1

GO

YAL

BRO

THER

S

PRAKASH

AN

QUADRILATERALS

A. SUMMATIVE ASSESSMENT

8

8.1 PROPERTIES OF PARALLELOGRAM1. Sum of the angles of a quadrilateral is

360°.

2. A diagonal of a parallelogram divides itinto 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

(ii) opposite angles are equal or

(iii) diagonals bisect each other or

(iv) a pair of opposite sides is equal andparallel

5. Diagonals of a rectangle bisect each otherand are equal and vice-versa.

6. Diagonals of a rhombus bisect each otherat right angles and vice-versa.

7. Diagonals of a square bisect each other atright angles and are equal, and vice-versa.

TEXTBOOK’S EXERCISE 8.1

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)]

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 =360

30

= 12°

3x= 3 × 12° = 36°, 5x= 5 × 12° = 60°,

9x= 9 × 12° = 108°, 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.

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° 2ABC = 180° [From (i) and (ii)] ABC = BAD = 90°

This shows that ABCD is a parallelogramone 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.

Question Bank In Mathematics Class IX (Term II)

2

GO

YAL

BRO

THER

S

PRAKASH

AN

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.4. Show that the diagonals of a square are

equal and bisect each other at right angles.

[2011 (T-II)]

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]

Now in AOB and COD,

AB = DC [Sides of a square]

AOB = COD

[Vertically opposite angles]

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°]

2BAC = 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.

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.

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]

3

GO

YAL

BRO

THER

S

PRAKASH

AN

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]

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 = DCAHence, AC bisects DCBOr, AC bisects C Proved.

(ii) In ABC and CDAAC = AC [Common]BAC = DAC [Given]and BCA = DCA [Proved above] ABC ADC [ASA congruence] BC DC [CPCT]But AB = DC [Given]AB BC = DC = ADHence, ABCD is a rhombus Proved.

Q.7. ABCD is a rhombus. Show that diagonalAC bisects A as well as C and diagonal BDbisects 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

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]

BCA = DCA

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 bisectsA as well asC.

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 = ACABC ADC [ASA congruence] AB = AD and CB = CD [CPCT]But, in a rectangle opposite sides are equal,i.e., AB = DC and BC = ADAB = BC = CD = DA

Hence, ABCD is a square. Proved.

(ii) In ABD and CDB, we have

AD = CD [Sides of a square]

AB = BC

BD = BD [Common]

ABDCBD [SSS congruence]

So,ABD = CBD

ADB = CDB[CPCT]

Hence, diagonal BD bisects B as well as D.Proved.

[CPCT]

[CPCT]

[CPCT]

4

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.9. In parallelogram ABCD, two points P andQ 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

Sol. Given : ABCD is a parallelogram and Pand Q are points on diagonal BD such thatDP = 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 haveAD = 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.10. ABCD is a parallelogram and AP and CQare perpendiculars from vertices A and C ondiagonal BD (see Fig.). Show that

(i) APB CQD (ii) AP = CQ

[2011 (T-II)]

Sol. Given : ABCD is a parallelogram and APand CQ are perpendiculars from vertices A and Con BD.To Prove : (i) APB CQD (ii) AP = CQProof : (i) In APB and CQD, we haveABP = 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.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

(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, Band 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 isparallel and equal]

5

GO

YAL

BRO

THER

S

PRAKASH

AN

(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]

ABCDEF [SSS axiom] Proved.

Q.12. ABCD is a trapezium in which AB || CD

and AD = BC (see Fig.). [2011 (T-II)]

Show that(i) A = B

(ii) C = D(iii) ABC BAD(iv) diagonal AC = diagonal BD

Sol. Given : In trapezium ABCD, AB || CD andAD = 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

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]

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]

OrC = 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.

OTHER IMPORTANT QUESTIONS

Q.1. Two consecutive angles of a parallelogramare 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 consecutiveangles 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 aparallelogram for all of the given properties.

6

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.3. In the given figure, PQRS is aparallelogram in which PSR = 130°. RQT isequal to : [2010]

(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 angleis : [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 have100° + 75° + 105° + x° = 360° (Let the fourthangle be x°) 280° + x = 360° x = 360° – 280° = 80°

Q.5. In a parallelogram sum of its twoadjacent 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 :[2010]

(a) Rhombus (b) Rectangle(c) Trapezium (d) None of these

Sol. (a) The diagonals of a rhombus bisecteach other at right angles.

