RS Agarwal Solution | Class 9th | Chapter-10 |  Quadrilaterals | Edugrown

Exercise MCQ

Question 1

Three angles of quadrilateral are 80^°, 95° and 112°. Its fourth angle is

(a) 78°

(b) 73°

(c) 85°

(d) 100°Solution 1

Question 2

The angles of a quadrilateral are in the ratio 3:4:5:6. The smallest of these angles is

(a) 45°

(b) 60°

(c) 36°

(d) 48°Solution 2

Question 3

In the given figure, ABCD is a parallelogram in which ∠BAD = 75° and ∠CBD = 60°. Then, ∠BDC =?

(a) 60°

(b) 75°

(c) 45°

(d) 50°

Solution 3

Question 4

ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?

(a) 40° 

(b) 25° 

(c) 65°  

(d) 130° Solution 4

Correct option: (a)

ABCD is a rhombus.

⇒ AD ∥ BC and AC is the transversal.

⇒ ∠DAC = ∠ACB  (alternate angles)

⇒ ∠DAC = 50° 

In ΔAOD, by angle sum property,

∠AOD + ∠DAO + ∠ADO = 180° 

⇒ 90° + ∠50° + ∠ADO = 180° 

⇒ ∠ADO = 40° 

⇒ ∠ADB = 40° Question 5

In which of the following figures are the diagonals equal?

1. Parallelogram
2. Rhombus
3. Trapezium
4. Rectangle

Solution 5

Question 6

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a

a. trapezium

b. parallelogram

c. rectangle

d. rhombusSolution 6

Question 7

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is

1. 10 cm
2. 12 cm
3. 9cm
4. 8cm

Solution 7

Question 8

The length of each side of a rhombus is 10 cm and one of its diagonals is of length 16 cm. The length of the other diagonal is

Solution 8

Question 9

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is

(a) 55° 

(b) 70° 

(c) 45° 

(d) 50° Solution 9

Correct option: (b)

∠DAO + ∠OAB = ∠DAB

⇒ ∠DAO + 35° = 90° 

⇒ ∠DAO = 55° 

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

⇒ ∠ODA = ∠DAO (angles opposte to equal sides are equal)

⇒ ∠ODA = 55° 

In DODA, by angle sum property,

∠ODA + ∠DAO + ∠AOD = 180° 

⇒ 55° + ∠55° + ∠AOD = 180° 

⇒ ∠AOD = 70° Question 10

If ABCD is a parallelogram with two adjacent angles ∠A = ∠B, then the parallelogram is a

1. rhombus
2. trapezium
3. rectangle
4. none of these

Solution 10

Question 11

In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A and ∠B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB =?

1. 40°
2. 50°
3. 80°
4. 100°

Solution 11

Question 12

The bisectors of any adjacent angles of a parallelogram intersect at

1. 30°
2. 45°
3. 60°
4. 90°

Solution 12

Question 13

The bisectors of the angles of a parallelogram enclose a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 13

Correct option: (c)

The bisectors of the angles of a parallelogram enclose a rectangle.Question 14

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S then PQRS is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) quadrilateral whose opposite angles are supplementarySolution 14

Correct option: (d)

In ΔAPB, by angle sum property,

∠APB + ∠PAB + ∠PBA = 180° 

In ΔCRD, by angle sum property,

∠CRD + ∠RDC + ∠RCD = 180° 

Now, ∠SPQ + ∠SRQ = ∠APB + ∠CRD

= 360° – 180° 

= 180° 

Now, ∠PSR + ∠PQR = 360° – (∠SPQ + ∠SRQ)

= 360° – 180° 

= 180° 

Hence, PQRS is a quadrilateral whose opposite angles are supplementary. Question 15

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 15

Question 16

The figure formed by joining the mid-points of the adjacent sides of a square is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 16

Question 17

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 17

Question 18

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 18

Question 19

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 19

Question 20

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if

(a) ABCD is a parallelogram

(b) ABCD is a rectangle

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 20

Correct option: (d)

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

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

From (i) and (ii),

PQ ∥ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM ∥ NO and PN ∥ MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle if AC ⊥ BD. Question 21

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if

(a) ABCD is a parallelogram

(b) ABCD is a rhombus

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 21

Correct option: (c)

