CD
[Opposite sides of a || gm]
DP = BQ
[Given]
ABQ = CDP
[Alternate angles]
AQB CPD
[SAS congruence]
(iv)  AQ = CP
[CPCT]
(v) Since in APCQ, opposite sides are equal,
therefore it is a parallelogram. Proved.
Q.11. In ABC and DEF, AB = DE, AB || DE,
BC = EF and BC || EF. Vertices A, B and C are
joined to vertices D, E and F respectively
(see Fig.). Show that
Q.10. ABCD is a parallelogram and AP and CQ
are perpendiculars from vertices A and C on
diagonal BD (see Fig.). Show that
(i)  APB  CQD (ii) AP = CQ
[2011 (T-II)]
4
(iii) BE = CF and BE || CF
[BEFC is parallelogram]
AD = BE and AD || BE
[ABED is a parallelogram]
 AD = CF and AD || CF
(iv) ACFD is a parallelogram.
[One pair of opposite sides is
parallel and equal]
(v) AC = DF [Opposite sides of parallelogram
ACFD]
(vi) In ABC and DEF, we have
AB = DE
[Given]
BC = EF
[Given]
AC = DF
[Proved above]
ABC DEF [SSS axiom] Proved.
Q.12. ABCD is a trapezium in which AB || CD
and AD = BC (see Fig.).
[2011 (T-II)]
 AE || DC
…(i)
and AD || CE …(ii) [Construction]
ADCE is a parallelogram
[Opposite pairs of sides are parallel
A + E = 180°…(iii)
[Consecutive interior angles]
B + CBE = 180° …(iv)
[Linear pair]
AD = CE
…(v) [Opposite sides of a ||gm]
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AD = BC
…(vi) [Given]
BC = CE
[From (v) and (vi)]
E = CBE …(vii)
[Angles opposite to equal sides]
B + E = 180° …(viii) [From (iv) and (vii)]
Now, from (iii) and (viii) we have
A + E = B + E  A = B Proved.
(ii) A + D = 180° and B + C = 180°
 A + D = B + C D = C
[ A = B]
OrC = D Proved.
(iii) In ABC and BAD, we have
AD = BC
[Given]
A = B
[Proved]
AB = AB
[Common]
ABC BAD [SAS congruence]
(iv) diagonal AC = diagonal BD
[CPCT]
Proved.
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Show that
(i)  A =  B
(ii) C = D
(iii)  ABC   BAD
(iv) diagonal AC = diagonal BD
Sol. Given : In trapezium ABCD, AB || CD and
AD = BC.
To Prove :
(i) A = B
(ii) C = D
(iii) ABC BAD
(iv) diagonal AC = diagonal BD
Constructions : Join AC and
BD. Extend AB and draw a
line through C parallel to DA
meeting AB produced at E.
Proof : (i) Since AB || DC
OTHER IMPORTANT QUESTIONS
Q.1. Two consecutive angles of a parallelogram
are in the ratio 1 : 3. Then the smaller angle is :
(a) 50° (b) 90° (c) 60° (d) 45° [2011 (T-II)]
Sol. (d) Let the two consecutive angles are x°
and 3x°. Since, the sum of two consecutive
angles of a parallelogram is 180°.
Therefore, x° + 3x° = 180°
 4x° = 180°  x° = 45°
∴ the smaller angle is 45°.
Q.2. A quadrilateral is a parallelogram if :
(a) both pairs of opposite sides are equal
(b) both pairs of opposite angles are equal
(c) the diagonals bisect each other
(d) all of these
Sol. (d) A quadrilateral will be a
parallelogram for all of the given properties.
5
Sol. (a) x + 110° = 180°
Q.3. In the given figure, PQRS is a
parallelogram in which PSR = 130°. RQT is
equal to :
[2010]
Also, ADC = BCE = 110°
(Corresponding angles)
y
=
110°

∴ x = 70° and y = 110°
Q.8. In the given figure, ABCD is a rhombus.
If A = 80°, then CDB is equal to : [2010]
(a) 80°
(b) 90°
(c) 50° (d) 100°
Sol. (c) In ∆CDB, we have CD = CB
 CBD = CDB = x
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∴ CDB + CBD + DCB = 180°
 x + x + 80° = 180°  2x = 180° – 80°
 2x = 100°  x = 50°
Q.9. The diagonals of a rhombus are 12 cm
and 16 cm. The length of the side of the rhombus
is :
[2011 (T-II)]
(a) 10 cm (b) 12 cm (c) 8 cm (d) 16 cm
Sol. (a) In right ∆AOB, we have
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(a) 60°
(b) 50° (c) 70°
(d) 130°
Sol. (b) We have,
 S = Q = 130°
[Opposite angles of a parallelogram]
130° + RQT = 180°
[By linear pair]
    RQT = 180° – 130° = 50°
Q.4. If three angles of a quadrilateral are 100°,
75° and 105°, then the measure of the fourth angle
is :
[2010]
(a) 110° (b) 100° (c) 90° (d) 80°
Sol. (d) We know that the sum of the four
angles of a quadrilateral is 360°. Then, we have
100° + 75° + 105° + x° = 360° (Let the fourth
angle be x°)
 280° + x = 360°  x = 360° – 280° = 80°
Q.5. In a parallelogram sum of its two
adjacent angles is :
[2010]
(c) 90°
(b) 360° (c) 180° (d) 270°
Sol. (c) The sum of two adjacent angles of a
parallelogram is 180°.
Q.6. If the diagonals of a parallelogram
bisect each other at right angles then it is a :
(Linear pair)
 x = 180° – 110° = 70°
[2010]
(a) Rhombus
(b) Rectangle
(c) Trapezium
(d) None of these
Sol.
(a) The diagonals of a rhombus bisect
each other at right angles.
Q.7. In the figure, ABCD is a parallelogram.
The values of x and y are respectively :
(a) 70°, 110°
(c) 110°, 70°
AB2 = AO2 + OB2
= 82 + 62 = 100
 AB2 = 100
 AB = 10 cm
Q.10. If PQRS is a parallelogram, then
Q – S is equal to :
[2011 (T-II)]
(a) 90°
(b) 120° (c) 180°
(d) 0°
Sol. (d) Q and S are opposite angles of
parallelogram PQRS. Since opposite angles of a
parallelogram are equal therefore, Q – S = 0
(b) 70°, 70°
(d) 70°, 40°
6
Q.11. Two adjacent angles of a rhombus are
3x – 40° and 2x + 20°. The measurement of the
greater angle is :
[2011 (T-II)]
(a) 160° (b) 100° (c) 80° (d) 120°
Sol. (b) Since adjacent angles of a rhombus
are supplementary.
 3x – 40° + 2x + 20° = 180°
 5x – 20° = 180°  5x = 200°  x = 40°
 greater angle = 2x + 20° = 2 × 40° + 20°
= 100°
Q.12. In the given figure, ABCD is a
parallelogram in which DAC = 40°; BAC =
30°, DOC = 105°, then CDO equals :
(a) 360°
(b) 200° (c) 180° (d) 160°
Sol. (d) A = 180° – 100° = 80°
[adjacent angles of a parallelogram]
C = A = 80° [opposite angles]
 A + C = 80° + 80° = 160°
Q.15. Two adjacent angles of a parallelogram
are (2x + 30)° and (3x + 30)°. The value of x is :
[2011 (T-II)]
S
[2011 (T-II)]
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(a) 75°
(b) 70°
(c) 45°
(d) 85°
Sol. (c) In ODC, DOC = 105° (given)
OCD = 30°
(alternate angles)
 CDO + DOC + OCD = 180°
 CDO = 180° – 105° – 30° = 45°
Q.13. ABCD is a rhombus such that ACB =
40° then ADC is :
[2011 (T-II)]
(a) 40°
(b) 45°
(c) 100° (d) 60°
Sol. (c) DAC = 40°
(alternate angles)
Since, AD = CD, therefore,
[2011 (T-II)]
(a) 30°
(b) 60°
(c) 24°
(d) 36°
Sol. Since, adjacent angles of a parallelogram
are supplementary.
 2x + 30° + 3x + 30° = 180°
 5x = 180° – 60° = 120°
 5x = 120°   x = 24°
Q.16. ABCD is a quadrilateral and AP and DP
are bisectors of A and D. The value of x is :
(a) 60°
(b) 85°
Sol. (c) x = APD
= 180° –
= 180° –
(c) 95°
(d) 100°
1
{360° – (130° + 60°)}
2
1
(360° – 190°) = 180° – 85° = 95°
2
Q.17. In quadrilateral PQRS, if P = 60° and
Q : R : S = 2 : 3 : 7, then S is equal to :
DAC = ACD = 40°
 ADC = 180° – 40° – 40° = 100°
[2011 (T-II)]
(a) 175°
(b) 135° (c) 150° (d) 210°
Sol. (a) 60° + 2x + 3x + 7x = 360°
 12x = 300°  x = 25°
 S = 7x = 7 × 25° = 175°
Q.14. In the given figure, ABCD is a
parallelogram. If B = 100°, then A + C is
equal to :
[2011 (T-II)]
7
Sol. (c) In ∆AOB, AOB = 118°
( sum of the angles of triangle is 180°)
Also, AOB = COD = 118°
(Vertically opposite angles)
Q.23. In the given figure, PQRS is a rectangle.
If RPQ = 30°, then the value of (x + y) is :
Q.18. In the given figure, ABCD is a rhombus
in which diagonals AC and BD intersect at O.
Then mAOB is
[2011 (T-II)]
[2010, 2011 (T-II)]
(a) 60°
(b) 80°
(c) 90°
(d) 45°
Sol. (c) Since, diagonals of a rhombus bisect
each other at right angles.
Q.19. Two angles of a quadrilateral are 50°
and 80° and other two angles are in the ratio
8 : 15, then the remaining two angles are :
(a) 90° (b) 120° (c) 150° (d) 180°
Sol. (d) Since, diagonals of a rectangle bisect
each other, therefore,
R = S
and R = P = 30° [Alternate angles]
 S = 30°