Q.7. In the figure, ABCD is a parallelogram.The values of x and y are respectively :

(a) 70°, 110° (b) 70°, 70°

(c) 110°, 70° (d) 70°, 40°

Sol. (a) x + 110° = 180° (Linear pair) x = 180° – 110° = 70°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∴ CDB + CBD + DCB = 180° x + x + 80° = 180° 2x = 180° – 80° 2x = 100° x = 50°

Q.9. The diagonals of a rhombus are 12 cmand 16 cm. The length of the side of the rhombusis : [2011 (T-II)]

(a) 10 cm (b) 12 cm (c) 8 cm (d) 16 cmSol. (a) In right ∆AOB, we have

AB2 = AO2 + OB2

= 82 + 62 = 100 AB2 = 100 AB = 10 cm

Q.10. If PQRS is a parallelogram, thenQ – 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 aparallelogram are equal therefore, Q – S = 0

7

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.11. Two adjacent angles of a rhombus are3x – 40° and 2x + 20°. The measurement of thegreater 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 :

[2011 (T-II)]

(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,

DAC = ACD = 40°

ADC = 180° – 40° – 40° = 100°

Q.14. In the given figure, ABCD is aparallelogram. If B = 100°, then A + C isequal to : [2011 (T-II)]

(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 parallelogramare (2x + 30)° and (3x + 30)°. The value of x is :

[2011 (T-II)]

(a) 30° (b) 60° (c) 24° (d) 36°

Sol. Since, adjacent angles of a parallelogramare supplementary.

2x + 30° + 3x + 30° = 180°

5x = 180° – 60° = 120°

5x = 120° x = 24°

Q.16. ABCD is a quadrilateral and AP and DPare bisectors of A and D. The value of x is :

[2011 (T-II)]

(a) 60° (b) 85° (c) 95° (d) 100°Sol. (c) x = APD

= 180° –1

2{360° – (130° + 60°)}

= 180° –1

2(360° – 190°) = 180° – 85° = 95°

Q.17. In quadrilateral PQRS, if P = 60° andQ : R : S = 2 : 3 : 7, then S is equal to :

[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°

8

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.18. In the given figure, ABCD is a rhombusin which diagonals AC and BD intersect at O.Then mAOB is [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 ratio8 : 15, then the remaining two angles are :

[2011 (T-II)]

(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°

Q.22. In the given figure, the measure ofDOC is equal to :

(a) 90° (b) 180° (c) 118° (d) 62°

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 :

[2010, 2011 (T-II)]

(a) 90° (b) 120° (c) 150° (d) 180°

Sol. (d) Since, diagonals of a rectangle bisecteach other, therefore,

R = Sand 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 rectanglein which APB = 100°. The value of x is :

[2010]

(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 squareis 8 cm, then its area is : [2010]

(a) 64 cm2 (b) 32 cm2 (c) 16 cm2 (d) 48 cm2

9

GO

YAL

BRO

THER

S

PRAKASH

AN

Sol. (b) Let the side of square is a, then

diagonal a 2 = 8

a2 =64

2= 32 ( 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) 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 quadrilateralABCD are produced as shown in the figure. Thenwhich of the following relations is true?

[HOTS]

(a) x + y = ∠1 + ∠2 + ∠3 + ∠4(b) x – y = ∠1 + ∠2 + ∠3 + ∠4

(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 aparallelogram in which PT and QT are anglebisectors of P and Q respectively. Measure ofPTQ is : [2010]

(a) 90° (b) 60° (c) 80° (d) 100°

Sol. (a) Since adjacent angles of aparallelogram are supplementary.

P + Q = 180° 1

2(P + Q) = 90°

In PTQ,1

2P +

1

2Q + PTQ = 180°

PTQ = 90°

Q.30. If the two adjacent angles of aparallelogram are (3x – 20)° and (50 – x)° thenvalue 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°

Q.31. In the given figure, ABCD is a rhombus.The value of x is : [2010]

(a) 60° (b) 20° (c) 30° (d) 40°Sol. (c) In ODC, we have

x + 2x + 90° = 180°

10

GO

YAL

BRO

THER

S

PRAKASH

AN

[ diagonals of a rhombus bisect eachother at right angles]

3x + 90° = 180° 3x = 90° x = 30°

Q.32. Can the 95°, 70°, 110° and 80° be theangles of a quadrilateral ? Why or why not ?

Sol. We know that the sum of angles of aquadrilateral 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)]

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 anglesof 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. Provethat 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 + bQ.36. Two opposite angles of a parallelogram

are (3x – 2)° and (63 – 2x)°. Find all the anglesof the parallelogram. [2011 (T-II)]

Sol. Since opposite angles of a parallelogramare equal. 3x – 2 = 63 – 2x 3x + 2x = 63 + 2 5x = 65 x = 13Angles 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 measureof all the four angles of the parallelogram.

[2011 (T-II)]

Sol. Since opposite angles of a parallelogramare 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 thefour angles of the parallelogram. [2011 (T-II)]

Sol. Let angles are 2x° and 3x°. Sinceadjacent angles of a parallelogram aresupplementary. 2x° + 3x° = 180° 5x° = 180° x = 36° angles are 72°, 108°, 72°, 108°

11

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.39. In a parallelogram PQRS, if P =(3x – 5)°, Q = (2x + 15)°, find the value of x.

[2011 (T-II)]

Sol. Since adjacent angles of a parallelogramare 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]

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 linesegment DX cuts the side BC at X and thediagonal AC at O such that COD = 105°. Findthe value of x. [2011 (T-II)]

Sol. The angles of a square are bisected bythe diagonals.