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

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

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

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ ∥ RS and QR ∥ SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal. Question 22

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if

(a) ABCD is a rhombus

(b) diagonals of ABCD are equal

(c) diagonals of ABCD are perpendicular

(d) diagonals of ABCD are equal and perpendicularSolution 22

Correct option: (d)

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

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

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

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular. Question 23

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

1. 108°
2. 54°
3. 72°
4. 81°

Solution 23

Question 24

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angles of the parallelogram is

1. 68°
2. 102°
3. 112°
4. 136°

Solution 24

Question 25

If ∠A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3:7:6:4, then ABCD is a

1. rhombus
2. kite
3. trapezium
4. parallelogram

Solution 25

Question 26

Which of the following is not true for a parallelogram?

1. Opposite sides are equal.
2. Opposite angles are equal.
3. Opposite angles are bisected by the diagonals.
4. Diagonals bisect each other.

Solution 26

Question 27

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a

1. square
2. rhombus
3. rectangle
4. kite

Solution 27

Question 28

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?

1. 45°
2. 55°
3. 60°
4. 75°

Solution 28

Question 29

If area of a ‖gm with side ɑ and b is A and that of a rectangle with side ɑ and b is B, then

(a) A > B

(b) A = B

(c) A < B

(d) A ≥ BSolution 29

Question 30

In the given figure, ABCD is a ‖gm and E is the mid-point at BC, Also, DE and AB when produced meet at F. Then,

Solution 30

Question 31

P is any point on the side BC of a ΔABC. P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is

(a) a trapezium

(b) a parallelogram

(c) a rectangle

(d) a rhombusSolution 31

Correct option: (b)

In ΔABC, D and E are the mid-points of sides AB and AC respectively.

Hence, DENM is a parallelogram.Question 32

The parallel sides of a trapezium are ɑ and b respectively. The line joining the mid-points of its non-parallel sides will be

Solution 32

Question 33

In a trapezium ABCD, if E and F be the mid-points of the diagonals AC and BD respectively. Then, EF =?

Solution 33

Question 34

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB=?

1. 45°
2. 60°
3. 90°
4. 30°

Solution 34

Question 35

In the given figures, ABCD is a rhombus. Then

(a) AC² + BD² = AB²

(b) AC² + BD² = 2AB²

(c) AC² + BD² = 4AB²

(d) 2(AC² + BD²)=3AB²

Solution 35

Question 36

In a trapezium ABCD, if AB ‖ CD, then (AC² + BD²) =?

(a) BC² + AD² + 2BC. AD

(b) AB² +CD² + 2AB.CD

(c) AB² + CD² + 2AD. BC

(d) BC² + AD² + 2AB.CD

Solution 36

Question 37

Two parallelogram stand on equal bases and between the same parallels. The ratio of their area is

1. 1:2
2. 2:1
3. 1:3
4. 1:1

Solution 37

Question 38

In the given figure, AD is a median of ΔABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF =?

Solution 38

Question 39

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30°and ∠AOB = 70°. Then, ∠DBC =?

1. 40°
2. 35°
3. 45°
4. 50°

Solution 39

Question 40

Three statement are given below:

1. In a ‖gm, the angle bisectors of two adjacent angles enclose a right angle.
2. The angle bisectors of a ‖gm form a rectangle.
3. The triangle formed by joining the mid-point of the sides of an isosceles triangle is not necessarily an isosceles.

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 40

Question 41

Three statements are given below:

 I. In a rectangle ABCD, the diagonal AC bisects ∠A as well as ∠C.

 II. In a square ABCD, the diagonal AC bisects ∠A as well as ∠C

 III. In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C

Which is true?

1. I only
2. II and III
3. I and III
4. I and II

Solution 41

Question 42

In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ. Solution 42

Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.

⇒ PQ = SR (opposite sides of parallelogram are equal)

⇒ PQ = 2 cmQuestion 43

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.Solution 43

The given statement is false.

Diagonals of a parallelogram bisect each other. Question 44

What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 118°?Solution 44

In quadrilateral PQRS, ∠P and ∠S are adjacent angles.

Since the sum of adjacent angles ≠ 180°, PQRS is not a parallelogram.

