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Question 2

How does a trapezium differ from a parallelogram?

Solution:
A trapezium has only one pair of opposite sides parallel but a parallelogram has two pairs of opposite sides equal and parallel.

Question 3

ABCD is a parallelogram. What special name will you give it if the following additional 
        facts are known?
        (i) AB = AD.

        (ii) ∠ DAB = 900.

        (iii) AB = AD and ∠ DAB = 900.


Solution:
(i) AB = AD

  

ABCD is a rhombus.

AB || CD, AB = CD and AD || BC, AD = BC since ABCD is a parallelogram.

By the given data AB = AD.

∴ ABCD is a rhombus since the pairs of opposite sides are parallel and all sides are equal.

(ii) ∠ DAB = 90°   

ABCD is a rectangle. The pairs of opposite sides are parallel and ∠ DAB = 90° , then all the other angles are equal to 90° .

(iii) AB = AD and ∠ DAB = 900.

    


ABCD is a square since ∠ DAB = 90° and the pairs of opposite sides are parallel and a pair of adjacent sides are equal.

Question 4

ABCD is a trapezium in which AB || DC. If ∠ A = ∠ B = 400, what are the measures of the other two angles?


Solution:

Given:

ABCD is a trapezium in which AB || DC. ∠ A = ∠ B = 400.

To find: The measures of ∠ C and ∠ D.

Proof:

AB || DC. ∠ A = ∠ B = 400 …………. (i)

∠ A + ∠ D = 1800 (Sum of interior angles on the same side of the transversal is 180°)

⇒ 40° + ∠ D = 180° [From (i)]

⇒ ∠ D = 180° - 40°

∴ ∠ D = 140°

∠ B + ∠ C = 1800

⇒ 40° + ∠ C = 180° [From (i)]

⇒ ∠ C = 180° - 40°

∴ ∠ C = 140°

The measures of the other two angles are 140° each.

Question 5

The angles P,Q,R and S of a quadrilateral PQRS are in the ratio of 1 : 3 : 7 : 9.


        (i) Find the measure of each angle.

        (ii) Is PQRS a trapezium? Why?

        (iii) Is PQRS a parallelogram? Why?


Solution:

(i) ∠ P + ∠ Q + ∠ R + ∠ S = 360° [Sum of the angles of a quadrilateral is 360° ]

The angles are in the ratio 1 : 3 : 7 : 9.⇒ ∠ P = 360 × = 18°

∠ Q = 360 × = 54°

∠ R = 360 × = 126°

∠ S = 360 × = 162°

∴ ∠ P = 18° , ∠ Q = 54° , ∠ R = 126° and ∠ S = 162° .

(ii) Yes, PQ || SR.

∠ P + ∠ S = 18° + 162° = 180°

∠ Q + ∠ R = 54° + 126° = 180°

∴ PQ || SR.

∴ PQRS is a trapezium.

(iii) No, PS is not parallel to QR.

∠ P + ∠ R = 18° + 126° = 144°

∠ Q + ∠ S = 54° + 162° = 216°

⇒ PS is not parallel to QR.

∴ PQRS is not a parallelogram.

Question 6

For each of the following statements, state whether the statement is true(T) or false(F).


        (i) Every rectangle is a parallelogram.

        (ii) Every square is a rectangle.

        (iii) Every parallelogram is a rhombus.

        (iv) Every square is a rhombus.

        (v) Every rectangle is a square.

        (vi) Every parallelogram is a rectangle.

        (vii) Every square is a parallelogram.

        (viii) Every rhombus is a parallelogram.

        (ix) Every rhombus is a square.

        (x) Every parallelogram is a square.

        (xi) Every parallelogram is a trapezium.

        (xii) Every trapezium is a parallelogram.

        (xiii) Every square is a trapezium.

        (xiv) Every trapezium is a square.