OCX = 45° ( DCB = 90° and CA bisectsDCB)

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.42. Show that each angle of a rectangle is aright angle. [2010]

Sol. ABCD is a rectangle

ABCD is a parallelogram. AD || BC.Now, AD || BC and line AB intersects them at Aand B. A + B = 180°

[Sum of the interior angles on thesame side of transversal is 180°]

90° + B = 180° [A = 90° (Given)] B = 90°Similarly, we can show that C = 90° andD = 90°.

Hence, A = B = C = D = 90°.

Q.43. The angles of a quadrilateral are in theratio 2 : 3 : 5 : 8. Find the angles of thequadrilateral. [2010]

Sol. Let A = 2x°, B = 3x°, C = 5x° andD = 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° and8x° = 8 × 20° = 160°

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)

ADC = 58° ( ABC = 58°)

12

GO

YAL

BRO

THER

S

PRAKASH

AN

ODC = 29° ( ODC =1

2ADC)

In ∆OCD, OCD + ODC + COD = 180°

OCD + 29° + 90° = 180°

OCD = 61° ACD = 61°Q.45. Diagonals of a parallelogram are

perpandicular to each other. Is this statementtrue? Give reason for your answer.

Sol. This statement is false, becausediagonals of a parallelogram are bisect each otherbut 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 aquadrilateral is 360°.

Q.47. In the figure, BDEE and FDCE areparallelograms. Can you say that BD = CD?Given reason for your answer. [HOTS]

Sol. Since, BDEF and FDCE areparallelograms, therefore F, D and E are sure mid-points of AB, BC and AC respectively. Clearly,since D is the mid-point of BC, therefore, we cansay that BD = CD.Q.48. In the given figure, PQRS is a

parallelogram in which PL and RM are bisectorsof P and R respectively. Prove that PMRL isa parallelogram. [2010, 2011 (T-II)]

Sol. We are given a parallelogram PQRS inwhich PL and MR bisect P and R respectively

and we have to show PMRL is a parallelogram.

In PSL and RQM, we haveS = Q …(i) (Opp. s of a ||gm)

SPL =1

2P …(ii) (Given)

andQRM=1

2R …(iii)

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.

Q.49. Find the measure of each angle of aparallelogram, if one of its angles is 30° lessthan the twice the smaller angle.

[2010, 2011 (T-II)]

Sol. Let smallest angle of the parallelogram isx°, then adjacent angle = (2x – 30)°.Since adjacent angles of the parallelogram aresupplementary.x + 2x – 30 = 180°

3x = 180° + 30° x =210

3

= 70°

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 rightangles. [2011 (T-II)]

Sol. Given : A parallelogram ABCD such thatthe bisectors of adjacent angles A and B intersectat O.

To Prove : AOB = 90°.

13

GO

YAL

BRO

THER

S

PRAKASH

AN

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

2A +

1

2B = 90°

1 + 2 = 90° …(i)

AO is the bisector of A and OB is the bisector of B.

1 11 = A and 2 = B

2 2

In ∆AOB, we have 1 + AOB + 2 = 180°

90° + AOB = 180°

[From (i), 1 + 2 = 90°]

AOB = 90°Q.51. The angle between two altitudes of a

parallelogram through the vertex of an obtuseangle of the parallelogram is 60°. Find theangles of the parallelogram. [2011 (T-II)]

Sol. Let ABCD be a parallelogram in whichD is obtuse.Through vertex D altitudes DM and DN aredrawn.Also, we have MDN = 60° (Given)Now, in quad. BMDN,MDN + DMB + MBN + BND = 360°

(Angle sum property of a quadrilateral)

Or, 60° + 90° + MBN + 90° = 360°

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 aresupplementary)

Or, BAD = 180° – ABC = 180° – 120° = 60°

Also, BCD = BAD

(Opposite angles of a parallelogram are equal)

Or, BCD = 60°

Hence, the angles of parallelogram are :

120°, 60°, 120° and 60°.

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)]

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]

14

GO

YAL

BRO

THER

S

PRAKASH

AN

Since, QA is the bisector of Q. Therefore,

PQA = AQR = 60°

Now, PQ || RS, and transversal AQ intersects

them.

∴ RAQ = PQA

RAQ = 60° [ PQA = 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

A is the mid-point of RS. Proved.

Q.53. In the given figure, PQRS is aparallelogram in which PQ is produced to T suchthat QT = PQ. Prove that ST bisects RQ.

[2010, 2011 (T-II)]

Sol. Since PQRS is a parallelogram.

Therefore, PQ || RS

Now, PQ || RS and transversal QR intersectsthem.

1 = 2 …(i)

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)]

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 aparallelogram ABCD such that AE = CF. Showthat BFDE is a parallelogram. [2011 (T-II)]

Sol.

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.

