Chapter 8 Quadrilaterals solved by expert teachers as per NCERT (CBSE) Book guidelines. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations.

NCERT Exemplar Class 9 Maths Solutions Quadrilaterals

Exercise 8.1: Multiple Choice Questions (MCQs)

Question 1:
Three angles of a quadrilateral are 75°, 90° and 75°, then the fourth angle is
(a) 90°
(b) 95°
(c) 105°
(d) 120°
Solution:
(d) Given, ∠A = 75°, ∠B = 90° and ∠C = 75°
We know that, sum of all the1 angles of a quadrilateral is 360°.
∠A+ ∠B + ∠C + ∠D = 360°
=> 75° + 90°+ 75°+ ∠D = 360°
∠D = 360° – (75° + 90° + 75°)
= 360° -240° = 120°
Hence, the fourth angle of a quadrilateral is 120°.

Question 2:
A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(a) 55°
(b) 50°
(c) 40°
(d) 25°
Solution:

Question 3:
ABCD is a rhombus such that ∠ACB = 40°, then ∠ADB is
(a) 40°
(b) 45°
(c) 50°
(d) 60°
Solution:

Question 4:
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if
(a) PQRS is a rectangle
(b) PQRS is a parallelogram
(c) diagonals of PQRS are perpendicular
(d) diagonals of PQRS are equal
Solution:
(c) Since, the quadrilateral ABCD formed by joining the mid-points of quadrilateral PQRS is a rectangle.
AC = BD [since, diagonals of a rectangle are equal]
=> PQ = QR

Question 5:
The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if
(a) PQRS is a rhombus
(b) PQRS is a parallelogram
(c) diagonals of PQRS are perpendicular
(d) diagonals of PQRS are equal
Solution:
(d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

Question 6:
If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a
(a) rhombus
(b) parallelogram
(c) trapezium
(d) kite
Solution:
(c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4.
Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°.
3x + 7x + 6x + 4x = 360°
=> 20x = 360°
=> x=360°/20° = 18°

Question 7:
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) rhombus
(c) parallelogram
(d) quadrilateral whose opposite angles are supplementary
Solution:

Question 8:
If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
(a) a square                   (b) a rhombus
(c) a rectangle               (d) any other parallelogram
Solution:

Question 9:
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(a) a rhombus
(b) a rectangle
(c) a square
(d) any parallelogram
Solution:

Question 10:
D and E are the mid-points of the sides AB and AC of ΔABC and 0 is any point on side BC. 0 is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is
(a) a square
(b) a rectangle
(c) a rhombus
(d) a parallelogram
Thinking Process
Use the mid-point theorem i.e., the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Solution:
(d) In ΔABC, D and E are the mid-points of sides AB and AC, respectively.
By mid-point theorem,
DE || BC …(i)
DE = 1/2 BC

Question 11:
The figure formed by joining the mid-points 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 equal and perpendicular
(d) diagonals of ABCD are perpendicular
Solution:

Question 12:
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(a) 24°
(b) 86°
(c) 38°
(d) 32°
Solution:

Question 13:
Which of the following is not true for a parallelogram?
(a) Opposite sides are equal
(b) Opposite angles are equal
(c) Opposite angles are bisected by the diagonals
(d) Diagonals bisect each other
Solution:
(c) We know that, in a parallelogram, opposite sides are equal, opposite angles are equal, opposite angles are not bisected by the diagonals and diagonals bisect each other.

Question 14:
D and E are the mid-points of the sides AB and AC, respectively, of ΔABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(a) ∠DAE = ∠EFC
(b) AE = EF
(c) DE = EF
(d) ∠ADE = ∠ECF
Solution:

Exercise 8.2: Very Short Answer Type Questions

Question 1:
Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.
Solution:

Question 2:
Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.
Solution:
No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other.

Question 3:
Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral? Why or why not?
Solution:

• No, we know that, sum of all angles of a quadrilateral is 360°.
• Here, sum of the angles = 110°+ 80° + 70° + 95° = 355° ≠ 360°
• So, these angles cannot be the angles of a quadrilateral.

Question 4:
In quadrilateral ABCD, ∠A + ∠D = 180°. What special name can be given to this quadrilateral?
Solution:
It is a trapezium because sum of cointerior angles is 180°.

Question 5:
All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?
Solution:
We know that, sum of all angles in a quadrilateral is 360°.
If ABCD is a quadrilateral,
∠A+ ∠B+ ∠C + ∠D = 360° …(i)
But it is given all angles are equal.
∠A = ∠B = ∠C = ∠D From Eq. (i)
∠A + ∠A + ∠A + ∠A = 360°
=> 4 ∠A = 360°
∠A = 90°
So, all angles of a quadrilateral are 90°.
Hence, given quadrilateral is a rectangle.

Question 6:
Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give reason for your answer.
Solution:
No, diagonals of a rectangle are equal but need not be perpendicular.