y = 180° – (30° + 30°) = 120°
Similarly, in ORQ, 2x + 60° = 180°
 x = 60°
 x + y = 120° + 60° = 180°
Q.24. In the given figure, ABCD is a rectangle
in which APB = 100°. The value of x is :
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(a) 140°, 90°
(b) 100°, 130°
(c) 80°, 150°
(d) 70°, 160°
Sol. (c) 50° + 80° + 8x + 15x = 360°
 23x = 360° – 130°  23x = 230°  x = 10°
 remaining two angles are 80° and 150°
Q.20. In a quadrilateral ABCD, if A = 80°,
B = 70°, C = 130°, then D is : [2011 (T-II)]
(a) 80°
(b) 70°
(c) 130° (d) 150°
Sol. (a) D = 360° – (80 + 70° + 130°)
= 360° – 280° = 80°
Q.21. The diagonals of a parallelogram PQRS
intersect at O. If QOR = 90° and QSR = 50°,
then ORS is :
[2011 (T-II)]
(a) 90°
(b) 40°
(c) 70°
(d) 50°
Sol. (b) ORS = 180° – (90° + 50°)
= 180° – 140° = 40°
S
[2011 (T-II)]
(a) 40° (b) 50° (c) 60° (d) 70°
Sol. (b) In PBC, BPC = 180° – 100° = 80°
and B = C = x.
 2x + 80° = 180°  x = 50°
Q.25. If the length of the diagonal of a square
is 8 cm, then its area is :
[2010]
Q.22. In the given figure, the measure of
DOC is equal to :
(a) 90°
(b) 180°
(c) 118°
[2010]
(a) 64 cm2 (b) 32 cm2 (c) 16 cm2 (d) 48 cm2
(d) 62°
8
(c) x + y = 2 (∠1 + ∠2 + ∠3 + ∠4)
(d) none of these
Sol. (a) In ∆ABD, x = 1 + 2 … (i)
[exterior angles of a triangle]
In ∆BCD, y = 3 + 4 … (ii)
[exterior angles of a triangle]
Adding (i) and (ii), we get
x + y = 1 + 2 + 3 + 4
Q.29. In the given figure, PQRS is a
parallelogram in which PT and QT are angle
bisectors of P and Q respectively. Measure of
PTQ is :
[2010]
Sol. (b) Let the side of square is a, then
diagonal a 2 = 8
 a2 =
64
= 32
2
( area of the square = a2)
Q.26. In the given figure, ABCD is a rhombus.
If OAB = 35°, then the value of x is : [2010]
(a) 25° (b) 35° (c) 55° (d) 70°
Sol. (c) Clearly, AOB = 90°
( diagonals of a rhombus bisect
each other at right angles)
 x = 180°– (35° + 90°) = 180° – 125° = 55°
Q.27. In the given figure, ABCD is a rhombus.
If ABD = 40°, then the value of y is : [2010]
(a) 90° (b) 60° (c) 80° (d) 100°
Sol. (a) Since adjacent angles of a
parallelogram are supplementary.
ER
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 P + Q = 180° 
In PTQ,
1
(P + Q) = 90°
2
1
1
P + Q + PTQ = 180°
2
2
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 PTQ = 90°
Q.30. If the two adjacent angles of a
parallelogram are (3x – 20)° and (50 – x)° then
value of x is :
[2010]
(a) 55° (b) 75° (c) 20° (d) 80°
Sol. (b) (3x – 20)° + (50 – x)° = 180°
[Adjacent angles of a parallelogram are
supplementary]
 2x + 30° = 180°  2x = 150°  x = 75°
(a) 40° (b) 50° (c) 80° (d) 100°
Sol. (d) Since, AB = AD, therefore,
B = D = 40°
 y = 180° – (40° + 40°) = 100°
Q.28. The sides BA and DC of a quadrilateral
ABCD are produced as shown in the figure. Then
which of the following relations is true?
[HOTS]
Q.31. In the given figure, ABCD is a rhombus.
The value of x is :
[2010]
(a) 60°
(b) 20° (c) 30°
Sol. (c) In ODC, we have
x + 2x + 90° = 180°
(a) x + y = ∠1 + ∠2 + ∠3 + ∠4
(b) x – y = ∠1 + ∠2 + ∠3 + ∠4
9
(d) 40°
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Sol. Since ABCD is a parallelogram.
Therefore, AB || DC and AD || BC.
Now, AB || DC and transversal BD intersect
them.
 BDC = ABD
[ alternate angles are equal]
 8x = 32°  x = 4 and AD || BC and
transversal BD intersect them.
 ADB = DBC  9y = 27°  y = 3
Hence, x = 4, y = 3.
Q.34. If the angles of quadrilateral are in the
ratio 1 : 2 : 3 : 4. Find measure of all the angles
of quadrilateral.
[2011 (T-II)]
Sol. We have,
A : B : C : D = 1 : 2 : 3 : 4.
So, let A = x°, B = 2x°, C = 3x°, D = 4x°,
 A + B + C + D = 360°
 x° + 2x° + 3x° + 4x° = 360°
 10x° = 360°  x = 36
Thus, the angles are :
A = 36°, B = (2 × 36)° = 72°, C = (3 × 36)°
= 108° and D = (4x)° = (4 × 36)° = 144°
Q.35. The sides BA and DC of quadrilateral
ABCD are produced as shown in figure. Prove
that x + y = a + b.
[2011 (T-II)]
Sol. Join BD.
In ABD, we have
ABD + ADB = b
... (i)
In CBD, we have
CBD + CDB = a
... (ii)
Adding (i) and (ii), we get
(ABD + CBD) + (ADB + CDB) = a + b
 x° + y° = a + b
Hence, x + y = a + b
Q.36. Two opposite angles of a parallelogram
are (3x – 2)° and (63 – 2x)°. Find all the angles
of the parallelogram.
[2011 (T-II)]
Sol. Since opposite angles of a parallelogram
are equal.
 3x – 2 = 63 – 2x
 3x + 2x = 63 + 2
 5x = 65  x = 13
Angles are (3 × 13 – 2)° = 37°; 180° – 37° = 143°
 angles are 37°, 143°, 37°, 143°
Q.37. Two opposite angles of a parallelogram
are (3x – 10)° and (2x + 35)°. Find the measure
of all the four angles of the parallelogram.
ER
[ diagonals of a rhombus bisect each
other at right angles]