Q.55. In a triangle ABC, median AD isproduced to X such that AD = DX. Prove thatABXC 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)

CD

B

A

X

15

GO

YAL

BRO

THER

S

PRAKASH

AN

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 = CXand AB || CX

Hence, ABXC is a parallelogram. ProvedQ.56. Prove that bisectors of the angles of a

parallelogram form a rectangle. [2010]

Sol. Given : A parallelogram ABCD in whichbisectors 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 intersectsthem at A and B respectively. Therefore,

A + B = 180°[ Sum of consecutive interior angles is 180°]

1

2A +

1

2B = 90°

BAS + ABS = 90° ... (i) [AS and BSare bisectors of A and B respectively]

But, in ABS, we have

BAS + ABS + ASB = 180°

[Sum of the angles of a triangle is 180°]

90° + ASB = 180°

ASB = 90°

RSP = 90°

ASB and RSP are vertically opposite angles

RSP = ASB

Similarly, we can prove that SRQ = 90°,RQP = 90° and SPQ = 90°

Hence, PQRS is a rectangle.Q.57. Prove that the diagonals of a

parallelogram divide it into four triangles ofequal area. [2010]

Sol. Given : A parallelogram ABCD. Thediagonals AC and BD intersect at O.

To Prove : ar (OAB) = ar (OBC)

= ar (OCD) = ar (AOD)

Proof : Since the diagonals of a parallelogrambisect each other at the point of intersection.

OA = OC and OB = OD

Also, the median of a triangle divides it into twoequal 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 areperpendicular to each other. [2010, 2011 (T-II)]

Sol. Given : A rhombus ABCD whose diagonalsAC and BD intersect at O.

To Prove : BOC = DOC = AOD = AOB= 90°

Proof : We know that a parallelogram is arhombus, if all of its sides are equal. So ABCDis a rhombus ABCD is a ||gm such that

16

GO

YAL

BRO

THER

S

PRAKASH

AN

AB = BC = CD = DA …(i)Since the diagonals of a parallelogram bisecteach other.

OB = OD and OA = OC …(ii)

Now, in s BOC and DOC, we haveBO = OD [From (ii)]BC = DC [From (i)]OC = OC [Common]

So, by SSS criterion of congruenceBOC 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 diagonalBD. Prove that AP = CQ. [2010, 2011 (T-II)]

Sol. Given : ABCD is a parallelogram, inwhich AP BD and CQ BD.

To Prove : AP = CQ.

Proof : In APB and CDQ,

AB = CD [Opposite sides of a parallelogram]

P = Q [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

Q.60. In a quadrilateral ABCD, AO and BOare the bisectors of A and B respectively.

Prove that AOB =1

2(C + D).

[2010, 2011 (T-II)]

Sol. In AOB, we have

AOB + 1 + 2 = 180° AOB = 180° – (1 + 2)

AOB = 180° –1 1

A + B2 2

1 11 = A and 2 = B

2 2

AOB = 180° –1

2(A + B)

AOB = 180° –1

2[360° – (C + D)]

A + B + C + D = 360°A + B = 360° ( C + D)

AOB = 180° – 180° +1

2(C + D)

AOB =1

2(C + D)

Q.61. In a parallelogram ABCD, the bisectorof A also bisects BC at X. Prove thatAD = 2AB. [2010, 2011 (T-II)]

Sol. Since AX is the bisector of A.

1 =1

2A …(i)

Since ABCD is a parallelogram.Therefore, AD || BC and AB intersects them.

17

GO

YAL

BRO

THER

S

PRAKASH

AN

A + B = 180° B = 180° – AIn ABX, we have 1 + 2 + B = 180°

1

2A + 2 + 180° – A = 180°

2 –1

2A = 0 2

1

2A …(ii)

From (i) and (ii), we have 1 = 2

Thus, in ABX, we have 1 = 2

BX = AB [ Sides opposite to equal anglesin 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 aparallelogram divides it into two congruenttriangles. [2010, 2011 (T-II)]

Sol. Given : ABCD is a parallelogram inwhich AC is a diagonal.

To Prove : ∆ABC ∆CDA.Proof : In ∆ABC and ∆CDA since, ABCD isparallelogram.∴ 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 isoscelestriangle in whichAB = AC, CD || AB andAD is bisectors ofexterior ∠CAE of ∆ABC.Prove that ∠CAD =∠BCA and ABCD is aparallelogram.

[2010, 2011 (T-II)]

Sol. Given : An isosceles ∆ABC having AB =AC. AD is the bisector of exterior CAE andCD || AB

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]

Now, in ∆ABC, we have, ext. CAE = 1 + 2

[ An exterior angle is equal to the sum of

two opposite interior angles]

ext. CAE = 22 [ 1 = 2 from (i)]

23 = 22 [ AD is the bisector of

ext. CAE ∴ CAE = 23]

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

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 orRS is joined to Q and S and lines through Pparallel to QM and through R parallel to SMmeet at N. Prove that line segments MN and PQare 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°

18

GO

YAL

BRO

THER

S

PRAKASH

AN

(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

1 Mark Questions

1. In a rectangleABCD, diagonalsAC 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

2. Which of the following is not true for a

parallelogram ?

