Chapter 10 : Quadrilaterals  R S Aggarwal And V Aggarwal Solutions for Class 9 Maths CBSE
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Chapter 10  Quadrilaterals Excercise MCQ
Three angles of quadrilateral are 80^{°}, 95° and 112°. Its fourth angle is
(a) 78°
(b) 73°
(c) 85°
(d) 100°
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°
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°
ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?
(a) 40°
(b) 25°
(c) 65°
(d) 130°
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°
In which of the following figures are the diagonals equal?
 Parallelogram
 Rhombus
 Trapezium
 Rectangle
If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a
a. trapezium
b. parallelogram
c. rectangle
d. rhombus
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is
 10 cm
 12 cm
 9cm
 8cm
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
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°
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°
If ABCD is a parallelogram with two adjacent angles ∠A = ∠B, then the parallelogram is a
 rhombus
 trapezium
 rectangle
 none of these
In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A and ∠B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB =?
 40°
 50°
 80°
 100°
The bisectors of any adjacent angles of a parallelogram intersect at
 30°
 45°
 60°
 90°
The bisectors of the angles of a parallelogram enclose a
 rhombus
 square
 rectangle
 parallelogram
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 supplementary
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.
The figure formed by joining the midpoints of the adjacent sides of a quadrilateral is a
 rhombus
 square
 rectangle
 parallelogram
The figure formed by joining the midpoints of the adjacent sides of a square is a
 rhombus
 square
 rectangle
 parallelogram
The figure formed by joining the midpoints of the adjacent sides of a parallelogram is a
 rhombus
 square
 rectangle
 parallelogram
The figure formed by joining the midpoints of the adjacent sides of a rectangle is a
 rhombus
 square
 rectangle
 parallelogram
The figure formed by joining the midpoints of the adjacent sides of a rhombus is a
 rhombus
 square
 rectangle
 parallelogram
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 other
Correct option: (d)
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔADC, R and S are the midpoints 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 midpoints 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.
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 other
Correct option: (c)
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔBCD, Q and R are the midpoints of sides BC and CD respectively.
In ΔADC, S and R are the midpoints of sides AD and CD respectively.
In ΔABD, P and S are the midpoints 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.
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 perpendicular
Correct option: (d)
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔBCD, Q and R are the midpoints of sides BC and CD respectively.
In ΔADC, S and R are the midpoints of sides AD and CD respectively.
In ΔABD, P and S are the midpoints 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.
If an angle of a parallelogram is twothird of its adjacent angle, the smallest angle of the parallelogram is
 108°
 54°
 72°
 81°
If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angles of the parallelogram is
 68°
 102°
 112°
 136°
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
 rhombus
 kite
 trapezium
 parallelogram
Which of the following is not true for a parallelogram?
 Opposite sides are equal.
 Opposite angles are equal.
 Opposite angles are bisected by the diagonals.
 Diagonals bisect each other.
If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a
 square
 rhombus
 rectangle
 kite
In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?
 45°
 55°
 60°
 75°
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 ≥ B
In the given figure, ABCD is a ‖gm and E is the midpoint at BC, Also, DE and AB when produced meet at F. Then,
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 rhombus
Correct option: (b)
In ΔABC, D and E are the midpoints of sides AB and AC respectively.
Hence, DENM is a parallelogram.
The parallel sides of a trapezium are ɑ and b respectively. The line joining the midpoints of its nonparallel sides will be
In a trapezium ABCD, if E and F be the midpoints of the diagonals AC and BD respectively. Then, EF =?
In the given figure, ABCD is a parallelogram, M is the midpoint of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB=?
 45°
 60°
 90°
 30°
In the given figures, ABCD is a rhombus. Then
(a) AC^{2} + BD^{2} = AB^{2}
(b) AC^{2} + BD^{2} = 2AB^{2}
(c) AC^{2} + BD^{2} = 4AB^{2}
(d) 2(AC^{2} + BD^{2})=3AB^{2}
In a trapezium ABCD, if AB ‖ CD, then (AC^{2} + BD^{2}) =?