3x + 90° = 180°

3x = 90°  x = 30°
Q.32. Can the 95°, 70°, 110° and 80° be the
angles of a quadrilateral ? Why or why not ?
Sol. We know that the sum of angles of a
quadrilateral is 360°.
But, 95° + 70° + 110° + 80° = 355° < 360°.
So, 95°, 70°, 110° and 80° cannot be the angles
of a quadrilateral.
Q.33. In the figure, ABCD is a parallelogram.
Find the values of x and y.
[2011 (T-II)]
[2011 (T-II)]
Sol. Since opposite angles of a parallelogram
are equal.
 3x – 10 = 2x + 35
 3x – 2x = 35 + 10  x = 45°
Angles are 3 × 45° – 10° = 125°; 180° – 125°
= 55°
 angles are 125°, 55°, 125°, 55°
Q.38. Two adjacent angles of a parallelogram
are in the ratio 2 : 3. Find the measure of all the
four angles of the parallelogram. [2011 (T-II)]
Sol. Let angles are 2x° and 3x°. Since
adjacent angles of a parallelogram are
supplementary.
 2x° + 3x° = 180°
 5x° = 180°  x = 36°
 angles are 72°, 108°, 72°, 108°
10
Q.39. In a parallelogram PQRS, if  P =
(3x – 5)°, Q = (2x + 15)°, find the value of x.
Q.42. Show that each angle of a rectangle is a
right angle.
[2010]
Sol. ABCD is a rectangle
[2011 (T-II)]
Sol. Since adjacent angles of a parallelogram
are supplementary.
 (3x – 5)° + (2x + 15)° = 180°
 5x + 10 = 180°  5x = 170°  x = 34°
Q.40. ABCD is a parallelogram. If DAB =
60° and DBC = 80°, find CDB. [2011 (T-II)]
Sol. We have, DAB = DCB = 60°
[opposite angles of a ||gm]
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Q.43. The angles of a quadrilateral are in the
ratio 2 : 3 : 5 : 8. Find the angles of the
quadrilateral.
[2010]
Sol. Let A = 2x°, B = 3x°, C = 5x° and
D = 8x°
 A + B + C + D = 360°
 2x + 3x + 5x + 8x = 360°  18x = 360°
 x = 20°
Thus, the angles are : 2x° = 2 × 20° = 40°,
3x° = 3 × 20° = 60°, 5x° = 5 × 20° = 100° and
8x° = 8 × 20° = 160°
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In DCB, we have,
CDB + DCB + DBC = 180°
[angles of a triangle]
 CDB + 60° + 80° = 180°
 CDB = 180° – 140° = 40°
Q.41. In the figure, ABCD is a square. A line
segment DX cuts the side BC at X and the
diagonal AC at O such that COD = 105°. Find
the value of x.
[2011 (T-II)]
 ABCD is a parallelogram.
 AD || BC.
Now, AD || BC and line AB intersects them at A
and B.
 A + B = 180°
[ Sum of the interior angles on the
same side of transversal is 180°]
 90° + B = 180° [ A = 90° (Given)]
 B = 90°
Similarly, we can show that C = 90° and
D = 90°.
Hence, A = B = C = D = 90°.
Sol. The angles of a square are bisected by
the diagonals.
OCX = 45° ( DCB = 90° and CA bisects
DCB)
Also, COD + COX = 180° (linear pair)
 105° + COX = 180°
 COX = 180° – 105° = 75°
Now, in COX, we have,
OCX + COX + OXC = 180°
 45° + 75° + OXC = 180°
 OXC = 180° – 120° = 60°
Hence, x = 60°
Q.44. ABCD is a rhombus with ABC = 58°.
Find ACD.
[2011 (T-II)]
Sol. ABCD is a rhombus.
 ABCD is a parallelogram.
 ABC = ADC (opposite angles)
( ABC = 58°)
 ADC = 58°
11
 ODC = 29°
( ODC =
1
ADC)
2
In ∆OCD, OCD + ODC + COD = 180°
and we have to show PMRL is a parallelogram.
In PSL and RQM, we have
S = Q
…(i) (Opp. s of a ||gm)
SPL =
 OCD + 29° + 90° = 180°
 OCD = 61°  ACD = 61°
Q.45. Diagonals of a parallelogram are
perpandicular to each other. Is this statement
true? Give reason for your answer.
Sol. This statement is false, because
diagonals of a parallelogram are bisect each other
but not perpendicular.
Q.46. Can all the angles of a quadrilateral be
acute ? Give reason for your answer.
Sol. No, because sum of the angles of a
quadrilateral is 360°.
Q.47. In the figure, BDEE and FDCE are
parallelograms. Can you say that BD = CD?
Given reason for your answer.
[HOTS]
1
P …(ii)
2
andQRM=
(Given)
1
R …(iii)
2
But P = R (Opp. s of a ||gm)
Using (ii) and (iii), we get
SPL = QRM …(iv)
Also,
PS = QR
…(v)
(Opp. sides of a ||gm)
From (i), (iv) and (v), we get
PSL  RQM (ASA)

SL = MQ
(CPCT)
But PQ = RS
(Opp. sides of a ||gm)
 PQ – MQ = SR – SL
 PM = RL
But PM || RL ( PQRS is a parallelogram)
 PMRL is a parallelogram.
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Q.49. Find the measure of each angle of a
parallelogram, if one of its angles is 30° less
than the twice the smaller angle.
Sol. Since, BDEF and FDCE are
parallelograms, therefore F, D and E are sure midpoints of AB, BC and AC respectively. Clearly,
since D is the mid-point of BC, therefore, we can
say that BD = CD.
Q.48. In the given figure, PQRS is a
parallelogram in which PL and RM are bisectors
of P and R respectively. Prove that PMRL is
a parallelogram.
[2010, 2011 (T-II)]
Sol. We are given a parallelogram PQRS in
which PL and MR bisect P and R respectively
[2010, 2011 (T-II)]
Sol. Let smallest angle of the parallelogram is
x°, then adjacent angle = (2x – 30)°.
Since adjacent angles of the parallelogram are
supplementary.
x + 2x – 30 = 180°
 3x = 180° + 30°  x =
210
= 70°
3
 smallest angle = 30° and adjacent angle
= 2 × 70° – 30°
= 140° – 30° = 110°
Hence, angles are 70°, 110°, 70°, 110°.
Q.50. In a parallelogram show that the angle
bisectors of two adjacent angles intersect at right
angles.
[2011 (T-II)]
Sol. Given : A parallelogram ABCD such that
the bisectors of adjacent angles A and B intersect
at O.
To Prove : AOB = 90°.
12
Or, 240° + MBN = 360°
Or, MBN = 360° – 240° = 120°
Or, ABC = 120°
( MBN = ABC)
Now, ADC = ABC = 120° (Opposite angles
of a parallelogram are equal)
Again, BAD  ABC = 180°
(Adjacent angles of a parallelogram are
supplementary)
Or, BAD = 180° – ABC = 180° – 120° = 60°
Also, BCD = BAD
(Opposite angles of a parallelogram are equal)
Or, BCD = 60°
Proof : Since ABCD is a parallelogram.
Therefore, AD || BC.
Now, AD || BC and transversal AB intersect
them.
∴ A + B = 180°
( sum of adjacent interior angles is 180°)


1
1
A +
B = 90°
2
2
1 + 2 = 90°
Hence, the angles of parallelogram are :
120°, 60°, 120° and 60°.
…(i)
Q.52. In the figure, PQRS is a parallelogram
and SPQ = 60°. If the bisectors of P and Q
meet at A on RS, prove that A is the mid-point of
RS.
[2011 (T-II)]
 AO is the bisector of A and OB is the bisector of B.


1 = 1 A and 2 = 1 B



2
2
In ∆AOB, we have 1 + AOB + 2 = 180°
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AOB = 90°
Q.51. The angle between two altitudes of a
parallelogram through the vertex of an obtuse
angle of the parallelogram is 60°. Find the
angles of the parallelogram.
[2011 (T-II)]
Sol. Let ABCD be a parallelogram in which
D is obtuse.
Through vertex D altitudes DM and DN are
drawn.
Also, we have MDN = 60° (Given)
Now, in quad. BMDN,
MDN + DMB + MBN + BND = 360°
(Angle sum property of a quadrilateral)
ER
[From (i), 1 + 2 = 90°]
S
 90° + AOB = 180°
Sol. We have, SPQ = 60° and P + Q = 180°
∴ 60° + Q = 180°  Q = 120°
Now, PQ || SR and transversal PA intersects
them.
∴ APQ = APS  APS = 30°
[ APQ = 30°]
Thus, in ∆APS, we have APS = PAS
[Each equal to 30°]
 AS = PS …(i) [ sides opposite to equal
angles are equal]
Or, 60° + 90° + MBN + 90° = 360°
13
Since, QA is the bisector of Q. Therefore,
PQA = AQR = 60°
Now, PQ || RS, and transversal AQ intersects
them.
∴ RAQ = PQA
[ PQA = 60°]
 RAQ = 60°
Thus, in ∆RQA, we have
RQA = RAQ
[Each equal to 60°]
RA
=
QR
[ sides opposite