(a) opposite sides are equal

(b) opposite angles are equal

(c) opposite angles are bisected by the diago-

nals

(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 inter-

sect at O. If ∠BOC = 90° and ∠BDC = 50°, then

∠OAB is :

(a) 90° (b) 50° (c) 40° (d) 10°

6. IfAPB 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

7. In a parallelogramABCD, bisectors of twoadjacent angles A and B meet at O. The measure ofthe angle AOB is equal to :

(a) 90° (b) 180° (c) 60° (d) 360°

8. Lengths of two adjacent sides of a paral-lelogram are in the ratio 2 : 7. If its perimeter is180 cm, then the adjacent sides of the parallelo-gram 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 quadri-lateral such that a = 2b, b = 2c and c = 2d, then thevalue of d is :

(a) 36° (b) 24°

(c) 30° (d) none of these

2 Marks Questions

10. Diagonals of a quadrilateral PQRS bisecteach other. If ∠P = 35°, find ∠Q.

11. The angles of a quadrilateral are in the ra-tio 2 : 4 : 5 : 7. Find the angles.

12. The lengths of the diagonals of a rhombusare 24 cm and 18 cm. Find the length of each sideof the rhombus.

13. Diagonals AC and BD of a parallelogramABCD intersect each other at O. If OA = 3 cm andOD = 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 ofAB and BC, determine the length of DE.

19

GO

YAL

BRO

THER

S

PRAKASH

AN

15. ABCD is a rhombus AC = 16 cm, BC =10 cm. Find the length of the diagonal BD.

[2011 (T-II)]

16. In the given figure, ABCD is a rhombus.AO = 5 cm, area of the rhombus is 25 sq. cm. Findthe length of BD. [2011 (T-II)]

17. In the given figure, ABCD is a parallelo-gram. Compute DCA, ACB and ADC, givenDAC = 60° and ABC = 75°. [2011 (T-II)]

18. ABCD is a parallelogram. The angle bisec-tors of A and D intersect at O. Find the mea-sure of AOD. [2011 (T-II)]

19. If PQRS is a rhombus with PQR = 55°,find PRS. [2011 (T-II)]

3 Marks Questions

20. Show that the quadrilateral formed by join-ing the mid-points of sides of a square is also square.

21. E is the mid-point of the side AD of trape-ziumABCD withAB | | DC.Aline through E drawnparallel to AB intersect BC at F. Show that F is themid-point of BC.

22. If one angle of a parallelogram is a rightangle, it is a rectangle. Prove.

23. Two parallel lines ‘l’and ‘m’are intersectedby a transversal ‘p’. Show that the quadrilateralformed by the bisectors of interior angles is a rect-angle. [2011 (T-II)]

24. In the given figure, ABCD is a square, ifPQR = 90° and PB = QC = DR, prove thatQB = RC, PQ = QR, QPR = 45°. [2011 (T-II)]

4 Marks Questions

25. ABCD is a rectangle in which diagonal BDbisects ∠B. Show that ABCD is a square.

26. If ABCD is a trapezium in which AB | | CDand AD = BC, prove that ∠A = ∠B.

27. P, Q, R, and S are respectively the mid-points of the sides AB, BC, CD and DA of a quad-rilateral ABCD in which AD = BC. Prove thatPQRS is a rhombus.

8.2 THE MID-POINT THEOREM1. 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.

TEXTBOOK’S EXERCISE 8.2

Q.1. ABCD is a quadrilateral in which P, Q, Rand S are mid-points of the sides AB, BC, CDand DA respectively. (see Fig.). AC is adiagonal. Show that : [2011 (T-II)]

20

GO

YAL

BRO

THER

S

PRAKASH

AN

(i) SR || AC and SR =1

2AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Given : ABCD is a quadrilateral in which P, Q, Rand S are mid-points of AB, BC, CD and DA. ACis a diagonal.To Prove :

(i) SR || AC and SR =1

2AC

(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.

PQ || AC and PQ =1

2AC …(1)

[Mid-point theorem]In ADC, R is the mid-point of CD and Sis the mid-point of AD

SR || AC and SR =1

2AC …(2)

[Mid-point theorem](ii) From (1) and (2), we get

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 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 || AC and PQ =1

2AC …(i)

[Mid-point theorem]Similarly, in DAC,

SR || AC and SR =1

2AC …(ii)

From (i) and (ii), we have PQ || SR and PQ = SRPQRS 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.

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 =1

2AC …(i)

[Mid-point theorem]Similarly, in ADC,

21

GO

YAL

BRO

THER

S

PRAKASH

AN

SR || AC and SR =1

2AC …(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

2AD =

1

2BC AS = BQ

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)]

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

Fig.). Show that the line segments AF and EC

trisect the diagonal BD. [2010, 2011 (T-II)]

Sol. Given : A parallelogram ABCD, in which Eand F are mid-points of sides AB and DCrespectively.

