# Important Questions for CBSE Class 9 Maths Chapter 8 - Quadrilaterals

## CBSE Class 9 Maths Chapter-8 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 9 Maths Chapter 8 - Quadrilaterals prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your examination.

1 Marks Quetions

1. A quadrilateral ABCD is a parallelogram if

(a) AB = CD

(b) ABBC

(c)

Ans. (c)

2. In figure, ABCD and AEFG are both parallelogram if

(a)

(b)

(c)

(d)

Ans. (c)

3. In a square ABCD, the diagonals AC and BD bisects at O. Then is

(a) acute angled

(b) obtuse angled

(c) equilateral

(d) right angled

Ans. (d) right angled

4. ABCD is a rhombus. If

(a)

(b)

(c)

(d)

Ans. (c)

5. In fig ABCD is a parallelogram. If and then is

Ans.

6. If the diagonals of a quadrilateral bisect each other, then the quadrilateral must be.

(a) Square

(b) Parallelogram

(c) Rhombus

(d) Rectangle

Ans. (b) Parallelogram

7. The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular bisector of each other then quadrilateral ABCD is a

(a) Kite

(b) Square

(c) Trapezium

(d) Rectangle

Ans. (b) Square

8. The quadrilateral formed by joining the mid points of the sides of a quadrilateral ABCD taken in order, is a rectangle if

(a) ABCD is a parallelogram

(b) ABCD is a rut angle

(c) Diagonals AC and BD are perpendicular

(d) AC=BD

Ans. (a) ABCD is a parallelogram

9. In the fig ABCD is a Parallelogram. The values of and are

(a) 30, 35

(b) 45, 30

(c) 45, 45

(d) 55, 35

Ans. (b) 45, 30

10. In fig if DE=8 cm and D is the mid-Point of AB, then the true statement is

(a) AB=AC

(b) DE||BC

(c) E is not mid-Point of AC

(d) DEBC

Ans. (c) E is not mid-Point of AC

11. The sides of a quadrilateral extended in order to form exterior angler. The sum of these exterior angle is

(a)

(b)

(c)

(d)

Ans. (d)

12. ABCD is rhombus with The measure of is

(a)

(b)

(c)

(d)

Ans. b)

13. In fig D is mid-point of AB and DEBC then AE is equal to

(b) EC

(c) DB

(d) BC

Ans. (b) EC

14. In fig D and E are mid-points of 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

Ans. (d) 4.1 cm

15. A diagonal of a parallelogram divides it into

(a) two congruent triangles

(b) two similes triangles

(c) two equilateral triangles

(d) none of these

Ans. (a) two congruent triangles

16. A quadrilateral is a _________, if its opposite sides are equal:

(a) Kite

(b) trapezium

(d) parallelogram

Ans. (d) parallelogram

17. In the adjoining Fig. AB = AC. CD||BA and AD is the bisector of prove that

(a) and

Ans. In

[Opposite angle of equal sides are equal]

[Exterior angle]

Now

Also CD||BA Given)

is a parallelogram

(ii) ABCD is a parallelogram

18. Which of the following is not a parallelogram?

(a) Rhombus

(b) Square

(c) Trapezium

(d) Rectangle

Ans. (c) Trapezium

19. The sum of all the four angles of a quadrilateral is

(a) 1800

(b) 3600

(c) 2700

(d) 900

Ans. (b) 3600

20. In Fig ABCD is a rectangle P and Q are mid-points of AD and DC respectively. Then length of PQ is

(a)5 cm

(b) 4 cm

(c) 2.5 cm

(d) 2 cm

Ans. (c) 2.5 cm

21. In Fig ABCD is a rhombus. Diagonals AC and BD intersect at O. E and F are mid points of AO and BO respectively. If AC = 16 cm and BD = 12 cm then EF is

(a)10 cm

(b) 5 cm

(c) 8 cm

(d) 6 cm

Ans. (b) 5 cm

2 Marks Quetions

1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all angles of the quadrilateral

Ans. Let in quadrilateral ABCD, A = B = C = and D =

Since, sum of all the angles of a quadrilateral =

A + B + C + D =

Now A =

B =

C =

And D =

Hence angles of given quadrilateral are and

2. If the diagonals of a parallelogram are equal, show that it is a rectangle.

Ans. Given: ABCD is a parallelogram with diagonal AC = diagonal BD

To prove: ABCD is a rectangle.