to equal angles are equal]
 RA = PS … (ii)
[ PQRS is a ||gm ∴ PS = QR]
From (i) and (ii), we get AS = RA
3 = 4
[Vertically opp. angles]
and
QT = RS
[From (ii)]
So, by AAS criterion of congruence,
QOT  ROS
 QO = OR
[CPCT]
 O is the mid-point of QR.
 ST bisects RQ.
Q.54. E and F are points on diagonal AC of a
parallelogram ABCD such that AE = CF. Show
that BFDE is a parallelogram.
[2011 (T-II)]
Sol.
 A is the mid-point of RS. Proved.
Q.53. In the given figure, PQRS is a
parallelogram in which PQ is produced to T such
that QT = PQ. Prove that ST bisects RQ.
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[2010, 2011 (T-II)]
Construction : Join BD to meet AC at O.
Proof : We know that the diagonals of a
parallelogram bisect each other.
Therefore, AC and BD bisect each other at O.
∴ OA = OC
But, AE = CF  OA – AE = OC – CF
 OE = OF
Thus, in quadrilateral BFDE diagonals BD and
EF are such that OF = OE and OD = OB, i.e., the
diagonals BD and EF bisect each other.
Hence, BFDE is a parallelogram. Proved.
Sol. Since PQRS is a parallelogram.
Therefore, PQ || RS
Now, PQ || RS and transversal QR intersects
them.
 1 = 2 …(i)
Q.55. In a triangle ABC, median AD is
produced to X such that AD = DX. Prove that
ABXC is a parallelogram.
[2010, 2011 (T-II)]
Sol. Since AD is the median of ∆ABC.
Therefore, D is the mid-point of BC.
So, BD = DC
… (i)
A
B
Now, PQRS is a parallelogram.
 PQ = RS
 QT = RS …(ii)
[ QT = PQ (given)]
Thus, in s QOT and ROS, we have
1 = 2
[From (i)]
C
D
X
14
 ASB and RSP are vertically opposite angles 
 RSP = ASB



Similarly, we can prove that SRQ = 90°,
RQP = 90° and SPQ = 90°
ER
S
Hence, PQRS is a rectangle.
Q.57. Prove that the diagonals of a
parallelogram divide it into four triangles of
equal area.
[2010]
Sol. Given : A parallelogram ABCD. The
diagonals AC and BD intersect at O.
To Prove : ar (OAB) = ar (OBC)
= ar (OCD) = ar (AOD)
Proof : Since the diagonals of a parallelogram
bisect each other at the point of intersection.
 OA = OC and OB = OD
Also, the median of a triangle divides it into two
equal parts.
Now, in ABC, BO is the median.
 ar (OAB) = ar (OBC) …(i)
In BCD, CO is the median
 ar (OBC) = ar (OCD) …(ii)
In ACD, DO is the median
 ar (OCD) = ar (AOD) …(iii)
From (i), (ii) and (iii), we get
ar (OAB) = ar (OBC) = ar (OCD)
= ar (AOD).
Q.58. Show that the diagonals of a rhombus are
perpendicular to each other. [2010, 2011 (T-II)]
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In ∆s ABD and ∆XDC, we have
BD = DC
[From (i)]
ADB = XDC
[Vertically opp. angles]
and AD = DX
[Given]
∆ABD  ∆XDC
[By SAS]
and ABD = XCD
[ corresponding parts of
congruent triangles are equal]
The transversal BC intersects AB and CX at B
and C respectively such that ABC = XCD
(i.e., alternate interior angles are equal)
Thus, in quadrilateral ABXC, we have AB = CX
and AB || CX
Hence, ABXC is a parallelogram. Proved
Q.56. Prove that bisectors of the angles of a
parallelogram form a rectangle.
[2010]
Sol. Given : A parallelogram ABCD in which
bisectors of angles A, B, C, D intersect at P, Q,
R, S to form a quadrilateral PQRS.
To Prove : PQRS is a rectangle.
Proof : Since ABCD is a parallelogram.
Therefore, AD || BC
Now, AD || BC and transversal AB intersects
them at A and B respectively. Therefore,
A + B = 180°
[ Sum of consecutive interior angles is 180°]

1
1
A + B = 90°
2
2
 BAS + ABS = 90°
are bisectors of A
But, in ABS, we have
BAS + ABS + ASB
[Sum of the angles



... (i) [ AS and BS
and B respectively]
Sol. Given : A rhombus ABCD whose diagonals
AC and BD intersect at O.
To Prove : BOC = DOC = AOD = AOB
= 90°
Proof : We know that a parallelogram is a
rhombus, if all of its sides are equal. So ABCD
is a rhombus  ABCD is a ||gm such that
= 180°
of a triangle is 180°]
90° + ASB = 180°
ASB = 90°
RSP = 90°
15
 AB = BC = CD = DA …(i)
Since the diagonals of a parallelogram bisect
each other.
 OB = OD and OA = OC …(ii)
Q.60. In a quadrilateral ABCD, AO and BO
are the bisectors of A and B respectively.
Prove that AOB =
1
(C + D).
2
[2010, 2011 (T-II)]
Sol. In  AOB, we have
AOB + 1 + 2 = 180°
  AOB = 180° – (1 + 2)
1
1
(A + B)
2
1
 AOB = 180° –
[360° – (C + D)]
2
 A + B + C + D = 360° 
 A + B = 360°  (C + D) 
ER
S
 AOB = 180° –
 AOB = 180° – 180° +
 AOB =
P = Q
1
(C + D)
2
1
(C + D)
2
Q.61. In a parallelogram ABCD, the bisector
of  A also bisects BC at X. Prove that
AD = 2AB.
[2010, 2011 (T-II)]
Sol. Since AX is the bisector of A.
 1 =
AB = CD

1
1


 1 = 2 A and 2 = 2 B


G
P R OY
A K AL
AS BR
H A OT
N H
Now, in s BOC and DOC, we have
BO = OD
[From (ii)]
BC = DC
[From (i)]
OC = OC
[Common]
So, by SSS criterion of congruence
BOC  DOC
 BOC = DOC [CPCT]
But, BOC + DOC = 180° [Linear pair axiom]
BOC = DOC = 90° [  BOC = DOC]
Similarly, AOB = AOD = 90°
Hence, AOB = BOC = COD = DOA
= 90°.
Q.59. In a parallelogram ABCD, AP and CQ
are perpendiculars from A and C on its diagonal
BD. Prove that AP = CQ.
[2010, 2011 (T-II)]
Sol. Given : ABCD is a parallelogram, in
which AP  BD and CQ  BD.
To Prove : AP = CQ.
Proof : In APB and CDQ,
1
 AOB = 180° –  A + B 
2
2

1
A
2
…(i)
[Opposite sides of a parallelogram]
[Each 90°]
ABP = CDQ
[Alternate interior angles of a parallelogram]
BAP = DCQ
[Third angle of a triangle]
 ABP  CDQ [ASA axiom]
AP = CQ
[CPCT] Proved
Since ABCD is a parallelogram.
Therefore, AD || BC and AB intersects them.
16
Sol. Given : An isosceles ∆ABC having AB =
AC. AD is the bisector of exterior CAE and
CD || AB
 A + B = 180°  B = 180° – A
In ABX, we have 1 + 2 + B = 180°

1
A + 2 + 180° – A = 180°
2
 2 –
1
1
A = 0  2 A
2
2
To Prove : CAD = BCA and ABCD is a
parallelogram.
Proof : In ∆ABC, we have AB = AC [Given]
 1 = 2 … (i) [ Angles opposite to equal
sides in a  are equal]
…(ii)
From (i) and (ii), we have 1 = 2
Thus, in ABX, we have 1 = 2
 BX = AB [ Sides opposite to equal angles
in a triangle are equal]
 2BX = 2AB
[Multiplying both sides by 2]
 BC = 2AB
[ X is the mid-point of BC
 2 BX = BC]
AD = 2AB [ ABCD is a ||gm  AD = BC]
Q.62. Prove that the diagonal of a
parallelogram divides it into two congruent
triangles.
[2010, 2011 (T-II)]
Sol. Given : ABCD is a parallelogram in
which AC is a diagonal.
G
P R OY
A K AL
AS BR
H A OT
N H
ER
S
Now, in ∆ABC, we have, ext. CAE = 1 + 2
[ An exterior angle is equal to the sum of
two opposite interior angles]
[ 1 = 2 from (i)]
 ext. CAE = 22
[ AD is the bisector of
 23 = 22
ext. CAE ∴ CAE = 23]
 3 = 2
 CAD = BCA
Thus, AC intersects lines AD and BC at A and C
respectively such that 3 = 2, i.e., alternate
interior angles are equal. Therefore, AD || BC.
But, CD || AB
To Prove : ∆ABC  ∆CDA.
Proof : In ∆ABC and ∆CDA since, ABCD is
parallelogram.
∴ AB || DC and BC || AD.
Since, BAC = DCA [Alternate angles]
BCA = DAC
[Alternate angles]
AC = AC
[Common]
∴ ∆ABC  ∆CDA Proved. [ASA axiom]
Q.63. In the figure,
ABC is an isosceles
triangle in
which
AB = AC, CD || AB and
AD is bisectors of
exterior ∠CAE of ∆ABC.
Prove that ∠ CAD =
∠BCA and ABCD is a
parallelogram.
Thus, ABCD is a quadrilateral such that
AD || BC and CD || AB.
Hence, ABCD is a parallelogram. Proved.
Q.64. PQ and RS are two equal and parallel
line segments. Any point M not lying on PQ or
RS is joined to Q and S and lines through P
parallel to QM and through R parallel to SM
meet at N. Prove that line segments MN and PQ
are equal and parallel to each other. [HOTS]
Sol. It is given that PQ = RS and PQ || RS.
Therefore, PQSR is a parallelogram.
So, PR = QS and PR || QS
… (i)
Now, PR || QS
Therefore, RPQ + PQS = 180°
[2010, 2011 (T-II)]
17
(Interior angles on the same side of the
transversal)
i.e., RPQ + PQM + MQS = 180° … (ii)
Also, PN || QM
(By construction)
Therefore, NPQ + PQM = 180°
i.e., NPR + RPQ + PQM = 180° … (iii)
So, NPR = MQS [From (ii) and (iii)] … (iv)
Similarly, NRP = MSQ
… (v)