To Prove : DP = PQ = QB

Proof : Since E and F are mid-points of AB andDC respectively.

AE =1

2AB and CF =

1

2DC …(i)

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 = QBor line segments AF and EC trisect the diagonal

BD. Proved.

Q.6. Show that the line segments joining themid-points of the opposite sides of aquadrilateral bisect each other. [V. Imp.]

Sol. Given : ABCD is a quadrilateral in whichEG and FH are the line segments joining the mid-points of opposite sides.

22

GO

YAL

BRO

THER

S

PRAKASH

AN

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.

EF =1

2AC and EF || AC …(i)

In ADC, H and G are mid-points of AD and CD

respectively.

HG =1

2AC and HG || AC …(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.

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.

(ii) MD AC (iii) CM = MA =1

2AB

[2010, 2011 (T-II)]

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]

(ii) DM BC (iii) CM = MA =1

2AB

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,

MA =1

2AB …(3)

Hence, CM = MA =1

2AB Proved.

[From (2) and (3)]

OTHER IMPORTANT QUESTIONS

Q.1. In the figure, D, E and F are the mid-points of the sides AB, BC and CA respectively.If AC = 8.2 cm, then value of DE is :

(a) 8.2 cm (b) 4.1 cm(c) 2.05 cm (d) none of these

Sol. (b) DE =1

2AC =

1

2× 8.2 = 4.1 cm

Q.2. In the figure, D and E are mid-points ofthe sides AB and AC respectively of ABC.Measure of B is : [2010]

23

GO

YAL

BRO

THER

S

PRAKASH

AN

(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 andS are the mid-points of the sides AB, BC, CD, DArespectively. If BD = 12 cm, then length of QRis : [2010]

(a) 6 cm (b) 8 cm (c) 3 cm (d) 4 cmSol. (d) By mid-point theorem,

QR =1

3BR =

1

3× 12 = 4 cm

Q.4. In the figure, P and Q are mid-points ofsides AB and AC respectively of ABC. If PQ =3.5 cm and AB = AC = 9 cm, then the perimeterof ABC is : [Imp.]

(a) 20 cm (b) 23 cm(c) 25 cm (d) 27 cm

Sol. (c) We have, AB = AC = 9 cm.

By mid-point theorem, PQ =1

2BC

BC = 2PQ∴ BC = 2 × 3.5 = 7 cm

∴ Perimeter of ∆ ABC = 9 cm + 9 cm + 7 cm

= 25 cm

Q.5. In the ABC, B is a right angle, D andE are the mid-points of the sides AB and ACrespectively. If AB = 6 cm and AC = 10 cm, thenthe length of DE is : [Imp.]

(a) 3 cm (b) 5 cm (c) 4 cm (d) 6 cm

Sol. (c) BC = 2AC 2– AB = 2(10) 2– (6)

= 100 – 36 64 = 8 cmAlso, D and E the mid-points of AB and ACrespectively.

DE =1

2BC (By mid-point theorem)

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 thefigure. The perimeter of ∆DEF is : [V. Imp.]

(a)1

2(AB +BC + CA)

(b) AB + BC + CA(c) 2 (AB + BC + CA)

(d) none of these

Sol. (a)

1We have, EF = BC,

2

1DF = AC (By mid-point theorem)

2

1and DE = AB

2

24

GO

YAL

BRO

THER

S

PRAKASH

AN

Therefore, perimeter of ∆DEF = DE + EF + DF

=1

2(AB + BC + CA)

Q.7. In the ABC, E and F are the mid-pointsof AB and AC respectively. The altitude APintersects EF at Q. The correct relation betweenAQ and QP is : [HOTS]

(a) AQ > QP (b) AQ = QP(c) AQ < QP (d) none of these

Sol. (b) In ABC, E and F are the mid-pointof 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 = QPQ.8. D and E are the mid-points of the sides

AB and AC of ABC and O is any point on sideBC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQPis :

(a) a square (b) a rectangle(c) a rhombus (d) a parallelogram

[HOTS]

Sol. (d) In ABC, D and E are respectivelythe mid-points of the sides AB and AC, then bymid-point theorem we have, DE || BC or,DE || PQ … (i)

Also, in ∆AOB, D and P are respectively the

mid-points of AB and BO, then by mid-point

theorem

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 ofAC.

Now, AD is the median in ABCD is the mid-point of BC.

Since DE is a line drawn through the mid-pointof 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 mid-points of AB and BC, determine the length of DE.

[2011 (T-II)]

Sol. Since, the line segment joining the mid-points of any two sides of a triangle is parallel tothe third side and is half of it.

∴ DE =1

2AC ⇒ DE =

1

2× 7 = 3.5 cm

25

GO

YAL

BRO

THER

S

PRAKASH

AN

Q.11. In ∆ABC, P, Q and R are mid-points ofsides BC, CA and AB respectively. If AC = 21 cm,BC = 29 cm and AB = 30 cm, find the perimeterof the quadrilateral ARPQ.