Hence, PQRS is a trapezium. Question 45

All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.Solution 45

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are acute, the sum will be less than 360°. Question 46

All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.Solution 46

The given statement is true.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are right angles,

Sum of all angles of a quadrilateral = 4 × 90° = 360° Question 47

All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.Solution 47

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are obtuse, the sum will be more than 360°. Question 48

Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.Solution 48

We know that the sum of all the four angles of a quadrilateral is 360°.

Here,

70° + 115° + 60° + 120° = 365° ≠ 360° 

Hence, we cannot form a quadrilateral with given angles. Question 49

What special name can be given to a quadrilateral whose all angles are equal?Solution 49

A quadrilateral whose all angles are equal is a rectangle. Question 50

If D and E are respectively the midpoints of the sides AB and BC of ΔABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then determine the length of DE.Solution 50

D and E are respectively the midpoints of the sides AB and BC of ΔABC.

Thus, by mid-point theorem, we have

Question 51

In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.Solution 51

Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.

Now, adjacent angles of parallelogram are supplementary.

⇒ ∠Q + ∠R = 180° 

⇒ 56° + ∠R = 180° 

⇒ ∠R = 124° Question 52

In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?

Solution 52

AFDE is a parallelogram

⇒ AF = ED …(i)

BDEF is a parallelogram.

⇒ FB = ED …(ii)

From (i) and (ii),

AF = FB Question 53

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together.

Is quadrilateral ABCD a ‖gm?

1. Diagonal AC and BD bisect each other.
2. Diagonal AC and BD are equal.

The correct answer is : (a)/ (b)/ (c)/ (d).Solution 53

Correct option: (a)

If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.

So, I gives the answer.

If the diagonals are equal, then the quad. ABCD is a parallelogram.

So, II gives the answer.Question 54

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quadrilateral ABCD a rhombus?

1. Quad. ABCD is a ‖gm.
2. Diagonals AC and BD are perpendicular to each other.

The correct answer is: (a) / (b)/ (c)/ (d).Solution 54

Correct option: (c)

If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.

So, statement I is not sufficient to answer the question.

If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.

So, statement II is not sufficient to answer the question.

However, if the statements are combined, then the quad. ABCD is a rhombus.Question 55

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is ‖gm ABCD a square?

1. Diagonals of ‖gm ABCD are equal.
2. Diagonals of ‖gm ABCD intersect at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 55

Correct option: (c)

If the diagonals of a ‖gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.

If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.

However, if both the statements are combined, then ‖gm ABCD will be a square.Question 56

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quad. ABCD a parallelogram?

1. Its opposite sides are equal.
2. Its opposite angles are equal.

The correct answer is: (a)/ (b)/ (c)/ (d)Solution 56

Correct option: (b)

If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.

If the opposite angles are equal, then the quad. ABCD is a parallelogram.Question 57

 Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angle is 100°.	The sum of all the angle of a quadrilateral is 360°.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 57

Question 58

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.Then, PQRS is a parallelogram.	The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 58

The Reason (R) is true and is the correct explanation for the Assertion (A).Question 59

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.	The diagonals of a rhombus bisect each other at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 59

Question 60

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
Every parallelogram is a rectangle.	The angle bisectors of a parallelogram form a rectangle.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 60

Question 61

Each question consists of two statement, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The diagonals of a ‖gm bisect each other.	If the diagonals of a ‖gm are equal and intersect at right angles, then the parallelogram is a square.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 61

Question 62

Column I	Column II
(a) Angle bisectors of a parallelogram form a	(p) parallelogram
(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a	(q) rectangle
(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent side of a rectangle is a	(r) square
(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is	(s) rhombus

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 62

Question 63

Column I	Column II
(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7cm. If P and Q are the mid-points of AD and BC respectively, then PQ =	(p) equal
(b) In the given figure, PQRS is a ‖gm whose diagonal intersect at O. If PR = 13 cm, then OR=	(q) at right angle
(c) The diagonals of a square are	(r) 8.5 cm
(d) The diagonals of a rhombus bisect each other	(s) 6.5 cm

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 63

Exercise Ex. 10B

Question 1

In the adjoining figure, ABCD is a parallelogram in which =72⁰. Calculate _,and .

Solution 1

Question 2

In the adjoining figure , ABCD is a parallelogram in which

 and . Calculate _.

Solution 2

Question 3

In the adjoining figure, M is the midpoint of side BC of parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.