Proof: In triangles ABC and ABD,

AB = AB [Common]

AC = BD [Given]

AD = BC [opp. Sides of a gm]

DAB = CBA [By C.P.C.T.] ……….(i)

But DAB + CBA = ……….(ii)

[ ADBC and AB cuts them, the sum of the interior angles of the same side of transversal is ]

From eq. (i) and (ii),

DAB = CBA =

Hence ABCD is a rectangle.

3. Diagonal AC of a parallelogram ABCD bisects A (See figure). Show that:

(i) It bisects C also.

(ii) ABCD is a rhombus.

Ans. Diagonal AC bisects A of the parallelogram ABCD.

(i) Since AB DC and AC intersects them.
1 = 3 [Alternate angles] ……….(i)
Similarly 2 = 4 ……….(ii)
But 1 = 2 [Given] ……….(iii)
3 = 4 [Using eq. (i), (ii) and (iii)]
Thus AC bisects C.

(ii) 2 = 3 = 4 = 1
AD = CD [Sides opposite to equal angles]
AB = CD = AD = BC
Hence ABCD is a rhombus.

4. ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See figure). Show that:

(i) APB CQD

(ii) AP = CQ

Ans. Given: ABCD is a parallelogram. AP BD and CQ BD

To prove: (i) APB CQD (ii) AP = CQ

Proof: (i) InAPB and CQD,

1 = 2 [Alternate interior angles]

AB = CD [Opposite sides of a parallelogram are equal]

APB = CQD =

APB CQD [By ASA Congruency]

(ii) Since APB CQD

AP = CQ [By C.P.C.T.]

5. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (See figure). AC is a diagonal. Show that:

(i) SR AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Ans. In ABC, P is the mid-point of AB and Q is the mid-point of BC.

Then PQ AC and PQ = AC

(i) In ACD, R is the mid-point of CD and S is the mid-point of AD.

Then SR AC and SR = AC

(ii) Since PQ = AC and SR = AC

Therefore, PQ = SR

(iii) Since PQ AC and SR AC

Therefore, PQ SR [two lines parallel to given line are parallel to each other]

Now PQ = SR and PQ SR

Therefore, PQRS is a parallelogram.

6. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Ans. Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x

[sum of angles of a quadrilateral is ]

7. Show that each angle of a rectangle is a right angle.

Ans. We know that rectangle is a parallelogram whose one angle is right angle.

Let ABCD be a rectangle.

To prove

Proof: and AB is transversal

8. A transversal cuts two parallel lines prove that the bisectors of the interior angles enclose a rectangle.

Ans. and EF cuts them at P and R.

[alternate interior angles]

[alternate]

9. Prove that diagonals of a rectangle are equal in length.

Ans. ABCD is a rectangle and AC and BD are diagonals.

To prove AC = BD

Proof:

AD = BC [In a rectangle opposite sides are equal]

AB = AB common [common]

10. If each pair of opposite sides of a quadrilateral is equal, then prove that it is a parallelogram.

Ans. Given A quadrilateral ABCD in which AB = DC and AD = BC

To prove: ABCD is a parallelogram

Construction: Join AC

Proof: In and

AB=DC

AC=AC [common]

is a parallelogram.

11.
Ans. ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each other

But [given]

Or OX=OY

Now in quadrilateral AYCX, the diagonals AC and XY bisect each other

is a parallelogram.

In fig ABCD is a parallelogram and x, y are the points on the diagonal BD such that Dx<By show that AYCX is a parallelogram.

12. Show that the line segments joining the mid points of opposite sides of a quadrilateral bisect each other.

Ans. Given ABCD is quadrilateral E, F, G, H are mid points of the side AB, BC, CD and DA respectively

To prove: EG and HF bisect each other.

In , E is mid-point of AB and F is mid-point of BC

And

Similarly, and

From (i) and (ii), and

is a parallelogram and EG and HF are its diagonals

Diagonals of a parallelogram bisect each other

Thus, EG and HF bisect each other.

13. ABCD is a rhombus show that diagonal AC bisects as well as and diagonal BD bisects as well as

Ans. ABCD is a rhombus

In and

[Sides of a rhombus]

[Sides of a rhombus]

[Common]

[By SSS Congruency]

And

Hence AC bisects as well as

Similarly, by joining B to D, we can prove that

Hence BD bisects as well as

14. In fig AD is a median of is mid-Point of AD.BE produced meet AC at F. Show that

Ans. Let M is mid-Point of CF Join DM

In is mid- Point of AD and

is mid-point of AM

FM=MC

Hence Proved.