Sol. Clearly, AQ =1

2AC =

1

2× 21 cm = 10.5 cm

AR =1

2AB =

1

2× 30 cm = 15 cm

RP =1

2AC =

1

2× 21 cm = 10.5 cm

(By mid-point theorem)

PQ =1

2AB =

1

2× 30 cm = 15 cm

(By mid-point theorem)

Now, perimeter of ARPQ = AR + RP + PQ + AQ

= 15 cm + 10.5 cm + 15 cm + 10.5 cm = 51 cm

Q.12. In there are three or more parallel linesand the intercepts made by them on thetransversal are equal the correspondingintercepts on any other transversal are alsoequal. [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, Crespectively such that AB = BC. Also, q isanother transversal which cuts l, m, n in P, Q, Rrespectively.To Prove : PQ = QRConstruction : Through Q, draw a line r suchthat 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]

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 linesRQ, PQ and PR have been drawn respectivelyparallel to sides BC, CA and AB of a ABC.

Show that BC =1

2QR. [HOTS]

Sol. Given : ABC, lines are drawn throughA, B and C parallel respectively to the sides BC,CA and AB forming ∆PQR.

To prove : BC =1

2QR.

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

26

GO

YAL

BRO

THER

S

PRAKASH

AN

∴ ARCB is a parallelogram.

BC = AR … (ii) [opposite sides of a ||gm]

From (i) and (ii), QA = AR =1

2QR. … (iii)

From (i) and (iii), BC =1

2QR. Proved.

Q.14. Show that the line segments joining themid-points of opposite sides of a quadrilateralbisect 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, CDand DA respectively.

We have to show that EG and HF bisect eachother. Join EF, FG, GH, HE and AC.

In ABC, E is the mid point of AB and F is themid point of BC.

EF || AC and EF =1

2AC …(i)

Similarly, HG || AC and HG =1

2AC …(ii)

From (i) and (ii), EF || HG and EF = HG

EFGH is a parallelogram and EG and HF areits diagonals. We know that the diagonals of aparallelogram bisect each other. Thus, EG andHF bisect each other.Q.15. In the given figure, AD and BE are

medians of ABC and BE || DF. Prove that

CF =1

4AC. [2010, 2011 (T-II)]

Sol. In BEC, DF is a line through the mid-point D of BC and parallel to BE intersecting CEat F. Therefore, F is the mid-point of CE.Because the line drawn through the mid-point ofone side of a triangle and parallel to another sidebisects the third side. Now, F is the mid-point ofCE.

CF =1

2CE

CF =1

2

1AC

2

E is the mid-point of AC1

EC = AC2

CF =1

4AC

Q.16. In ABC, AD is the median through Aand E is the mid-point of AD. BE produced meets

AC in F. Prove that AF =1

3AC.

[2010, 2011 (T-II)]

Sol. Through D, draw DK || BF. In ADK, Eis 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

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

3AC

Q.17. Prove that the straight line joining themid-points of the diagonals of a trapezium isparallel to the parallel sides. [2010]

27

GO

YAL

BRO

THER

S

PRAKASH

AN

Sol. We are given a trapezium ABCD inwhich AB || DC and M, N are the mid-points ofthe diagonals AC and BD.We need to prove that MN || AB || DC.Join CN and let it meet AB at E.

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 ofthe sides AC and CE respectively.

MN || AE or MN || ABAlso AB || DC

MN || AB || DC

Q.18. In the given figure, D, E and F are,respectively the mid-points of sides BC, CA andAB of an equilateral triangle ABC. Prove thatDEF is also an equilateral triangle.

[2010]

Sol. Since the segment joining the mid-pointsof two sides of a triangle is half of the third side.Therefore, D and E are mid-points of BC and ACrespectively.

DE1

2AB …(i)

E and F are the mid-points of AC and ABrespectively.

EF =1

2BC …(ii)

F and D are the mid-points of AB and BCrespectively.

FD =1

2AC …(iii)

Now, ABC is an equilateral triangle. AB = BC = CA

1

2AB =

1

2BC =

1

2CA

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 ofa triangle, are concurrent to each other.

[2011 (T-II)]

Sol. Given : A triangle ABC and D, E, F arethe mid-points of sides BC, CA and ABrespectively.

To Prove : AFE FBD EDC DEF

Proof : Since the segment joining the mid-pointsof the sides of a triangle is half of the third side.Therefore,

DE =1

2AB DE = AF = BF …(i)

EF =1

2BC EF = BD = CD …(ii)

DF =1

2AC DF = AE = EC …(iii)

Now, in s DEF and AFE, we have

DE = AF [From (i)]

DF = AE [From (ii)]

and EF = FE [Common]

28

GO

YAL

BRO

THER

S

PRAKASH

AN

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 segmentjoining the mid-points of any two sides is parallelto third side and is half of it. [2011 (T-II)]

Sol. Given : A triangle ABC in which D andE are the mid-points of AB and AC respectively.DE is joined.