Solution:

(i) True

(ii) True

(iii) False, In a parallelogram only pairs of opposite sides are parallel and equal but in a rhombus pairs of opposite sides are parallel and all the sides are equal.

(iv) True

(v) False, since in a rectangle the pairs of opposite sides are parallel and equal, but in a square all the sides are equal.

(vi) False, since in a parallelogram only pairs of opposite sides are parallel, and equal but in a rectangle the pairs of opposite sides are parallel and measure of each angle should be 90° .

(vii) True

(viii) True

(ix) False, since pairs of opposite sides are parallel and all sides are equal in a rhombus. But in a square all the angles are 90° .

(x) False, since in a parallelogram only pairs of opposite sides are parallel and equal, all the sides are not equal. But in a square, all the sides are equal and parallel and the angles are all 90° each.

(xi) True

(xii) False, since in a trapezium only one pair of opposite sides are parallel and but in a parallelogram both pairs of opposite sides are equal and parallel.

(xiii) True

(xiv) False, since in a trapezium only one pair of opposite sides are parallell but in a square both pairs of opposite sides are parallel.

Question 7

Lengths of two adjacent sides of a parallelogram are 4cm and 3cm. Find its perimeter.


Solution:


Let ABCD be the parallelogram.

Side AB = 4 cm and side BC = 3 cm.

The opposite sides of a parallelogram are equal.

⇒ Side CD = 4 cm and side DA = 3 cm.

⇒ The perimeter = (4 + 3 + 4 + 3) cm = 14 cm.

∴ The perimeter = 14 cm.

Question 8

One angle of a parallelogram is of measure 700. Find the measures of the remaining angles of the parallelogram.


Solution:


Let PQRS be the parallelogram.

Let ∠ P = 70° …………..(1)

∠ P = ∠ R [Opposite angles of a parallelogram are equal.]

∴ ∠ R = 70° .

∠ P + ∠ Q = 180° [Interior angles on the same side of the transversal.]

⇒ 70° + ∠ Q = 180° [From(1)]

⇒ ∠ Q = 180° - 70°

∴ ∠ Q = 110° ……………………(2)

∠ Q = ∠ S [Opposite angles of a parallelogram are equal.]

∴ ∠ S = 110° [From (2)]

∴ ∠ Q = 110° , ∠ R = 70° and ∠ S = 110°

Question 9

Two adjacent angles of a parallelogram are equal. Find the measure of each angle of the parallelogram.


Solution:
Let the parallelogram be ABCD.

Let the equal angles be ∠ A and ∠ B.

⇒ ∠ A = ∠ B………………………(1)

∠ A + ∠ B = 180° [Sum of the adjacent angles of a parallelogram is 180° .]

But ∠ A = ∠ B [From (1)]

⇒ 2 ∠ A = = 90° .

Similarly, ∠ B = 90° .

∠ C = 90° [Opposite angle to ∠ A]

∠ D = 90° [Opposite angle to ∠ B]

∴ The measure of each angle of the parallelogram is 90° .

Hence the parallelogram is a rectangle.

Question 10

The ratio of two sides of a parallelogram is 3 : 5 and its perimeter is 48 cm. Find the sides of the parallelogram.

          [Hint: Let two sides be 3x and 5x]


Solution:


Let the parallelogram be PQRS.

The sides are in the ratio 3 : 5.

Let PQ = 5x and QR = 3x.

The perimeter = 48 cm

⇒ 2 (5x + 3x) = 48 cm

⇒ 5x + 3x = 24 cm

8x = 24 cm

∴ x = = 3 cm.

∴ PQ = 5 × 3 = 15 cm

QR = 3 × 3 = 9 cm.

RS = PQ [Opposite sides of a parallelogram are equal]

∴ RS = 15 cm.

QR = SP [Opposite sides of a parallelogram are equal]

∴ SP = 9 cm.

Therefore the sides of the parallelogram are 15m, 9cm, 15cm and 9cm.