15. Prove that a quadrilateral is a parallelogram if the diagonals bisect each other.

Ans. ABCD is a quadrilateral in which diagonals AC and BD intersect each other at O

In and

[Given]

[Given]

And [Vertically apposite angle

[By SAS]

[By C.P.C.T]

But this is Pair of alternate interior angles

16. In fig ABCD is a Parallelogram. AP and CQ are Perpendiculars from the Vertices A and C on diagonal BD.

Show that

(i)

(ii)

Ans. (I) in and

AB=DC [opposite sides of a Parallelogram]

And

[ASA]

(II) (By C.P.C.T)

17. ABCD is a Parallelogram E and F are the mid-Points of BC and AD respectively. Show that the segments BF and DE trisect the diagonal AC.

Ans. FD||BE and FD=BE

BEDF Is a Parallelogram

EG||BH and E is the mid-Point of BC

G is the mid-point of HC

Or HG=GC…………..(i)

Similarly AH=HG………….(ii)

From (i) and (ii) we get

AH=HG=GC

Thus the segments BF and DE bisects the diagonal AC.

18. Prove that if each pair of apposite angles of a quadrilateral is equal, then it is a parallelogram.

Ans. Given: ABCD is a quadrilateral in which and

To Prove: ABCD is a parallelogram

Proof: [Given]

[Given]

[By….(i)]

These are sum of interior angles on the same side of transversal

and

ABCD is a parallelogram.

19. In Fig. ABCD is a trapezium in which AB||DC E is the mid-point of AD. A line through E is parallel to AB show that bisects the side BC

Ans. Join AC

In

E is mid-point of AD and EO||DC

O is mid point of AC [A line segment joining the midpoint of one side of a parallel to second side and bisect the third side]

In

O is mid point of AC

OF||AB F is mid point of BC

Bisect BC

20. In Fig. ABCD is a parallelogram in which X and Y are the mid-points of the sides DC and AB respectively. Prove that AXCY is a parallelogram

Ans. In the given fig

ABCD is a parallelogram

AB||CD and AB = CD

And

And

[X and Y are mid-point of DC and AB respectively]

is a parallelogram

21. The angles of quadrilateral are in the ratio 3:5:10:12 Find all the angles of the quadrilateral.

Ans. Suppose angles of quadrilaterals are

30x=360

22. In fig D is mid-points of AB. P is on AC such that and DE||BP show that

Ans. In ABP
D is mid points of AB and DE||BP
E is midpoint of AP
AE = EP also PC = AP

2PC = AP
2PC = 2AE
PC = AE
AE = PE = PC
AC = AE + EP + PC
AC = AE + AE + AE
AE =AC
Hence Proved.

23. Prove that the bisectors of the angles of a Parallelogram enclose a rectangle. It is given that adjacent sides of the parallelogram are unequal.

Ans. ABCD is a parallelogram

Or [Sum of angle of a ]

Similarly, and

. Thus each angle of quadrilateral PQRS is
Hence PQRS is a rectangle.

24. Prove that a quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal

To Prove: ABCD is a parallelogram

Construction: Join AC and BD intersect each other at O.
Proof: [By AAA
Because

AO=OC
And BO=OD
ABCD is a parallelogram
Diagonals of a parallelogram bisect each other.

3 Marks Quetions

1. Show that is diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans. Given: Let ABCD is a quadrilateral.

Let its diagonal AC and BD bisect each other at right angle at point O.

OA = OC, OB = OD

AndAOB = BOC = COD = AOD =

To prove: ABCD is a rhombus.

Proof: In AOD and BOC,

OA = OC[Given]

AOD = BOC[Given]

OB = OD[Given]

AODCOB [By SAS congruency]

Again, In AOB and COD,

OA = OC[Given]

AOB = COD[Given]

OB = OD[Given]

AOBCOD [By SAS congruency]

Now In AOD and BOC,

OA = OC[Given]

AOB = BOC[Given]

OB = OB[Common]

AOBCOB [By SAS congruency]

AB = BC [By C.P.C.T.]……….(iii)

From eq. (i), (ii) and (iii),

AD = BC = CD = AB

And the diagonals of quadrilateral ABCD bisect each other at right angle.