(a) BC^{2} + AD^{2} + 2BC. AD
(b) AB^{2} +CD^{2} + 2AB.CD
(c) AB^{2} + CD^{2} + 2AD. BC
(d) BC^{2} + AD^{2} + 2AB.CD
Two parallelogram stand on equal bases and between the same parallels. The ratio of their area is
 1:2
 2:1
 1:3
 1:1
In the given figure, AD is a median of ΔABC and E is the midpoint of AD. If BE is joined and produced to meet AC in F, then AF =?
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 =?
 40°
 35°
 45°
 50°
Three statement are given below:
 In a ‖gm, the angle bisectors of two adjacent angles enclose a right angle.
 The angle bisectors of a ‖gm form a rectangle.
 The triangle formed by joining the midpoint of the sides of an isosceles triangle is not necessarily an isosceles.
Which is true?
 I only
 II only
 I and II
 II and III
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?
 I only
 II and III
 I and III
 I and II
In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ.
Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.
⇒ PQ = SR (opposite sides of parallelogram are equal)
⇒ PQ = 2 cm
Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.
The given statement is false.
Diagonals of a parallelogram bisect each other.
What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 118°?
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.
All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.
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°.
All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.
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°
All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.
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°.
Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.
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.
What special name can be given to a quadrilateral whose all angles are equal?
A quadrilateral whose all angles are equal is a rectangle.
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.
D and E are respectively the midpoints of the sides AB and BC of ΔABC.
Thus, by midpoint theorem, we have
In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.
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°
In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?
AFDE is a parallelogram
⇒ AF = ED …(i)
BDEF is a parallelogram.
⇒ FB = ED …(ii)
From (i) and (ii),
AF = FB
In each of the questions are question is followed by two statements I and II. The answer is
 if the question can be answered by one of the given statements alone and not by the other;
 if the question can be answered by either statement alone;
 if the question can be answered by both the statements together but not by any one of the two;
 if the question cannot be answered by using both the statement together.
Is quadrilateral ABCD a ‖gm?
 Diagonal AC and BD bisect each other.
 Diagonal AC and BD are equal.
The correct answer is : (a)/ (b)/ (c)/ (d).
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.
In each of the questions are question is followed by two statements I and II. The answer is
 if the question can be answered by one of the given statements alone and not by the other;
 if the question can be answered by either statement alone;
 if the question can be answered by both the statements together but not by any one of the two;
 if the question cannot be answered by using both the statement together
Is quadrilateral ABCD a rhombus?
 Quad. ABCD is a ‖gm.
 Diagonals AC and BD are perpendicular to each other.
The correct answer is: (a) / (b)/ (c)/ (d).
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.
In each of the questions are question is followed by two statements I and II. The answer is
 if the question can be answered by one of the given statements alone and not by the other;
 if the question can be answered by either statement alone;
 if the question can be answered by both the statements together but not by any one of the two;
 if the question cannot be answered by using both the statement together
Is ‖gm ABCD a square?
 Diagonals of ‖gm ABCD are equal.
 Diagonals of ‖gm ABCD intersect at right angles.
The correct answer is: (a)/ (b)/ (c)/ (d).
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.
In each of the questions are question is followed by two statements I and II. The answer is
 if the question can be answered by one of the given statements alone and not by the other;
 if the question can be answered by either statement alone;
 if the question can be answered by both the statements together but not by any one of the two;
 if the question cannot be answered by using both the statement together
Is quad. ABCD a parallelogram?
 Its opposite sides are equal.
 Its opposite angles are equal.
The correct answer is: (a)/ (b)/ (c)/ (d)
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.
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
 Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
 Assertion (A) is true and Reason (R) is false.
 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).
Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
 Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
 Assertion (A) is true and Reason (R) is false.
 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 midpoints of AB, BC, CD and DA respectively. Then, PQRS is a parallelogram. 
The line segment joining the midpoints 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).
The Reason (R) is true and is the correct explanation for the Assertion (A).
Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
 Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
 Assertion (A) is true and Reason (R) is false.
 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).
Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
 Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
 Assertion (A) is true and Reason (R) is false.
 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).
Each question consists of two statement, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
 Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
 Assertion (A) is true and Reason (R) is false.
 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).