Question 11

Two adjacent angles of a parallelogram are in the ratio of 2 : 3. Find all the angles of the parallelogram.


Solution:

Let the parallelogram be PQRS.

Two adjacent angles of a parallelogram are in the ratio of 2 : 3.

Let ∠ P be equal to 2x and ∠ Q be equal to 3x.

⇒ 2x + 3x = 180°(Sum of the adjacent angles of the parallelogram is 180°)

⇒ 5x = 180°

⇒ x = = 36°

∴ ∠ P = 2x = 2 × 36 = 72°

∴ ∠ P = 72° …………………………………..(1)

∴ ∠ Q = 3x = 3 × 36 = 108°

∴ ∠ Q = 108° ………………………………..(2)

∠ P and ∠ R are equal. [Opposite angles of a parallelogram are equal.]

∴ ∠ R = 72° .

∠ Q and ∠ S are equal. [Opposite angles of a parallelogram are equal.]

∴ ∠ S = 108° [From (2)]

∴The angles of the parallelogram are 72°
, 108°, 72° and 108°.

Question 12

The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the lengths of all the sides of the parallelogram.


Solution:


Let ABCD be a parallelogram.

Let the side BC be x cm.

⇒ The side AB = (x + 25) cm

The perimeter of the parallelogram = 150 cm

⇒ 2 [x + (x + 25)] = 150

⇒ [x + (x + 25)] = 75

⇒ 2x + 25 = 75

⇒ 2x = 75 – 25

⇒2x=50

x == 25

∴ x = 25cm

∴ BC = 25 cm……………………………(1)

∴ AB = (x + 25) cm = 25 + 25 = 50 cm.

∴ AB = 50 cm……………………………(2)

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

⇒ CD = 50 cm [From (2)]

BC = DA [Opposite sides of a parallelogram are equal]

⇒ DA = 25 cm [From (1)]

∴ AB = 50cm, BC = 25 cm, CD = 50 cm and DA = 25 cm.

Question 13

PR is a diagonal of the parallelogram PQRS(Figure).


           (i) Is PS = RQ? Why?

           (ii) Is SR = PQ? Why?

           (iii) Is PR = RP? Why?

           (iv) Is Δ PSR ≅ Δ RQP? Why?

               


Solution:
(i) Yes, PS = RQ, since PS and RQ are opposite sides of the parallelogram PQRS they are equal.

(ii) Yes, SR = PQ, since SR and PQ are opposite sides of the parallelogram PQRS they are equal.

(iii) Yes, PR = RP since both are the same line segment.

(iv) Yes, Δ PSR ≅ Δ RQP.

PQ = SR [Opposite sides of the parallelogram PQRS are equal.]

QR = SP [Opposite sides of the parallelogram PQRS are equal.]

PR is common.

∴ Δ PSR ≅ Δ RQP [By SSS congruence]

Question 14

The point of intersection of the diagonals of a quadrilateral divides one diagonal in the ratio 2 : 3 Can it be a parallelogram? Why?


Solution:

No, ABCD is not a parallelogram.

Let ABCD be the quadrilateral. AC and BD be the diagonals intersecting at O.

= [Given]

If ABCD is a parallelogram, the diagonals will bisects each other.

=

But it is given that = .

∴ ABCD is not a parallelogram.

Question 15

Diagonals of a parallelogram ABCD intersect at the point O(Figure). XY is a line segment passing through O such that X lies on AD and Y lies on BC. Give reasons for each of the following statements.


          (i) OB = OD.

          (ii) ∠ OBY = ∠ ODX

          (iii) ∠ BOY = ∠ DOX

          (iv) Δ BOY ≅ Δ DOX

          Now state whether or not XY is bisected at O.

           


Solution:
(i) Given: ABCD is a parallelogram. Diagonals AC and BD intersect at O. XY is a line segment passing through O such that X lies on AD and Y lies on BC.