Therefore, ABCD is a rhombus.

2. Show that the diagonals of a square are equal and bisect each other at right angles.

Ans.  Given: ABCD is a square. AC and BD are its diagonals bisect each other at point O.

To prove: AC = BD and AC  BD at point O.

Proof: In triangles ABC and BAD,

AB = AB[Common]

BC = AD [Sides of a square]

AC = BD [By C.P.C.T.] Hence proved.

Now in triangles AOB and AOD,

AO = AO[Common]

AB = AD[Sides of a square]

OB = OD[Diagonals of a square bisect each other]

AOBAOD[By SSS congruency]

AOB = AOD[By C.P.C.T.]

ButAOB + AOD = [Linear pair]

AOB = AOD =

OA  BD or AC  BD

Hence proved.

3. ABCD is a rhombus. Show that the diagonal AC bisects A as well as C and diagonal BD bisects B as well as D.

Ans.  ABCD is a rhombus. Therefore, AB = BC = CD = AD

Let O be the point of bisection of diagonals.

OA = OC and OB = OD

In AOB and AOD,

OA = OA[Common]

AB = AD[Equal sides of rhombus]

OB = OD(diagonals of rhombus bisect each other]

AOBAOD[By SSS congruency]

OA bisects A……….(i)

SimilarlyBOC DOC[By SSS congruency]

OCB = OCD[By C.P.C.T.]

OC bisects C……….(ii)

From eq. (i) and (ii), we can say that diagonal AC bisects A and C.

Now in AOB and BOC,

OB = OB[Common]

AB = BC[Equal sides of rhombus]

OA = OC(diagonals of rhombus bisect each other]

AOBCOB[By SSS congruency]

OBA = OBC[By C.P.C.T.]

OB bisects B……….(iii)

SimilarlyAODCOD[By SSS congruency]

ODA = ODC[By C.P.C.T.]

BD bisects D……….(iv)

From eq. (iii) and (iv), we can say that diagonal BD bisects B and D

4. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (See figure). Show that:

(i) APDCQB

(ii) AP = CQ

(iii) AQBCPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Ans. (i)In APD and CQB,

DP = BQ[Given]

AD = CB[Opposite sides of parallelogram]

APDCQB[By SAS congruency]

(ii) SinceAPDCQB

AP = CQ[By C.P.C.T.]

(iii) In AQB and CPD,

BQ = DP[Given]

ABQ = PDC[Alternate angles (ABCD and BD is transversal)]

AB = CD[Opposite sides of parallelogram]

AQBCPD[By SAS congruency]

(iv) SinceAQBCPD

AQ = CP[By C.P.C.T.]

AP = CQ[proved in part (i)]

AQ = CP[proved in part (iv)]

Since opposite sides of quadrilateral APCQ are equal.

Hence APCQ is a parallelogram.

5. ABCD is a rhombus and P, Q, R, S are mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.

Ans. Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ  AC and PQ =  AC ……….(i)

In ADC, R is the mid-point of CD and S is the mid-point of AD.

SR  AC and SR =  AC……….(ii)

From eq. (i) and (ii),PQ  SR and PQ = SR

PQRS is a parallelogram.

Now ABCD is a rhombus. [Given]

AB = BC

AB =  BCPB = BQ

1 = 2[Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

AP = CQ[P and Q are the mid-points of AB and BC and AB = BC]

Similarly AS = CR and PS = QR[Opposite sides of a parallelogram]

APS CQR[By SSS congreuancy]

3 = 4[By C.P.C.T.]

Now we have1 + SPQ + 3 =

And2 + PQR + 4 = [Linear pairs]

1 + SPQ + 3 = 2 + PQR + 4

Since1 = 2 and 3 = 4[Proved above]

SPQ = PQR……….(iii)

Now PQRS is a parallelogram [Proved above]

SPQ + PQR = ……….(iv)[Interior angles]

Using eq. (iii) and (iv),

SPQ + SPQ = 2SPQ =

SPQ =

Hence PQRS is a rectangle.

6. ABCD is a rectangle 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 rhombus.

Ans. Given: A rectangle ABCD in which P, Q, R and S are the mid-points of the sides 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 sides AB, BC respectively.

PQ  AC and PQ =  AC……….(i)

In ADC, R and S are the mid-points of sides CD, AD respectively.

SR  AC and SR =  AC……….(ii)

From eq. (i) and (ii), PQ  SR and PQ = SR……….(iii)

PQRS is a parallelogram.