To prove : DE || BC and DE =1

2BC.

Construction : Draw CF || BA, meeting DEproduced in F.Proof : In s AED and CEF, we haveAE = CE (given), AED = CEF

(vert. opp. angles)DAE = FCE (Alt. Int. angles)AED CEFAD = 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

2DF =

1

2BC

[ DE = EF]

Hence, DE || BC and DE =1

2BC.

PRACTICE EXERCISE 8.2A

1 Mark Questions

1. In the given figure, D is the mid-point ofAB and DE || BC, then AE is equal to :

(a) AD (b) EC (c) DB (d) BC

2. In the given figure, D and E are mid-pointsof AB and AC respectively. The length of DE is :

(a) 8.2 cm (b) 5.1 cm(c) 4.9 cm (d) 4.1 cm

3. In the given figure, if DE = 4 cm, BC =8 cm and D is the mid-point of AB, then the truestatement is :

(a) AB = AC (b) DE || BC(c) E is not the mid-point of AC(d) DE BC

4. In the givenfigure, ABCD is arectangle. P and Q aremid-points of AD andDC respectively. Thenlength of PQ is :

(a) 5 cm (b) 4 cm (c) 2.5 cm (d) 2 cm

29

GO

YAL

BRO

THER

S

PRAKASH

AN

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, geometrybox, etc.

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.

[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 andBE meet at P and CF and DE meet at Q. Prove

that PQ =1

4BC.

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

EF =1

2(AB + CD).

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.

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)]

12. In the given figure, PQRS is a square. M isthe mid-point of PQ and AB RM. Prove thatRA = RB.

13. Show that the quadrilateral formed byjoining the mid-points of the sides of rhombus,taken in order, form a rectangle. [2011 (T-II)]

14. The diagonals of a quadrilateral ABCD areperpendicular, show that quadrilateral formed byjoining the mid-points of its sides, is rectangle.

[2011 (T-II)]

15. In the given figure, points A and B are onthe same side of a line m, AD m and BE mand meet m at D and E respectively. If C is themid-point of AB, prove that CD = CE.

[2011 (T-II)]

30

GO

YAL

BRO

THER

S

PRAKASH

AN

Procedure :

1. On a white sheet of paper, draw a ABC and cut it out.

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.

3. Cut out the triangular piece AXY and superimpose AY over YC such that YX falls along CBas 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 isparallel to the third side and is equal to half of it.

Figure-2

Figure-1

31

GO

YAL

BRO

THER

S

PRAKASH

AN

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. Verifythe 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, geometrybox, etc.

Procedure :

1. On a white sheet of paper, draw a parallelogram ABCD and cut it out. Draw the diagonal ACof the parallelogram and cut it along AC to get two triangular cut outs.

Figure-1

2. Now, superimpose one triangle over the other as shown below.

Observations : In figure 2, we see that the two triangles exactly cover each other. Hence, thetriangles are congruent.

Conclusion : From the above activity, it is verified that a diagonal of a parallelogram divides itinto two congruent triangles.

Do Yourself : Draw three different parallelograms and verify the above property by paper cuttingand pasting.

Figure-2

32

Figure-2

GO

YAL

BRO

THER

S

PRAKASH

AN

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, geometrybox, etc.

Procedure :

1. On a white sheet of paper, draw a parallelogram ABCD and both its diagonals AC and BDintersecting at O. Cut out the four triangles so formed.

Figure-1

2. Superimpose OAB over OCD and OBC over ODA as shown..

Observations : In figure 2, we see that OAB exactly covers OCD and OBC exactly coversODA.

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 eachother.

Activity-4

Objective :To show that the figure obtained by joining the mid points of consecutive sides of aquadrilateral is a parallelogram.

33

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 andcut it out.

2. By paper folding, find the mid points of AB, BC, CD andDA and mark them as P, Q, R and S respectively. Join P toQ, Q to R, R to S and S to P.

3. Cut out the quadrilateral PQRS. Join PR.

4. Cut the quadrilateral PQRS along

PR into two triangles. Super-

impose the triangles PQR and

PSR such that PQ falls along RS

as shown.

Figure-1

Figure-2

Figure-3

Figure-4

GO

YAL

BRO

THER

S

PRAKASH

AN

34

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.

GO

YAL

BRO

THER

S

PRAKASH

AN

ANSWERS

Practice Exercise 8.1A

1. (c) 2. (c) 3. (d) 4. (c) 5. (c) 6. (c) 7. (a)

8. (b) 9. (b) 10. 145° 11. 40°, 80°, 100°, 140° 12. 15 cm

13. 6 cm, 4 cm 14. 3.5 cm 15. 12 cm 16. 5 cm 17. 45°, 60°, 75°

18. 90° 19. 62.5°

Practice Exercise 8.2A

1. (b) 2. (d) 3. (b) 4. (c) 5. 20 cm 6. 17 cm

7. 17.5 cm