To Prove: OB = OD

Proof: ABCD is a parallelogram, and the diagonals intersect at O.

The diagonals of a parallelogram bisects each other.

∴ OB = OD.

(ii) AB || CD and BD is the transversal.

The alternate angles formed are equal.

∠ OBY = ∠ ODX [Alternate angles]

(iii) BD and XY intersect at O.

The vertically opposite angles formed are equal.

∴ ∠ BOY = ∠ DOX [Vertically opposite angles]

(iv) In Δ BOY and Δ DOX,

OB = OD [From (i)]

∠ OBY = ∠ ODX [From (ii)]

∠ BOY = ∠ DOX [From (iii)]

∴ Δ BOY ≅ Δ DOX.(ASA)

Yes, XY is bisected at O.

Question 16

In figure, ABCD is a parallelogram, CE bisects ∠ C and AF bisects ∠ A. Give reasons for each of the following statements.


 
        

          (vi) ∠ CEB = ∠ FAB.

          (vii) CE || FA.

          (viii) AE|| FC.

          (ix) AECF is a parallelogram.


Solution:
(i) ABCD is a parallelogram.BAD = ∠ BCD [Opposite angles of the parallelogram ABCD are equal]

(ii) ∠ DAF = ∠ FAB  [AF bisects ∠ BAD.]

     ∠ FAB  = ∠ BAD…………………..(1)

(iii) ∠ DCE = ∠ BCE [CE bisects ∠ BCD.]

∠ DCE = ∠ BCD…………………..(2)

(iv) ABCD is a parallelogram.

⇒ ∠ A =∠ C [Opposite angles of a parallelogram are equal.]

∠ A = ∠ C……………………….(3)

⇒ ∠ FAB = ∠ DCE

(v) AB || DC [Opposite sides of a parallelogram are parallel.]

And CE is the transversal passing through AB and DC.

∴ ∠ DCE = ∠ CEB [Alternate angles]

(vi) ∠ FAB = ∠ BCE ……………[From (iv)]

∠ DCE = ∠ CEB…………….[From (v)]

∴ ∠ CEB = ∠ FAB.[From (iv) and (v)]

(vii) ∠ CEB = ∠ FAB [ from (vi)]

.1.e., The corresponding angles formed by lines CE and FA out by transversal AB are equal.

∴ CE || FA.

(viii) AB || CD [Opposite sides of a parallelogram are parallel]

AE lies on AB and CF lies on CD.

∴ AE || FC.

(ix) AE ||FC [From (viii)]

CE || FA [From (vii)]

Two pairs of opposite sides are parallel.

Lines AF and EC are between parallel sides AB and CD so they are of equal length.
i.e., AF = EC
AE and FC are lines joining two parallel sides of same length and hence they are also equal.
i.e.,AE = FC
∴ AECF is a parallelogram.


Question 18

Which of the following statements are true for a rectangle?


          (i) It has two pairs of opposite sides of equal length.

          (ii) It has all its sides of equal length.

          (iii) Its diagonals are equal.

          (iv) Its diagonals bisect each other.

          (v) Its diagonals are perpendicular to each other.

          (vi) Its diagonals are equal and perpendicular to each other.

          (vii) Its diagonals are perpendicular and bisect each other.

          (viii) Its diagonals are equal and bisect each other.

          (ix) Its diagonals are equal, perpendicular and bisect each other.

          (x) All of its angles are equal


Solution:

(i) True

(ii) False, since length and breadth of a rectangle need not be equal always.

(iii) True

(iv) True

(v) False, since diagonals need not be perpendicular to each other always.

(vi) False, since diagonals need not be perpendicular to each other always.

(vii) False, since diagonals need not be perpendicular to each other always.

(viii) True

(ix) False, since diagonals need not be perpendicular to each other always.

(x) True

Question 20

The diagonals of a parallelogram are not perpendicular to each other. Is it a rhombus? Why?