Now ABCD is a rectangle.[Given]

In triangles APS and BPQ,

AP = BP[P is the mid-point of AB]

PAS = PBQ[Each  ]

And AS = BQ[From eq. (iv)]

APS BPQ[By SAS congruency]

PS = PQ[By C.P.C.T.]………(v)

From eq. (iii) and (v), we get that PQRS is a parallelogram.

PS = PQ

Hence, PQRS is a rhombus.

7. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (See figure). Show that the line segments AF and EC trisect the diagonal BD.

Ans. Since E and F are the mid-points of AB and CD respectively.

AE =  AB and CF =  CD……….(i)

But ABCD is a parallelogram.

AB = CD and AB  DC

AB =  CD and AB  DC

AE = FC and AE  FC[From eq. (i)]

AECF is a parallelogram.

FA  CEFP CQ[FP is a part of FA and CQ is a part of CE] ……...(ii)

Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.

In DCQ, F is the mid-point of CD and FP CQ

P is the mid-point of DQ.

DP = PQ……….(iii)

Similarly, In ABP, E is the mid-point of AB and EQ AP

Q is the mid-point of BP.

BQ = PQ……….(iv)

From eq. (iii) and (iv),

DP = PQ = BQ………(v)

Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ

BQ =  BD……….(vi)

From eq. (v) and (vi),

DP = PQ = BQ =  BD

Points P and Q trisects BD.

So AF and CE trisects BD.

8.  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.

Ans. (i) In ABC, M is the mid-point of AB[Given]

MD  BC

AD = DC[Converse of mid-point theorem]

Thus D is the mid-point of AC.

(ii)  BC (given) consider AC as a transversal.

1 = C[Corresponding angles]

1 = [C = ]

Thus MD  AC.

(iii)  In AMD and CMD,

1 = 2 = [proved above]

MD = MD[common]

AMD CMD[By SAS congruency]

AM = CM[By C.P.C.T.]……….(i)

Given that M is the mid-point of AB.

AM =  AB……….(ii)

From eq. (i) and (ii),

CM = AM =  AB

9. In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that

Ans. Given ABCD is a parallelogram is and bisectors of intersect each other at P.

To prove

Proof:

But ABCD is a parallelogram and ADBC

Hence Proved

10. In figure diagonal AC of parallelogram ABCD bisects show that

(i) if bisects

ABCD is a rhombus

Ans.(i) ABDC and AC is transversal

(Alternate angles)

And (Alternate angles)

But,

(ii)

AC=AC [common]

[given]

[proved]

[By CPCT]

11. In figure ABCD is a parallelogram. AX and CY bisects angles A and C. prove that AYCX is a parallelogram.

Ans. Given in a parallelogram AX and CY bisects  respectively and we have to show that AYCX in a parallelogram.

…(i) [opposite angles of parallelogram]

[Given] …(ii)

And [give] …..(iii)

But

By (2) and (3), we get

Also, [opposite sides of parallelogram] ….(v)

From (i), (iv) and (v), we get

But, AB =CD [opposite sides of parallelogram]

AB-BY=CD-DX

Or

Ay=CX

But

is a parallelogram

12. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Ans. Given ABC in which E and F are mid points of side AB and AC respectively.

To prove: EF||BC

Construction: Produce EF to D such that EF = FD. Join CD

Proof:

AF=FC
[vertically opposite angles]
EF=FD [By construction]

And
AE= BE[is the mid-point]
And

13. Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles.

Ans. Given ABCD is a quadrilateral diagonals AC and BD bisect each other at O at right angles

To Prove: ABCD is a rhombus

Proof: diagonals AC and BD bisect each other at O

And

Now In And

Given

[Common]

And (Given)

(SAS)

(C.P.C.T.)

Similarly, BC=CD, CD=DA and DA=AB,

Hence, ABCD is a rhombus.

14. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides.

Ans. Given a trapezium ABCD in which and M,N are the mid Points of the diagonals AC and BD.

We need to prove that

Join CN and let it meet AB at E

Now in  and

[Alternate angles]

[Alternate angles]

And [given]

[ASA]

[By C.P.C.T]

Now in and are the mid points of the sides AC and CE respectively.