Therefore, ∆PNR
∆QMS
[By ASA, using (i), (iv) and (v)]
So, PN = QM and NR = MS
[CPCT]
As, PN = QM and PN || QM, we have PQMN is
a parallelogram.
So, MN = PQ and NM || PQ.
Proved.
PRACTICE EXERCISE 8.1A
2. Which of the following is not true for a
parallelogram ?
(a) opposite sides are equal
(b) opposite angles are equal
G
P R OY
A K AL
AS BR
H A OT
N H
(c) opposite angles are bisected by the diagonals
S
1. In a rectangle ABCD, diagonals AC and BD
intersect at O. If AO = 3 cm, then the length of the
diagonal BD is equal to :
(a) 3 cm (b) 9 cm (c) 6 cm (d) 12 cm
7. In a parallelogram ABCD, bisectors of two
adjacent angles A and B meet at O. The measure of
the angle AOB is equal to :
(a) 90°
(b) 180° (c) 60°
(d) 360°
8. Lengths of two adjacent sides of a parallelogram are in the ratio 2 : 7. If its perimeter is
180 cm, then the adjacent sides of the parallelogram are :
(a) 10 cm, 20 cm
(b) 20 cm, 70 cm
(c) 41 cm, 140 cm (d) none of these
9. If a, b, c and d are four angles of a quadrilateral such that a = 2b, b = 2c and c = 2d, then the
value of d is :
(a) 36°
(b) 24°
(c) 30°
(d) none of these
ER
1 Mark Questions
(d) diagonals bisect each other.
3. Three angles of a quadrilateral are 75°, 90°
and 75°.The fourth angle is :
(a) 90°
(b) 95°
(c) 105° (d) 120°
4. ABCD is a rhombus such that ∠ACB = 40°.
Then ∠ADB is :
(a) 40°
(b) 45°
(c) 50°
(d) 60°
5. Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then
∠OAB is :
(a) 90°
(b) 50°
(c) 40°
(d) 10°
6. If APB and CQD are two parallel lines, then
the bisectors of the angles APQ, BPQ, CQP and
PQD form
(a) a square
(b) a rhombus
(c) a rectangle
(d) any other parallelogram
2 Marks Questions
10. Diagonals of a quadrilateral PQRS bisect
each other. If ∠P = 35°, find ∠Q.
11. The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the angles.
12. The lengths of the diagonals of a rhombus
are 24 cm and 18 cm. Find the length of each side
of the rhombus.
13. Diagonals AC and BD of a parallelogram
ABCD intersect each other at O. If OA = 3 cm and
OD = 2 cm, find the lengths of AC and BD.
14. In ∆ABC, AB = 5 cm, BC = 8 cm and AC =
7 cm. If D and E are respectively the mid-points of
AB and BC, determine the length of DE.
18
15. ABCD is a rhombus AC = 16 cm, BC =
10 cm. Find the length of the diagonal BD.
22. If one angle of a parallelogram is a right
angle, it is a rectangle. Prove.
23. Two parallel lines ‘l’and ‘m’are intersected
by a transversal ‘p’. Show that the quadrilateral
formed by the bisectors of interior angles is a rectangle.
[2011 (T-II)]
[2011 (T-II)]
16. In the given figure, ABCD is a rhombus.
AO = 5 cm, area of the rhombus is 25 sq. cm. Find
the length of BD.
[2011 (T-II)]
17. In the given figure, ABCD is a parallelogram. Compute DCA, ACB and ADC, given
DAC = 60° and ABC = 75°.
[2011 (T-II)]
24. In the given figure, ABCD is a square, if
PQR = 90° and PB = QC = DR, prove that
QB = RC, PQ = QR, QPR = 45°. [2011 (T-II)]
ER
19. If PQRS is a rhombus with PQR = 55°,
find PRS.
[2011 (T-II)]
S
18. ABCD is a parallelogram. The angle bisectors of A and D intersect at O. Find the measure of AOD.
[2011 (T-II)]
4 Marks Questions
25. ABCD is a rectangle in which diagonal BD
bisects ∠B. Show that ABCD is a square.
26. If ABCD is a trapezium in which AB | | CD
and AD = BC, prove that ∠A = ∠B.
27. P, Q, R, and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AD = BC. Prove that
PQRS is a rhombus.
8.2 THE MID-POINT THEOREM
1. The line segment joining the mid-points of
any two sides of a triangle is parallel to the third
side and is half of it.
2. A line through the mid-point of a side of a
triangle parallel to another side bisects the third
side.
3. The quadrilateral formed by joining the
mid-points of the sides of a quadrilateral, in
order, is a parallelogram.
G
P R OY
A K AL
AS BR
H A OT
N H
3 Marks Questions
20. Show that the quadrilateral formed by joining the mid-points of sides of a square is also square.
21. E is the mid-point of the side AD of trapezium ABCD with AB | | DC. A line through E drawn
parallel to AB intersect BC at F. Show that F is the
mid-point of BC.
TEXTBOOK’S EXERCISE 8.2
Q.1. ABCD is a quadrilateral in which P, Q, R
and S are mid-points of the sides AB, BC, CD
and DA respectively. (see Fig.). AC is a
diagonal. Show that :
[2011 (T-II)]
19
1
(i) SR || AC and SR = AC
2
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Given : ABCD is a quadrilateral in which P, Q, R
and S are mid-points of AB, BC, CD and DA. AC
is a diagonal.
To Prove :
SR || AC and SR =
…(1)
[Mid-point theorem]
In ADC, R is the mid-point of CD and S
is the mid-point of AD
SR || AC and SR =
1
AC
2
…(2)
[Mid-point theorem]
1
AC
2
…(ii)
From (i) and (ii), we have PQ || SR and PQ = SR
PQRS is a parallelogram.
[One pair of opposite sides is parallel and equal]
Since ABQS is a parallelogram.
AB = SQ
[Opposite sides of a || gm]
Similarly, since PBCR is a parallelogram.
BC = PR
Thus, SQ = PR
[AB = BC]
Since SQ and PR are diagonals of parallelogram
PQRS, which are equal.
PQRS is a rectangle. Proved.
Q.3. ABCD is a rectangle and P, Q, R and S are
mid-points of the sides AB, BC, CD and DA
respectively. Show that the quadrilataral PQRS
is a rhombus.
Sol. Given : A rectangle ABCD in which P, Q, R,
S are the mid-points of AB, BC, CD and DA
respectively, PQ, QR, RS and SP are joined.
G
P R OY
A K AL
AS BR
H A OT
N H
[Mid-point theorem]
(ii) From (1) and (2), we get
…(i)
S
(ii) PQ = SR
(iii) PQRS is a parallelogram
Proof :
(i) In ABC, P is the mid-point of AB and Q
is the mid-point of BC.
1
AC
2
1
AC
2
Similarly, in DAC,
1
AC
2
 PQ || AC and PQ =
PQ || AC and PQ =
ER
(i) SR || AC and SR =
To Prove : PQRS is a rectangle.
Construction : Join AC, PR and SQ.
Proof : In ABC,
P is mid-point of AB and Q is mid-point of BC.
[Given]
PQ || SR and PQ = SR
(iii) Now in quadrilateral PQRS, its one pair
of opposite sides PQ and SR is equal and
parallel.
 PQRS is a parallelogram. Proved.
Q.2. ABCD is a rhombus and P, Q, R and S are
the mid-points of the sides AB, BC, CD and DA
respectively. Show that the quadrilateral PQRS
is a rectangle.
[2010]
Sol. Given : ABCD is a rhombus in which P, Q,
R and S are mid-points of sides AB, BC, CD and
DA respectively.
To Prove : PQRS is a rhombus.
Construction : Join AC.
Proof : In ABC, P and Q are the mid-points of
the sides AB and BC.
PQ || AC and PQ =
Similarly, in ADC,
20
1
AC
2
…(i)
[Mid-point theorem]
SR || AC and SR =
1
AC
2
Fig.). Show that the line segments AF and EC
trisect the diagonal BD.
[2010, 2011 (T-II)]
…(ii)
From (i) and (ii), we get
PQ || SR and PQ = SR
…(iii)
Now in quadrilateral PQRS, its one pair of
opposite sides PQ and SR is parallel and equal.
[From (iii)]
PQRS is a parallelogram.
Now AD = BC
…(iv)
[Opposite sides of a rectangle ABCD]
1
1
AD = BC  AS = BQ
2
2
AE =
1
1
AB and CF = DC …(i)
2
2
S
But, AB = DC and AB || DC
…(ii)
[Opposite sides of a parallelogram]
 AE = CF and AE || CF.
AECF is a parallelogram.
[One pair of opposite sides is parallel and equal]
In BAP, E is the mid-point of AB
 EQ || AP
 Q is mid-point of PB
[Converse of mid-point theorem]
PQ = QB
…(iii)
Similarly, in DQC, P is the mid-point of DQ