Solution:
No, it is not a rhombus, because the diagonals of a rhombus are perpendicular to each other.


Question 21

AC is the diagonal of rectangle ABCD.

(i) Is BC = DA ? Why?

(ii) Is AB = CD? Why?

(iii) Is ∠ B = ∠ D? Why?

(iv) Is Δ ABC ≅ Δ CDA? By which congruence condition?


Solution:

(i) Yes, since BC and DA are the opposite sides of the rectangle.

(ii) Yes, since AB and CD are the opposite sides of the rectangle.

(iii) Yes, ∠ B = ∠ D, since each is equal to 90° .

(iv) Yes, Δ ABC ≅ Δ CDA.

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

BC = AD [Opposite sides of a rectangle are equal]

∠ B =∠ D = 90° [Each angles of the rectangle is 90° ]

∴ Δ ABC ≅ Δ CDA [By SAS congruence property]

Question 22

ABCD is a rhombus and its diagonals intersect each other at O(figure).

(i) Is OB = OD? Why?

(ii)Is BC = DC? Why?

(iii)Is Δ BOC ≅ Δ DOC? By which congruence condition?

(iv)Is ∠ BCO = ∠ DCO? Why?

(v)Is Δ BAO ≅ Δ DAO? By which congruence condition?

(vi)Is ∠ BAO = ∠ DAO? Why?

(vii)Does diagonal AC of the rhombus bisect ∠ A and ∠ C? Why?

         


Solution:
(i) Yes, OB = OD, since the diagonals bisect each other.

(ii) Yes, BC = DC, since BC and DC are the sides of the rhombus.

(iii) Yes, Δ BOC ≅ Δ DOC.

In Δ BOC and Δ DOC,

BC = DC [Sides of the rhombus are of equal length]

OB = OD [Diagonals bisect each other]

OC is common.

∴ Δ BOC ≅ Δ DOC [by SSS congruence property]

(iv) Yes, ∠ BCO = ∠ DCO [Corresponding parts of congruent triangles BOC and DOC are congruent]

(v) Yes, Δ BAO ≅ Δ DAO.

In Δ BAO and Δ DAO,

AB = AD [Sides of the rhombus are equal]

OB = OD [Diagonals bisect each other]

 OA is common

∴ Δ BAO ≅ Δ DAO [by SSS congruence property]

(vi) Yes, ∠ BAO = ∠ DAO [Corresponding parts of congruent triangles BAO and DAO are congruent]

(vii) ∠ BCO = ∠ DCO [From (iv)]

∠ BAO = ∠ DAO [From (vi)]

But ∠ A =∠ C [Opposite angles of a rhombus]

∠ A = ∠ C

Yes, the diagonal AC of the rhombus bisects ∠ A and ∠ C.

Question 23

Diagonal AC of a rhombus ABCD is equal to one of its sides BC (figure). Find all the angles of the rhombus.


Solution:

Given: ABCD is a rhombus and AC = BC.

To find: The angles of the rhombus.

Proof: ABCD is a rhombus.

AC = BC [Given]…………………………(1)

But BC = AB [Sides of a rhombus are equal.]………………..(2)

From (1) and (2),

AC =BC = AB

⇒ Δ ABC is an equilateral triangle.

⇒ ∠ ABC = 60° ,

∠ BCA = 60° ………………….(3)

∠ CAB = 60° …………………..(4)

Now, from Δ ADC, AD = DC [Sides of the rhombus are equal.]

AD = BC [The sides of the rhombus are equal.]

But from (1), BC = AC

∴ AD = AC

∴ AD=DC = AC

∴ Δ ADC is an equilateral triangle.

∠ CAD = 60° ………………………..(5)

∠ ADC = 60°

∠ DCA = 60° …………………………(6)

From (3) and (6), ∠ BCA + ∠ DCA = 60 + 60° = 120°

∴ ∠ C = 120°

From (4) and (5), ∠ CAB + ∠ CAD = 60° + 60° = 120°

∴ ∠ A = 120° .