Or

Also

15. In fig is a right angle in is the mid-point of intersects BC at E. show that

(i) E is the mid-point of BC

(ii) DEBC

Ans. Proof:and D is mid points of AC

In  and

CE=BE

DE= DE

And

16. ABC is a triangle and through vertices A, B and C lines are drawn parallel to BC, AC and AB respectively intersecting at D, E and F. prove that perimeter of is double the perimeter of .

Ans. Is a parallelogram

Is a parallelogram

Or

Similarly, ED = 2AB and FD = 2AC

Perimeter of

Perimeter of

= 2AB+2BC+2AC

= 2[AB+BC+AC]

= 2 Perimeter of

Hence Proved.

17. In fig ABCD is a quadrilateral P, Q, R and S are the mid Points of the sides AB, BC, CD and DA, AC is diagonal. Show that

(i) SR||AC

(ii) PQ=SR

(iii) PQRS is a parallelogram

(iv) PR and SQ bisect each other

Ans. In ABC, P and Q are the mid-points of the sides AB and BC respectively

(i) PQ||AC and PQ=AC

(ii) Similarly SR||AC and SR=AC

PQ||SR and PQ=SR

(iii) Hence PQRS is a Parallelogram.

(iv) PR and SQ bisect each other.

18. In are respectively the mid-Points of sides AB,DC and CA. show that  is divided into four congruent triangles by Joining D,E,F.

Ans. D and E are mid-Points of sides AB and BC of ABC

DE||AC {A line segment joining the mid-Point of any two sides of a triangle parallel to third side}

Similarly, DF||BC and EF||AB

ADEF, BDEF and DFCE are all Parallelograms.

DE is diagonal of Parallelogram BDFE

Similarly, DAFFED

And EFCFED

So all triangles are congruent

19. ABCD is a Parallelogram is which P and Q are mid-points of opposite sides AB and CD. If AQ intersect DP at S BQ intersects CP at R, show that

(i) APCQ is a Parallelogram

(ii) DPBQ is a parallelogram

(iv) PSQR is a parallelogram

AP||QC [AB||CD]……..(i)

AP=AB, CQ=CD (Given)

Also AB= CD

So AP=QC……….(ii)

Therefore, APCQ is a parallelogram

[It any two sides of a quadrilateral equal and parallel then quad is a parallelogram]

(ii) Similarly, quadrilateral DPBQ is a Parallelogram because DQ||PB and DQ=PB

SP||QR [SP is a part of DP and QR is a Part of QB]

Similarly, SQ||PR

So. PSQR is also parallelogram.

20.are three parallel lines intersected by transversals P and q such that and cut off equal intercepts AB and BC on P In fig Show that cut off equal intercepts DE and EF on q also.

Ans. In fig are 3 parallel lines intersected by two transversal P and Q.

To Prove DE=EF

Proof: In

B is mid-point of AC

And BG||CF

G is mid-point of AF [By mid-point theorem]

Now In

G is mid-point of AF and GE || AD

E is mid-point of FD [By mid-point theorem]

DE=EF

Hence Proved.

21. ABCD is a parallelogram in which E is mid-point of AD. DF||EB meeting AB produced at F and BC at L prove that DF = 2DL

Ans. In

BE||DF (Given)

By converse of mid-point theorem B is mid-point of AF

ABCD is parallelogram

From (i) and (ii)

CD = BF

Consider and

DC = FB [Proved above]

[Alternate angles]

[Vertically opposite angles]

[ASA]

DL=LF

DF=2DL

22. PQRS is a rhombus if find

Ans. [opposite angles of a parallelogram are equal]

Let

In we have RS=RQ

[ opposite Sides of equal angles are equal]

In

[By angle sum property]

23. ABCD is a trapezium in which AB||CD and AD = BC show that

(i)

(ii)

(iii)

Ans. Produce AB and Draw a line Parallel to DA meeting at E

….(i) [Sum of interior angles on the some side of transversal is ]

In

BC=CE (given)

…..(2)  [in a equal side to opposite angles are equal]

……(3)

By (i) and (3)

(i)

(ii)

AB=AB [common]

[Proved above]

[By SAS]

24. Show that diagonals of a rhombus are perpendicular to each other.

Ans. Given: A rhombus ABCD whose diagonals AC and BD intersect at a Point O

To Prove:

Proof: clearly ABCD is a Parallelogram in which

AB=BC=CD=DA

We know that diagonals of a Parallelogram bisect each other

OA=OC and OB=OD

Now in BOC and DOC, we have

OB=OD

BC=DC

OC=OC

BOC DOC [By SSS]

[By C.P.C.T]

But

Similarly,

Hence diagonals of a rhombus bisect each other at

25. Prove that the diagonals of a rhombus bisect each other at right angles

Ans. We are given a rhombus ABCD whose diagonals AC and BD intersect each other at O.