DP = PQ
…(iv)
From (iii) and (iv), we have DP = PQ = QB
or line segments AF and EC trisect the diagonal
BD. Proved.
Q.6. Show that the line segments joining the
mid-points of the opposite sides of a
quadrilateral bisect each other.
[V. Imp.]
Sol. Given : ABCD is a quadrilateral in which
EG and FH are the line segments joining the midpoints of opposite sides.
G
P R OY
A K AL
AS BR
H A OT
N H
In APS and BPQ,
AP = BP
[ P is the mid-point of AB]
AS = BQ
[Proved above]
PAS = PBQ
[Each = 90°]
APS BPQ
[SAS axiom]
 PS = PQ
…(v)
[CPCT]
From (iii) and (v), we have PQRS is a
rhombus Proved.
Q.4. ABCD is a trapezium in which AB || DC,
BD is a diagonal and E is the mid-point of AD.
A line is drawn through E parallel to AB
intersecting BC at F (see Fig.). Show that F is
the mid-point of BC.
[2011 (T-II)]
ER

Sol. Given : A parallelogram ABCD, in which E
and F are mid-points of sides AB and DC
respectively.
To Prove : DP = PQ = QB
Proof : Since E and F are mid-points of AB and
DC respectively.
Sol. Given : A trapezium ABCD with AB || DC,
E is the mid-point of AD and EF || AB.
To Prove : F is the mid-point of BC.
Proof : AB || DC and EF || AB AB, EF and DC
are parallel.
Intercepts made by parallel lines AB, EF and DC
on transversal AD are equal.
 Intercepts made by those parallel lines on
transversal BC are also equal.
i.e., BF = FC F is the mid-point of BC.
Q.5. In a parallelogram ABCD, E and F are the
mid-points of sides AB and CD respectively (see
21
To Prove : EG and FH bisect each other.
Construction : Join EF, FG, GH, HE and AC.
Proof : In ABC, E and F are mid-points of AB
and BC respectively.
1
AC and EF || AC
2
…(i)
 HG =
1
AC and HG || AC
2
…(ii)
From (i) and (ii), we get EF = HG and EF || HG
 EFGH is a parallelogram.
[ a quadrilateral is a parallelogram
if its one pair of opposite sides is equal
and parallel]
Now, EG and FH are diagonals of the
parallelogram EFGH.
 EG and FH bisect each other.
[Diagonal of a parallelogram bisect each other]
Proved.
1
AB
2
[2010, 2011 (T-II)]
(ii) MD  AC (iii) CM = MA =
Sol. Given : A triangle ABC, in which C = 90°
and M is the mid-point of AB and BC || DM.
To Prove : (i) D is the mid-point of AC [Given]
1
AB
2
Construction : Join CM.
Proof : (i) In ABC, M is the mid-point of AB.
[Given]
BC || DM
[Given]
D is the mid-point of AC.
[Converse of mid-point theorem] Proved.
(ii) ADM = ACB [ Corresponding angles]
But ACB = 90°
[Given]

ADM = 90°
But ADM + CDM = 180° [Linear pair]

CDM = 90°
Hence, MD  AC Proved.
(iii) AD = DC …(1)
[ D is the mid-point of AC]
Now, in ADM and CMD, we have
ADM = CDM
[Each = 90°]
AD = DC
[From (1)]
DM = DM
[Common]
ADM CMD
[SAS congruence]
 CM = MA …(2)
[CPCT]
Since M is mid-point of AB,
G
P R OY
A K AL
AS BR
H A OT
N H
Q.7. ABC is a triangle right angled at C. A line
through the mid-point M of hypotenuse AB and
parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC.
(iii) CM = MA =
S
In ADC, H and G are mid-points of AD and CD
respectively.
(ii) DM  BC
ER
EF =
 MA =
1
AB
2
Hence, CM = MA =
…(3)
1
AB
2
Proved.
[From (2) and (3)]
OTHER IMPORTANT QUESTIONS
Q.1. In the figure, D, E and F are the midpoints of the sides AB, BC and CA respectively.
If AC = 8.2 cm, then value of DE is :
(a) 8.2 cm
(c) 2.05 cm
(b) 4.1 cm
(d) none of these
1
1
AC =
× 8.2 = 4.1 cm
2
2
Q.2. In the figure, D and E are mid-points of
the sides AB and AC respectively of ABC.
Measure of B is :
[2010]
Sol. (b) DE =
22
Q.5. In the ABC, B is a right angle, D and
E are the mid-points of the sides AB and AC
respectively. If AB = 6 cm and AC = 10 cm, then
the length of DE is :
[Imp.]
(a) 50° (b) 30° (c) 80° (d) 40°
Sol. (a) In ADE, ADE
= 180° – (100° + 30°) = 50°
 By mid-point theorem, DE || BC
 B = ADE = 50°
[Alternate angles]
Q.3. ABCD is a quadrilateral and P, Q, R and
S are the mid-points of the sides AB, BC, CD, DA
respectively. If BD = 12 cm, then length of QR
is :
[2010]
(a) 3 cm (b) 5 cm (c) 4 cm (d) 6 cm
Sol. (c) BC =
S
G
P R OY
A K AL
AS BR
H A OT
N H
Q.4. In the figure, P and Q are mid-points of
sides AB and AC respectively of ABC. If PQ =
3.5 cm and AB = AC = 9 cm, then the perimeter
of ABC is :
[Imp.]
1
BC (By mid-point theorem)
2
 DE = 4 cm.
Q.6. In a ∆ABC, D, E and F are respectively
the mid-points of BC, CA and AB as shown in the
figure. The perimeter of ∆DEF is :
[V. Imp.]
(d) 4 cm
1
1
BR =
× 12 = 4 cm
3
3
(10) 2 – (6)2
ER
QR =
=
= 100 – 36  64 = 8 cm
Also, D and E the mid-points of AB and AC
respectively.
 DE =
(a) 6 cm (b) 8 cm (c) 3 cm
Sol. (d) By mid-point theorem,
AC2 – AB2
(a) 20 cm
(b) 23 cm
(c) 25 cm
(d) 27 cm
Sol. (c) We have, AB = AC = 9 cm.
(a)
1
(AB +BC + CA)
2
(b) AB + BC + CA
(c) 2 (AB + BC + CA)
(d) none of these
Sol. (a)
1