Solution 3

ABCD is a parallelogram.

Hence, AD || BC.

⇒ ∠DAM = ∠AMB (alternate angles)

⇒ ∠BAM = ∠AMB (since ∠BAM = ∠DAM)

⇒ BM = AB (sides opposite to equal angles are equal)

But, AB = CD (opposite sides of a parallelogram)

⇒ BM = AB = CD ….(i)

Question 4

In a adjoining figure, ABCD is a parallelogram in which =60^o. If the parallelogram in which and meet DC at P, prove that (i) PB=90^o, (ii) AD=DP and PB=PC=BC, (iii) DC=2AD.

Solution 4

Question 5

_{In the adjoining figure, ABCD is a parallelogram in which}

_Calculate 

Solution 5

Question 6

_{In a ||gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.}Solution 6

Question 7

_{If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .}Solution 7

Question 8

_{Find the measure of each angle of parallelogram , if one of its angles is  less than twice the smallest angle.}Solution 8

Question 9

_{ABCD is a parallelogram in which AB=9.5 cm and its parameter is 30 cm. Find the length of each side of the parallelogram.}Solution 9

Question 10

_{In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.}

Solution 10

Question 11

_{The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.}Solution 11

Question 12

_{Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.}Solution 12

Question 13

_{In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.}

Solution 13

Question 14

In a rhombus ABCD, the altitude from D to the side AB bisect AB. Find the angle of the rhombusSolution 14

Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In ΔAMD and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMD = ∠BMD (Each 90°)

MD = MD (common)

∴ ΔAMD ≅ ΔBMD (by SAS congruence criterion)

⇒ AD = BD (c.p.c.t.)

But, AD = AB (sides of a rhombus)

⇒ AD = AB = BD

⇒ ΔADB is an equilateral triangle.

⇒ ∠A = 60° 

⇒ ∠C = ∠A = 60° (opposite angles are equal)

⇒ ∠B = 180° – ∠A = 180° – 60° = 120° 

⇒ ∠D = ∠B = 120° 

Hence, in rhombus ABCD, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.Question 15

_{In the adjoining figure , ABCD is square . A line segment CX cuts AB at X and the diagonal BD at O, such that . Find the value of x.}

*Back answer incorrectSolution 15

Question 16

In a rhombus ABCD show that diagonal AC bisect ∠A as well as ∠C and diagonal BD bisect ∠B as well as ∠D.Solution 16

In ΔABC and ΔADC,

AB = AD (sides of a rhombus are equal)

BC = CD (sides of a rhombus are equal)

AC = AC (common)

∴ ΔABC ≅ ΔADC (by SSS congruence criterion)

⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)

⇒ AC bisects ∠A as well as ∠C.

Similarly,

In ΔBAD and ΔBCD,

AB = BC (sides of a rhombus are equal)

AD = CD (sides of a rhombus are equal)

BD = BD (common)

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

⇒ BD bisects ∠B as well as ∠D.Question 17

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.

Solution 17

In ΔAMO and ΔCNO

∠MAO = ∠NCO (AB ∥ CD, alternate angles)

AM = CN (given)

∠AOM = ∠CON (vertically opposite angles)

∴ ΔAMO ≅ ΔCNO (by ASA congruence criterion)

⇒ AO = CO and MO = NO (c.p.c.t.)

⇒ AC and MN bisect each other.Question 18

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that  and , prove that AQCP is a parallelogram.

Solution 18

Question 19

In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O.A line segment EOF is drawn to meet AB at E and DC at F . Prove that OE=OF.

Solution 19

Question 20

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.Solution 20

∠DCM = ∠DCN + ∠MCN

⇒ 90° = ∠DCN + 60° 

⇒ ∠DCN = 30° 

In ΔDCN,

∠DNC + ∠DCN + ∠D = 180° 

⇒ 90° + 30° + ∠D = 180° 

⇒ ∠D = 60° 

⇒ ∠B = ∠D = 60° (opposite angles of parallelogram are equal)

⇒ ∠A = 180° – ∠B = 180° – 60° = 120° 

⇒ ∠C = ∠A = 120° 

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.Question 21

ABCD is rectangle in which diagonal AC bisect ∠A as well as ∠C. Show that (i) ABCD is square, (ii) diagonal BD bisect ∠B as well as ∠D.Solution 21

(i) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

⇒ ∠BAC = ∠DAC ….(i)

And ∠BCA = ∠DCA ….(ii)

Since every rectangle is a parallelogram, therefore

AB ∥ DC and AC is the transversal.