We need to prove that OA=OC, OB=OD and

In and

AB=CD [Sides of rhombus]

[vertically opposite angles]

And  [Alternate angles]

AOBCOD [By ASA]

OA=OC

And OB=OD [By C.P.C.T]

Also in AOB and COB

OA=OC [Proved]

AB=CB [sides of rhombus]

And OB=OB [Common]

AOB COB [By SSS]

[By C.P.C.T]

But [linear pair]

26. In fig ABCD is a trapezium in which AB||DC and AD=BC. Show that

Ans. To show that

Draw CP||DA meeting AB at P

AP||DC and CP||DA

APCD is a parallelogram

Again in  CPB

[Angles opposite to equal sides]

But [By linear pair]

Also  [APCD is a parallelogram]

Or

= CB

27. In fig ABCD and ABEF are Parallelogram, prove that CDFE is also a parallelogram.

Ans. ABCD is a parallelogram

AB=DC also AB||DC…………..(i)

Also ABEF is a parallelogram

AB=FE and AB||FE……….(ii)

By (i) and (ii)

AB=DC=FE

AB=FE

And AB||DC||FE

AB||FE

CDEF is a parallelogram.

Hence Proved.

4 Marks Quetions

1. 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 both B as well as D.

Ans. ABCD is a rectangle. Therefore AB = DC ……….(i)

Also A = B = C = D =

1 = 2 and 3 = 4

[AC bisects A and C (given)]

AC = AC [Common]

From eq. (i) and (ii), AB = BC = CD = AD

Hence ABCD is a square.

AB = BA [Since ABCD is a square]

AD = DC [Since ABCD is a square]

BD = BD [Common]

ABD CBD [By SSS congruency]

ABD = CBD [By C.P.C.T.] ……….(iii)

And ADB = CDB [By C.P.C.T.] ……….(iv)

From eq. (iii) and (iv), it is clear that diagonal BD bisects both B and D.

2. An 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 figure). Show that:

(i) Quadrilateral ABED is a parallelogram.

(ii) Quadrilateral BEFC is a parallelogram.

(iv) Quadrilateral ACFD is a parallelogram.

(v) AC = DF

(vi) ABC DEF

Ans. (i) In ABC and DEF

AB = DE [Given]

And AB DE [Given]

ABED is a parallelogram.

(ii) In ABC and DEF

BC = EF [Given]

And BC EF [Given]

BEFC is a parallelogram.

(iii) As ABED is a parallelogram.

Also BEFC is a parallelogram.

CF BE and CF = BE ……….(ii)

From (i) and (ii), we get

ACFD is a parallelogram.

(v) As ACFD is a parallelogram.

AC = DF

(vi) InABC and DEF,

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved]

ABC DEF [By SSS congruency]

3. 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 figure). Show that F is the mid-point of BC.

Ans. Let diagonal BD intersect line EF at point P.

In DAB,

E is the mid-point of AD and EP AB [EF AB (given) P is the part of EF]

P is the mid-point of other side, BD of DAB.

[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]

Now in BCD,

P is the mid-point of BD and PF DC [EF AB (given) and AB DC (given)]

EF DC and PF is a part of EF.

F is the mid-point of other side, BC of BCD. [Converse of mid-point of theorem]

4. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

Ans. Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove: EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE.

Proof: In ABC, E and F are the mid-points of respective sides AB and BC.

EF AC and EF AC ……….(i)

G and H are the mid-points of respective sides CD and AD.

HG AC and HG AC ……….(ii)

From eq. (i) and (ii),

EF HG and EF = HG

EFGH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Ans. Let ABCD be a quadrilateral in which equal diagonals AC and BD bisect each other at right angle at point O.