We have, EF = BC,
2

1

DF = AC  (By mid-point theorem)
2

1

and
DE = AB 
2

1
By mid-point theorem, PQ =
BC
2
 BC = 2PQ
∴ BC = 2 × 3.5 = 7 cm
∴ Perimeter of ∆ ABC = 9 cm + 9 cm + 7 cm
= 25 cm
23
of AB and AC respectively. The altitude AP
intersects EF at Q. The correct relation between
AQ and QP is :
[HOTS]
Now, AD is the median in ABC
 D is the mid-point of BC.
Since DE is a line drawn through the mid-point
of side BC of ABC and is parallel to AB
(given). Therefore, E is the mid-point of AC.
Hence, BE is the median of ABC.
Q.10. In ∆ABC, AB = 5 cm, BC = 8 cm and
AC = 7 cm. If D and E are respectively the midpoints of AB and BC, determine the length of DE.
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(a) AQ > QP
(b) AQ = QP
(c) AQ < QP
(d) none of these
Sol. (b) In ABC, E and F are the mid-point
of AB and AC respectively.
∴ EF || BC
(Mid-point theorem)
Or, EQ = QP
 Q is the mid-point of AP (By converse of
mid-point theorem)
 AQ = QP
Q.8. D and E are the mid-points of the sides
AB and AC of ABC and O is any point on side
BC. O is joined to A. If P and Q are the midpoints of OB and OC respectively, then DEQP
is :
(a) a square
(b) a rectangle
(c) a rhombus (d) a parallelogram
DP || AO
… (ii)
Similarly, in ∆AOC, EQ || AO … (iii)
From (ii) and (iii) we have, DP || EQ … (iv)
Thus, from (i) and (iv), we have DE || PQ and
DP || EQ
Hence, quad. DEQP is a parallelogram.
Q.9. In the given figure, AD is the median and
DE || AB. Prove that BE is the median. [Imp.]
Sol. In order to prove that BE is the median,
it is sufficient to show that E is the mid-point of
AC.
S
1
=
(AB + BC + CA)
2
Q.7. In the ABC, E and F are the mid-points
ER
Therefore, perimeter of ∆DEF = DE + EF + DF
[HOTS]
Sol. (d) In ABC, D and E are respectively
the mid-points of the sides AB and AC, then by
mid-point theorem we have, DE || BC or,
DE || PQ … (i)
[2011 (T-II)]
Sol. Since, the line segment joining the midpoints of any two sides of a triangle is parallel to
the third side and is half of it.
∴ DE =
Also, in ∆AOB, D and P are respectively the
mid-points of AB and BO, then by mid-point
theorem
24
1
1
AC ⇒ DE =
× 7 = 3.5 cm
2
2
Q.11. In ∆ABC, P, Q and R are mid-points of
sides BC, CA and AB respectively. If AC = 21 cm,
BC = 29 cm and AB = 30 cm, find the perimeter
of the quadrilateral ARPQ.
Sol. Clearly, AQ =
AR =
1
1
AC = × 21 cm = 10.5 cm
2
2
1
1
AB =
× 30 cm = 15 cm
2
2

RP =
BC = QY
... (ii)
[Opposite sides of a parallelogram are equal]
But AB = BC
[Given]
 XY = QY
... (iii) [From (i) and (ii)]
Now in s PQX and RQY
PQX = RQY [Vertically opposite angles]
XQ = QY
[From (iii)]
PXQ = RYQ [Alternate angles, l || n]
 PQX ≅ RQY
[ASA axiom]
 PQ = QR Proved.
[CPCT]
Q.13. In the figure, through A, B and C lines
RQ, PQ and PR have been drawn respectively
parallel to sides BC, CA and AB of a ABC.
1
1
AC =
× 21 cm = 10.5 cm
2
2
(By mid-point theorem)
(By mid-point theorem)
Now, perimeter of ARPQ = AR + RP + PQ + AQ
= 15 cm + 10.5 cm + 15 cm + 10.5 cm = 51 cm
Show that BC =
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Q.12. In there are three or more parallel lines
and the intercepts made by them on the
transversal are equal the corresponding
intercepts on any other transversal are also
equal.
[2011 (T-II)]
Sol. Given : Lines l, m and n such that l || m || n,
p is a transversal which cuts l, m, n in A, B, C
respectively such that AB = BC. Also, q is
another transversal which cuts l, m, n in P, Q, R
respectively.
To Prove : PQ = QR
Construction : Through Q, draw a line r such
that r || p.
Proof :
r || p
[By construction]
 ABQX is a parallelogram.
[Pairs of opposite sides are parallel]
 AB = XQ ... (i)
[Opposite sides of a parallelogram are equal]
Again BCYQ is a parallelogram
[Pairs of opposite sides are parallel]
S
1
1
AB =
× 30 cm = 15 cm
2
2
ER
PQ =
1
QR.
2
[HOTS]
Sol. Given : ABC, lines are drawn through
A, B and C parallel respectively to the sides BC,
CA and AB forming ∆PQR.
To prove : BC =
1
QR.
2
Proof : Since, AQ || CB and AC || QB
∴ AQBC is a parallelogram.
 BC = QA … (i) [opposite sides of a ||gm]
Since, AR || BC and AB || RC
25
∴ ARCB is a parallelogram.
 BC = AR … (ii) [opposite sides of a ||gm]
1
From (i) and (ii), QA = AR = QR. … (iii)
2
1
From (i) and (iii), BC = QR. Proved.
2
Q.14. Show that the line segments joining the
mid-points of opposite sides of a quadrilateral
bisect each other.
[2010, 2011 (T-II)]
Sol. We are given a quadrilateral ABCD. E, F,
G and H are the mid points of sides AB, BC, CD
and DA respectively.
We have to show that EG and HF bisect each
other. Join EF, FG, GH, HE and AC.
In ABC, E is the mid point of AB and F is the
mid point of BC.
Sol. In BEC, DF is a line through the midpoint D of BC and parallel to BE intersecting CE
at F. Therefore, F is the mid-point of CE.
Because the line drawn through the mid-point of
one side of a triangle and parallel to another side
bisects the third side. Now, F is the mid-point of
CE.
1
CE
2

1 1
 CF =  AC 
2 2

 CF =
 E is the mid-point of AC 


1
 EC = AC


2

1
AC
4
Q.16. In ABC, AD is the median through A
and E is the mid-point of AD. BE produced meets
 CF =
AC in F. Prove that AF =
1
AC.
3
[2010, 2011 (T-II)]
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Sol. Through D, draw DK || BF. In ADK, E
is the mid-point of AD and EF || DK.
 F is the mid-point of AK
 AF = FK
…(i)
In BCF, D is the mid-point of BC and DK || BF
1
AC
2
1
Similarly, HG || AC and HG = AC
2

EF || AC and EF =
…(i)
…(ii)
 From (i) and (ii), EF || HG and EF = HG
 EFGH is a parallelogram and EG and HF are
its diagonals. We know that the diagonals of a
parallelogram bisect each other. Thus, EG and
HF bisect each other.
Q.15. In the given figure, AD and BE are
medians of ABC and BE || DF. Prove that
1
CF = AC.
[2010, 2011 (T-II)]
4
 K is the mid-point of FC
 FK = KC
…(ii)
From (i) and (ii), we have
AF = FK = KC
…(iii)
Now, AC = AF + FK + KC
 AC = AF + AF + AF
[Using (iii)]
 AC = 3(AF)  AF =
1
AC
3
Q.17. Prove that the straight line joining the
mid-points of the diagonals of a trapezium is
parallel to the parallel sides.
[2010]
26
Sol. We are given a trapezium ABCD in
which AB || DC and M, N are the mid-points of
the diagonals AC and BD.
We need to prove that MN || AB || DC.
Join CN and let it meet AB at E.
E and F are the mid-points of AC and AB
respectively.
 EF =
1
BC
2
…(ii)
F and D are the mid-points of AB and BC
respectively.
 FD =
1
AC
2
…(iii)
Now, ABC is an equilateral triangle.
 AB = BC = CA
1
1
1
AB = BC = CA
2
2
2
 DE = EF = FD
[Using (i), (ii) and (iii)]
Hence, DEF is an equilateral triangle.
Q.19. Prove that the four triangles formed by
joining in pairs, the mid-points of three sides of
a triangle, are concurrent to each other.
[2011 (T-II)]
S
Sol. Given : A triangle ABC and D, E, F are
the mid-points of sides BC, CA and AB
respectively.
To Prove : AFE  FBD  EDC  DEF
Proof : Since the segment joining the mid-points
of the sides of a triangle is half of the third side.
Therefore,
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Q.18. In the given figure, D, E and F are,
respectively the mid-points of sides BC, CA and
AB of an equilateral triangle ABC. Prove that
DEF is also an equilateral triangle.

ER
Now in CDN and EBN,
DCN = BEN [Alternate angles]
CDN = EBN [Alternate angles]
and DN = BN
[Given]

CDN  EBN
[SAS]
 CN = EN
[CPCT]
Now in ACE, M and N are the mid-points of
the sides AC and CE respectively.