⇒ ∠BAC = ∠DCA (alternate angles)

⇒ ∠DAC = ∠DCA [From (i)]

Thus, in ΔADC,

AD = CD (opposite sides of equal angles are equal)

But, AD = BC and CD = AB (ABCD is a rectangle)

⇒ AB = BC = CD = AD

Hence, ABCD is a square.

(ii) In ΔBAD and ΔBCD,

AB = CD

AD = BC

BD = BD

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

Hence, diagonal BD bisects ∠B as well as ∠D.Question 22

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.

Solution 22

Question 23

In the adjoining figure , ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF=2AB.

Solution 23

Question 24

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.Solution 24

l ∥ m and t is a transversal.

⇒ ∠APR = ∠PRD (alternate angles)

⇒ ∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)

Thus, PR intersects PS and RQ at P and R respectively such that ∠SPR = ∠PRQ i.e., alternate angles are equal.

⇒ PS ∥ RQ

Similarly, we have SR ∥ PQ.

Hence, PQRS is a parallelogram.

Now, ∠BPR + ∠PRD = 180° (interior angles are supplementary)

⇒ 2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)

⇒ ∠QPR + ∠QRP = 90° 

In ΔPQR, by angle sum property,

∠PQR + ∠QPR + ∠QRP = 180° 

⇒ ∠PQR + 90° = 180° 

⇒ ∠PQR = 90° 

Since PQRS is a parallelogram,

∠PQR = ∠PSR

⇒ ∠PSR = 90° 

Now, ∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are supplementary)

⇒ ∠SPQ + 90° = 180° 

⇒ ∠SPQ = 90° 

⇒ ∠SRQ = 90° 

Thus, all the interior angles of quadrilateral PQRS are right angles.

Hence, PQRS is a rectangle.Question 25

K, L, M and N are point on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.Solution 25

AK = BL = CM = DN (given)

⇒ BK = CL = DM = AN (i)(since ABCD is a square)

In ΔAKN and ΔBLK,

AK = BL (given)

∠A = ∠B (Each 90°)

AN = BK [From (i)]

∴ ΔAKN ≅ ΔBLK (by SAS congruence criterion)

⇒ ∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)

But, ∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90° 

⇒ ∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90° 

⇒ 2∠AKN + 2∠BKL = 180° 

⇒ ∠AKN + ∠BKL = 90° 

⇒ ∠NKL = 90° 

Similarly, we have

∠KLM = ∠LMN = ∠MNK = 90° 

Hence, KLMN is a square.Question 26

A is given . if lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB forming  , as shown in the adjoining figure, show that 

Solution 26

Question 27

In the adjoining figure, is a triangle and through A,B, C lines are drawn , parallel respectively to BC, CA and AB, intersecting at P, Q and R. prove that the perimeter of  is double the perimeter of  .

Solution 27

Exercise Ex. 10A

Question 1

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.Solution 1

Let the measure of the fourth angle = x° 

For a quadrilateral, sum of four angles = 360° 

⇒ x° + 75° + 90° + 75° = 360° 

⇒ x° = 360° – 240° 

⇒ x° = 120° 

Hence, the measure of fourth angle is 120°. Question 2

The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.Solution 2

Question 3

In the adjoining figure , ABCD is a trapezium in which AB || DC. If =55⁰ and = 70⁰, find  _and .

Solution 3

Since AB || DC

Question 4

In the adjoining figure , ABCD is a square andis an equilateral triangle . Prove that

(i)AE=BE, (ii) =15⁰

Solution 4

Given:

Question 5

In the adjoining figure , BMAC and DNAC. If BM=DN, prove that AC bisects BD.

Solution 5

Question 6

In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects  and , (ii) BE=DE,

(iii) 

Solution 6

Question 7

In the given figure , ABCD is a square and _PQR=90⁰. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=45⁰.

Solution 7

Question 8

If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.Solution 8

Given: O is a point within a quadrilateral ABCD

Question 9

In the adjoining figure, ABCD is a quadrilateral and AC is one of the diagonals .Prove that:

(i)AB+BC+CD+DA> 2AC

(ii)AB+BC+CD>DA

(iii)AB+BC+CD+DA>AC+BD

Solution 9

Given: ABCD is a quadrilateral and AC is one of its diagonals.