We have AC = BD and OA = OC ……….(i)

And OB = OD ……….(ii)

Now OA + OC = OB + OD

OC + OC = OB + OB [Using (i) & (ii)]

2OC = 2OB

OC = OB ……….(iii)

From eq. (i), (ii) and (iii), we get, OA = OB = OC = OD ……….(iv)

Now in AOB and COD,

OA = OD [proved]

AOB = COD [vertically opposite angles]

OB = OC [proved]

AOBDOC [By SAS congruency]

AB = DC [By C.P.C.T.] ……….(v)

Similarly, BOC AOD [By SAS congruency]

BC = AD [By C.P.C.T.] ……….(vi)

From eq. (v) and (vi), it is concluded that ABCD is a parallelogram because opposite sides of a quadrilateral are equal.

AB = BA [Common]

AC = BD [Given]

ABC = BAD [By C.P.C.T.] ……….(vii)

But ABC + BAD = [ABCD is a parallelogram] ……….(viii)

AD BC and AB is a transversal.

ABC + ABC = [Using eq. (vii) and (viii)]

2ABC = ABC =

Opposite angles of a parallelogram are equal.

From eq. (x) and (xi), we get

Now in AOB and BOC,

OA = OC [Given]

AOB = BOC = [Given]

OB = OB [Common]

AOBCOB [By SAS congruency]

AB = BC ……….(xiii)

From eq. (v), (vi) and (xiii), we get,

AB = BC = CD = AD ……….(xiv)

Now, from eq. (xii) and (xiv), we have a quadrilateral whose equal diagonals bisect each other at right angle.

Also sides are equal make an angle of with each other.

ABCD is a square.

6. ABCD is a trapezium in which AB CD and AD = BC (See figure). Show that:

(i) A = B

(ii) C = D

(iv) Diagonal AC = Diagonal BD

Ans. Given: ABCD is a trapezium.

AB CD and AD = BC

To prove: (i) A = B

(ii) C = D

(iv) Diag. AC = Diag. BD

Construction: Draw CE AD and extend AB to intersect CE at E.

Proof: (i) As AECD is a parallelogram. [By construction]

BC = EC

3 = 4 [Angles opposite to equal sides are equal]

Now 1 + 4 = [Interior angles]

And 2 + 3 = [Linear pair]

1 + 4 = 2 + 3

1 = 2 [3 = 4 ]

A = B

(ii) 3 = C [Alternate interior angles]

And D = 4 [Opposite angles of a parallelogram]

But 3 = 4 [BCE is an isosceles triangle]

C = D

AB = AB [Common]

1 = 2 [Proved]

AC = BD [By C.P.C.T.]

7. Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square.

Ans. Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and

To prove: ABCD is a square.

Proof:

OA=OC

OB=OD [given]

And

[vertically opposite angles]

But these are alternate angles

ABCD is a parallelogram whose diagonals bisects each other at right angles

Again in

AB=BC [Sides of a rhombus]

And BD=CA [Given]

[By CPCT]

These are alternate angles of these same side of transversal

Hence ABCD is a square.

8. Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side.

Ans. Given: A in which D and E are mid-points of the side AB and AC respectively

To Prove:

Construction: Draw

Proof: In and

[Vertically opposite angles]

AE=CE [Given]

And [Alternate interior angles]

[By ASA]

DE=FE [By C.P.C.T]

But DA = DB

DB = FC

Now DBFC

DBCF is a parallelogram

DEBC

Also DE = EF = BC

9. 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 quadrilateral PQRS is a rectangle.

Ans. Join AC and BD which intersect at O let BD intersect RS at E and AC intersect RQ at F

IN ABD P and S are mid-points of sides AB and AD.

PS||BD and PS=

Similarly, RQ||DB and RQ=BD

RS||BD||RQ and PS =

PS=RQ and PS||RQ

PQRS is a parallelogram

Now RF||EO and RE||FO

OFRE is also a parallelogram.

Again, we know that diagonals of a rhombus bisect each other at right angles.

[opposite angles of a parallelogram]

Each angle of the parallelogram PQRS is

Hence PQRS is a rectangle.

10. In the given Fig ABCD is a parallelogram E is mid-point of AB and CE bisects Prove that:

(ii) DE bisects

(iii)

Ans. ABCD is a parallelogram

And EC cuts them

[Alternate interior angle]

[ Alternate interior angles]

(ii) DE bisects

(iii) Now

But, the sum of all the angles of the triangle is

11. 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 mid-point of AC
(ii) MDAC
(iii) CM = MA = AB

Ans. Given ABC is a right angle at C

(i) M is mid-point of AB

And MD||BC

D is mid-Point of AC [a line through midpoint of one side of a parallel to another side bisect the third side.

(ii). MD||BC

[Corresponding angles]