MN || AE or MN || AB
Also
AB || DC

MN || AB || DC
1
AB  DE = AF = BF …(i)
2
1
EF = BC  EF = BD = CD …(ii)
2
1
DF = AC  DF = AE = EC …(iii)
2
DE =
[2010]
Sol. Since the segment joining the mid-points
of two sides of a triangle is half of the third side.
Therefore, D and E are mid-points of BC and AC
respectively.
 DE
1
AB
2
Now, in s DEF and AFE, we have
DE = AF
[From (i)]
DF = AE
[From (ii)]
and EF = FE
[Common]
…(i)
27
So, by SSS criterion of congruence, DEF 
AFE
Similarly, DEF  FBD and DEF  EDC
Hence, AFE  FBD  EDC  DEF.
Q.20. Prove that in a triangle, the line segment
joining the mid-points of any two sides is parallel
to third side and is half of it.
[2011 (T-II)]
Sol. Given : A triangle ABC in which D and
E are the mid-points of AB and AC respectively.
DE is joined.
To prove : DE || BC and DE =
1
BC.
2
Construction : Draw CF || BA, meeting DE
produced in F.
Proof : In s AED and CEF, we have
AE = CE (given), AED = CEF
(vert. opp. angles)
DAE = FCE (Alt. Int. angles)
 AED  CEF
 AD = CF and DE = EF (CPCT)
But, AD = BD.
 BD = CF and BD || CF (by construction)
 BCFD is a ||gm.
 DF || BC and DF = BC.
 DE || BC and DE =
1
1
DF =
BC
2
2
[ DE = EF]
1
BC.
2
S
Hence, DE || BC and DE =
(a) AD
3. In the given figure, if DE = 4 cm, BC =
8 cm and D is the mid-point of AB, then the true
statement is :
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1 Mark Questions
1. In the given figure, D is the mid-point of
AB and DE || BC, then AE is equal to :
ER
PRACTICE EXERCISE 8.2A
(b) EC
(c) DB
(d) BC
2. In the given figure, D and E are mid-points
of AB and AC respectively. The length of DE is :
(a) 8.2 cm
(c) 4.9 cm
(a) AB = AC
(b) DE || BC
(c) E is not the mid-point of AC
(d) DE BC
4. In the given
figure, ABCD is a
rectangle. P and Q are
mid-points of AD and
DC respectively. Then
length of PQ is :
(b) 5.1 cm
(d) 4.1 cm
(a) 5 cm (b) 4 cm
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(c) 2.5 cm (d) 2 cm
2 Marks Questions
5. In ABC, D and E are mid-points of AB
and AC. AD = 3.5 cm, AE = 4 cm, DE = 2.5 cm.
Find the perimeter of ABC.
[2011 (T-II)]
6. In ABC, AB = 12 cm BC = 15 cm and
AC = 7 cm. Find the perimeter of the triangle
formed by joining the mid-points of the sides of
the triangle.
[2011 (T-II)]
7. D and E are the mid-points of sides AB and
AC respectively of triangle ABC. If the perimeter
of ABC = 35 cm, find the perimeter of ADE.
12. In the given figure, PQRS is a square. M is
the mid-point of PQ and AB  RM. Prove that
RA = RB.
[2011 (T-II)]
3/4 Marks Questions
8. D, E and F are mid points of sides BC, CA
and AB respectively of a triangle ABC. DF and
BE meet at P and CF and DE meet at Q. Prove
9. In a trapezium ABCD, side AB is parallel
to side DC and E is the mid-point of side AD. If
F is a point on the side BC such that the segment
EF is parallel to side DC, prove that
1
(AB + CD).
2
[2011 (T-II)]
15. In the given figure, points A and B are on
the same side of a line m, AD  m and BE  m
and meet m at D and E respectively. If C is the
mid-point of AB, prove that CD = CE.
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EF =
S
1
BC.
4
ER
that PQ =
13. Show that the quadrilateral formed by
joining the mid-points of the sides of rhombus,
taken in order, form a rectangle. [2011 (T-II)]
14. The diagonals of a quadrilateral ABCD are
perpendicular, show that quadrilateral formed by
joining the mid-points of its sides, is rectangle.
10. ABCD is a kite having AB = AD and
BC = CD. Prove that the quadrilateral formed by
joining the mid-points of the sides is a rectangle.
[2011 (T-II)]
11. In the given figure, ABCD is a trapezium
in which AB || DC. E is the mid-point of AD and
F is a point on BC such that EF || DC. Prove
that F is the mid-point of BC.
[2011 (T-II)]
B. FORMATIVE ASSESSMENT
Activity-1
Objective : To verify the mid-point theorem for a triangle using paper cutting and pasting.
Materials Required : White sheets of paper, a pair of scissors, colour pencils, gluestick, geometry
box, etc.
29
Procedure :
1. On a white sheet of paper, draw a ABC and cut it out.
Figure-1
2. Using paper folding method, find the mid-points of AB,
AC and BC and mark them as X, Y and Z respectively.
Join X to Y.
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Figure-2
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3. Cut out the triangular piece AXY and superimpose AY over YC such that YX falls along CB
as shown on below.
Figure-3
Observations :
1. In figure 3(b), we see that AYX exactly covers YCB or AYX = YCB.
But, AYX and YCB are corresponding angles made on XY and BC by the transversal AC.
Therefore, XY || BC.
But, X and Y are the mid-points of AB and AC respectively.
2. In figure 3(b), we also observe that X and Z coincide. It implies XY = CZ.
But, CZ is half of BC.
Thus, we can say that the line segment joining the mid-points of two sides of a triangle is
parallel to the third side and is equal to half of it.
30
Conclusion : From the above activity, the mid-point theorem is verified.
Do Yourself : Draw a right triangle, an acute angled triangle and an obtuse angled triangle. Verify
the mid-point theorem for each case.
Activity-2
Objective : To verify that a diagonal of a parallelogram divides it into two congruent triangles.
Materials Required : White sheets of paper, colour pencils, a pair of scissors, gluestick, geometry
box, etc.
Procedure :
ER
S
1. On a white sheet of paper, draw a parallelogram ABCD and cut it out. Draw the diagonal AC
of the parallelogram and cut it along AC to get two triangular cut outs.
Figure-1
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2. Now, superimpose one triangle over the other as shown below.
Figure-2
Observations : In figure 2, we see that the two triangles exactly cover each other. Hence, the
triangles are congruent.
Conclusion :
From the above activity, it is verified that a diagonal of a parallelogram divides it
into two congruent triangles.
Do Yourself : Draw three different parallelograms and verify the above property by paper cutting
and pasting.
31
Activity-3
Objective : To verify that the diagonals of a parallelogram bisect each other.
Materials Required : White sheets of paper, colour pencils, gluestick, a pair of scissors, geometry
box, etc.
Procedure :
1. On a white sheet of paper, draw a parallelogram ABCD and both its diagonals AC and BD
intersecting at O. Cut out the four triangles so formed.
S
Figure-1
ER
2. Superimpose OAB over OCD and OBC over ODA as shown.
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.
Figure-2
Observations : In figure 2, we see that OAB exactly covers OCD and OBC exactly covers
ODA.
Or OAB  OCD and OBC  ODA.
So, OA = OC and OB = OD.
Conclusion : From the above activity, it is verified that the diagonals of a parallelogram bisect each
other.
Activity-4
Objective :To show that the figure obtained by joining the mid points of consecutive sides of a
quadrilateral is a parallelogram.
32
Materials Required : White sheets of paper, a pair of scissors,
colour pencils, gluestick, geometry box, etc.
Procedure :
1. On a white sheet of paper, draw a quadrilateral ABCD and
cut it out.
Figure-1
ER
S
2. By paper folding, find the mid points of AB, BC, CD and
DA and mark them as P, Q, R and S respectively. Join P to
Q, Q to R, R to S and S to P.
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Figure-2
3. Cut out the quadrilateral PQRS. Join PR.
Figure-3
4. Cut the quadrilateral PQRS along
PR into two triangles. Superimpose the triangles PQR and
PSR such that PQ falls along RS
as shown.
Figure-4
33
Observations :
In figure 4(b), we see that PQR exactly covers RSP.
 PQ = RS and QR = SP
 PQRS is a parallelogram. [ Each pair of opposite sides of the quadrilateral are equal.]
Conclusion : From the above activity, we can say that the figure obtained by joining the mid-points
of consecutive sides of a quadrilateral is a parallelogram.
Do Yourself : Verify the above property by drawing three different quadrilaterals.
ANSWERS
Practice Exercise 8.1A
1. (c)
2. (c)
3. (d)
8. (b)
9. (b)
10. 145°
11. 40°, 80°, 100°, 140°
12. 15 cm
14. 3.5 cm
15. 12 cm
17. 45°, 60°, 75°
13. 6 cm, 4 cm
18. 90°
4. (c)
5. (c)
16. 5 cm
6. (c)
19. 62.5°
Practice Exercise 8.2A
1. (b)
2. (d)
3. (b)
4. (c)
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7. 17.5 cm
5. 20 cm
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6. 17 cm
7. (a)