Question 10

Prove that the sum of all the angles of a quadrilateral is 360^0.Solution 10

Given: ABCD is a quadrilateral.

Exercise Ex. 10C

Question 1

P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that

(i) PQ ∥ AC and PQ = 

(ii) PQ ∥ SR

(iii) PQRS is a parallelogram.

Solution 1

(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii), we have

PQ = SR and PQ ∥ SR

(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram. Question 2

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse.Solution 2

Let ΔABC be an isosceles right triangle, right-angled at B.

⇒ AB = BC

Let PBSR be a square inscribed in ΔABC with common ∠B.

⇒ PB = BS = SR = RP

Now, AB – PB = BC – BS

⇒ AP = CS ….(i)

In ΔAPR and ΔCSR

AP = CS  [From (i)

∠APR = ∠CSR (Each 90°)

PR = SR (sides of a square)

∴ ΔAPR ≅ ΔCSR (by SAS congruence criterion)

⇒ AR = CR (c.p.c.t.)

Thus, point R bisects the hypotenuse AC.Question 3

In the adjoining figure , ABCD is a ||gm in which E and F are the midpoints of AB and CD respectively . If GH is a line segment that cuts AD, EF, and BC at G, P and H respectively, prove that GP = PH.

Solution 3

Question 4

M and N are points on opposites sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. show that MN is bisected at O.Solution 4

In ΔAOM and ΔCON

∠MAO = ∠OCN  (Alternate angles)

AO = OC (Diagonals of a parallelogram bisect each other)

∠AOM = ∠CON  (Vertically opposite angles)

∴ ΔAOM ≅ ΔCON  (by ASA congruence criterion)

⇒ MO = NO (c.p.c.t.)

Thus, MN is bisected at point O. Question 5

In the adjoining figure, PQRS is a trapezium in which PQ ∥ SR and M is the midpoint of PS. A line segment MN ∥ PQ meets QR at N. Show that N is the midpoint of QR.

Solution 5

Construction: Join diagonal QS. Let QS intersect MN at point O.

PQ ∥ SR and MN ∥ PQ

⇒ PQ ∥ MN ∥ SR

By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side. 

Now, in ΔSPQ

MO ∥ PQ and M is the mid-point of SP

So, this line will intersect QS at point O and O will be the mid-point of QS.

Also, MN ∥ SR

Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.

So, by using converse of mid-point theorem, N is the mid-point of QR.Question 6

In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.Solution 6

PM is the bisector of ∠P.

⇒ ∠QPM = ∠SPM ….(i)

PQRS is a parallelogram.

∴ PQ ∥ SR and PM is the transversal.

⇒ ∠QPM = ∠MS (ii)(alternate angles)

From (i) and (ii),

∠SPM = ∠PMS ….(iii)

⇒ MS = PS = 9 cm (sides opposite to equal angles are equal)

Now, ∠RMT = ∠PMS (iv)(vertically opposite angles)

Also, PS ∥ QT and PT is the transversal.

∠RTM = ∠SPM

⇒ ∠RTM = ∠RMT

⇒ RT = RM (sides opposite to equal angles are equal)

RM = SR – MS = 12 – 9 = 3 cm

⇒ RT = 3 cmQuestion 7

In the adjoining figure , ABCD is a trapezium in which AB|| DC and P, Q are the mid points of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ=QE, (ii) PR||AB, (iii) AR=RC.

Solution 7

Question 8

In the adjoining figure, AD is a medium of  _{and DE|| BA. Show that BE is also a median of .}

Solution 8

Question 9

_{In the adjoining figure , AD and BE are the medians of  and DF|| BE. Show that .}

Solution 9

Question 10

_{Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.}Solution 10

Question 11

_{In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .}

Solution 11

Question 12

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.Solution 12

Question 13

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.Solution 13

Question 14

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.Solution 14

Question 15

Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.Solution 15

Question 16

The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.Solution 16

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

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

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

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP  [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus.Question 17

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.Solution 17

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

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

From (i) and (ii),

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle. Question 18

The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.Solution 18

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

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

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

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square.