∴ The four angles of the rhombus ABCD are 120° , 60° , 120° and 60° .

Question 24

A quadrilateral shaped window frame has one diagonal longer than the other. Is the window frame forms of the shape of a rectangle? Why?


Solution:
No, the window frame is not in the shape of a rectangle, since the diagonals of a rectangle are equal.


Question 25

In figure, ABCD is a rectangle. BM and DN are perpendicular to AC from B and D respectively.

         (i) Is AB = CD? Why?

         (ii) Is ∠ BMA = ∠ DNC? Why?

         (iii) Is ∠ BAM = ∠ DCN? Why?

         (iv) Is Δ BMA ≅ Δ DNC? By which congruence condition?

         (v) Is BM = DN? Why?

            


Solution:
(i) Yes, AB = CD, since AB and CD are the opposite sides of the rectangle ABCD.

(ii) Yes, ∠ BMA = ∠ DNC. Each angle is equal to 90° since BM and DN are ⊥ to AC from B and D respectively.

(iii) Yes, ∠ BAM = ∠ DCN are alternate angles formed by the parallel lines AB and CD intersected by the diagonal AC.

(iv) Yes, Δ BMA ≅ Δ DNC by ASA congruence condition.

∠ BMA = ∠ DNC, (from (ii))

AB = CD (from (i)) and

∠ BAM = ∠ DCN.(from (iii))

(v) Yes, BM = DN, BM and DN are the corresponding parts of congruent triangles BMA and DNC.

Question 26

The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? If your answer is No, draw a figure to justify your answer.


Solution:
No, a quadrilateral whose diagonals are perpendicular to each other need not be always a rhombus. It can also be square.



Question 27

Diagonals of a rhombus are equal. Is this rhombus also a square?

Solution:
Yes, a rhombus whose diagonals are equal is also a square.

Question 28

Diagonals of a quadrilateral are equal. Is such a quadrilateral always a rectangle? If your answer is No, draw a figure to justify your answer.

Solution:
No, a quadrilateral whose diagonals are equal need not be always a rectangle, it can also be a square.



Question 29

The diagonals of a quadrilateral are of lengths 10 cm and 24 cm. If the diagonals bisect each other at right angles, find the length of each side of the quadrilateral. What special name can you give to this quadrilateral?

Solution:


Given: AC and BD are the diagonals of the quadrilateral ABCD. AC = 10cm and BD = 24 cm. AC and BD bisects each other at right angles.

To find: The length of each side of the quadrilateral.

Proof:

The lengths of the diagonals of the quadrilateral ABCD is 10cm and 24 cm. AC and BD bisect each other at right angles at the point O.

BD = 24

⇒ OB = 12cm

AC = 10cm

⇒ OA = 5 cm

∠ AOB = 90°

In the rt. Δ AOB,

AB2 = OB2 + OA2

       = 122 + 52

∴ AB = = = = 13

Similarly, from Δ s COB, COD and AOD, the lengths of BC, CD and DA is equal to 13 cm each.

∴ AB = BC = CD = DA = 13 cm.

∴ This quadrilateral ABCD is a rhombus.

Question 30

The length of each diagonal of a quadrilateral is 12 cm. The diagonals also bisect each other at right angles. What special name can you give to this quadrilateral?

 


Solution:



Given: Let the quadrilateral be PQRS. The diagonals PR and QS are of length 12 cm each bisect each other at right angles at O.

To find: The length of the sides of the quadrilateral and thus find the name of the quadrilateral.

Proof:

PR = QS = 12 cm and O is the mid point of PR and QS.

⇒ OP = OQ = 6 cm

From the rt. Δ POQ,

PQ2 = OP2 + OQ2

= 62 + 62

= 36 + 36

= 72

⇒ PQ = == 6

Similarly QR = RS = SP = 6

∴ The figure PQRS is a square.





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