Column I 
Column II 
(a) Angle bisectors of a parallelogram form a 
(p) parallelogram 
(b) The quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a 
(q) rectangle 
(c) The quadrilateral formed by joining the midpoints of the pairs of adjacent side of a rectangle is a 
(r) square 
(d) The figure formed by joining the midpoints of the pairs of adjacent sides of a quadrilateral is 
(s) rhombus 
The correct answer is:
(a) …….,
(b) …….,
(c) …….,
(d)…….
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 midpoints 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)…….
Chapter 10  Quadrilaterals Excercise Ex. 10A
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
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°.
The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.
In the adjoining figure , ABCD is a trapezium in which AB  DC. If _{}=55^{0} and _{}= 70^{0}, find _{ and }.
Since AB  DC
In the adjoining figure , ABCD is a square and_{}is an equilateral triangle . Prove that
(i)AE=BE, (ii) _{}=15^{0}
Given:_{}
In the adjoining figure , BM_{}AC and DN_{}AC. If BM=DN, prove that AC bisects BD.
_{}
_{ }
In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects and , (ii) BE=DE,
(iii)
_{}
In the given figure , ABCD is a square and _{P}QR=90^{0}. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=45^{0}.
_{}
_{}
_{}
If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.
Given: O is a point within a quadrilateral ABCD
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
Given: ABCD is a quadrilateral and AC is one of its disgonals.
_{}
Prove that the sum of all the angles of a quadrilateral is 360^{0.}
Given: ABCD is a quadrilateral.
_{}
Chapter 10  Quadrilaterals Excercise Ex. 10B
In the adjoining figure, ABCD is a parallelogram in which _{}=72^{0}. Calculate _{,and }.
In the adjoining figure , ABCD is a parallelogram in which
and . Calculate _{. }
_{}
_{}
In the adjoining figure, M is the midpoint of side BC of parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.
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)
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.
_{}
_{In the adjoining figure, ABCD is a parallelogram in which}
_{ }
_{Calculate }
_{}
_{ }
_{}
_{In a gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.}
_{}
_{If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .}
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_{Find the measure of each angle of parallelogram , if one of its angles is less than twice the smallest angle.}
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_{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.}
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_{In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.}
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_{The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.}
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_{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.}
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_{In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.}
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In a rhombus ABCD, the altitude from D to the side AB bisect AB. Find the angle of the rhombus
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 midpoint 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°.
_{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 incorrect
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In a rhombus ABCD show that diagonal AC bisect ∠A as well as ∠C and diagonal BD bisect ∠B as well as ∠D.
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.
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.
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.
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.
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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.
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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.
∠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°.
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.
(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.
In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.
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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.
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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.
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.
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.
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.
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
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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 .
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Chapter 10  Quadrilaterals Excercise Ex. 10C
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.
(i) In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
(ii) In ΔADC, R and S are the midpoints 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.
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.
Let ΔABC be an isosceles right triangle, rightangled 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.
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.
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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.
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.
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.
Construction: Join diagonal QS. Let QS intersect MN at point O.
PQ ∥ SR and MN ∥ PQ
⇒ PQ ∥ MN ∥ SR
By converse of midpoint theorem a line drawn, through the midpoint of any side of a triangle and parallel to another side bisects the third side.
Now, in ΔSPQ
MO ∥ PQ and M is the midpoint of SP
So, this line will intersect QS at point O and O will be the midpoint of QS.
Also, MN ∥ SR
Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.
So, by using converse of midpoint theorem, N is the midpoint of QR.
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.
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 cm
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) PRAB, (iii) AR=RC.
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In the adjoining figure, AD is a medium of _{and DE BA. Show that BE is also a median of .}
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_{In the adjoining figure , AD and BE are the medians of and DF BE. Show that .}
_{Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.}
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_{In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .}_{}
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.
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Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.
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Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.
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The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔBCD, Q and R are the midpoints of sides BC and CD respectively.
In ΔADC, S and R are the midpoints of sides AD and CD respectively.
In ΔABD, P and S are the midpoints 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.
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.
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔADC, R and S are the midpoints 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 midpoints 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.
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.
In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
In ΔBCD, Q and R are the midpoints of sides BC and CD respectively.
In ΔADC, S and R are the midpoints of sides AD and CD respectively.
In ΔABD, P and S are the midpoints 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.
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