Question 7:
Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.
Solution:
No, all the four angles of a quadrilateral cannot be obtuse. As, the sum of the angles of a quadrilateral is 360°, then may have maximum of three obtuse angles.

Question 8:
In ΔABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.
Solution:
In ΔABC, we have AB = 5cm, BC = 8 cm and CA = 7 cm. Since, D and E are the mid-points of AB and BC, respectively.
By mid-point theorem, DE || AC
and

Question 9:
In the figure, it is given that BDEF and FDCE are parallelogram. Can you say that BD = CD? Why or why not?

Solution:
Yes, in the given figure, BDEF is a parallelogram..
∴ BD || EF and BD = EF …(i)
Also, FDCE is a parallelogram.
∴ CD||EF
and CD = EF …(ii)
From Eqs. (i) and (ii), BD = CD = EF

Question 10:
In figure, ABCD and AEFG are two parallelograms. If ∠C = 55°, then determine ∠F.

Solution:
We have, ABCD and AEFG are two parallelograms and ∠C = 55°. Since, ABCD is a parallelogram, then opposite angles of a parallelogram are equal.
∠A = ∠C = 55° ...(i)
Also, AEFG is a parallelogram.
∴ ∠A=∠F = 55° [from Eq. (i)]

Question 11:
Can all the angles of a quadrilateral be acute angles? Give reason for your answer.
Solution:
No, all the angles of a quadrilateral cannot be acute angles. As, sum of the angles of a quadrilateral is 360°. So, maximum of three acute angles will be possible.

Question 12:
Can all the angles of a quadrilateral be right angles? Give reason for your answer.
Solution:
Yes, all the angles of a quadrilateral can be right angles. In this case, the quadrilateral becomes rectangle or square.

Question 13:
Diagonals of a quadrilateral ABCD bisect each other. If ∠A= 35°, determine ∠B.
Solution:
Since, diagonals of a quadrilateral bisect each other, so it is a parallelogram.
Therefore, the sum of interior angles between two parallel lines is 180° i.e.,
∠A+∠B = 180°
=> ∠B = 180° – ∠A = 180°- 35°
[∴ ∠A = 35°, given]
=> ∠B = 145°

Question 14:
Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD.
Solution:
Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal.
∴ CD = AB = 4cm

Exercise 8.3: Short Answer Type Questions

Question 1:
One angle of a quadrilateral is of 108° and the remaining three angles are equal. Find each of the three equal angles.
Thinking Process
The sum of all the angles In a quadrilateral is 360°, use this result and simplify it.
Solution:
Let each of the three equal angles be x°.
Now, sum of angles of a quadrilateral = 360°
=> 108° + x° + x° + x° = 360° => 3x° = 360° – 108°
x° = 252°/3
=> x° = 84°
∴  x° = 84°
Hence, each of the three equal angles is 84°.

Question 2:
ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°. Find angles C and D of the trapezium.
Solution:
Given, ABCD is a trapezium and whose parallel sides in the figure are AB and DC.
Since, AB || CD and BC is transversal, then sum of two cointerior angles is 180°.

Question 3:
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:
Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Question 4:
ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.
Solution:

Question 5:
E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.
Solution:

Question 6:
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersects BC at F. Show thatf is the mid-point of BC.
Thinking Process
Use the mid-point theorem i.e., the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it. Further shown the required result.
Solution:
Given ABCD is a trapezium in which AB || DC and EF||AB|| CD.
Construction Join, the diagonal AC which intersects EF at O.
To show F is the mid-point of BC.

Proof Now, in ΔADC, E is the mid-point of AD and OE || CD. Thus, by mid-point theorem, O is mid-point of AC.
Now, in ΔCBA, 0 is the mid-point of AC and OF || AB.
So, by mid-point theorem, F is the mid-point of BC.

Question 7:
Through A, B and C lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a ΔABC as shown in figure. Show that  BC = ½ QR

Solution:
Given In ΔABC, PQ || AB and PR || AC and RQ || BC.
To show BC = ½ QR
Proof In quadrilateral BCAR, BR || CA and BC|| RA
So, quadrilateral, BCAR is a parallelogram.
BC = AR …(i)
Now, in quadrilateral BCQA, BC || AQ
and AB||QC
So, quadrilateral BCQA is a parallelogram,
BC = AQ …(ii)
On adding Eqs. (i) and (ii), we get
2 BC = AR+ AQ
=> 2 BC = RQ
=> BC = ½ QR
Now, BEDF is a quadrilateral, in which ∠BED = ∠BFD = 90°
∠FSE = 360° – (∠FDE + ∠BED + ∠BFD) = 360° – (60° + 90° + 90°)
= 360°-240° =120°

Question 8:
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral ΔABC. Show that ΔDEF is also an equilateral triangle.
Solution:
Given In equilateral ΔABC, D, E and F are the mid-points of sides BC, CA and AB, respectively.
To show ΔDEF is an equilateral triangle.

Proof Since in ΔABC, E and F are the mid-points of AC and AB respectively, then EF || BC and
EF =½ BC ,,.(i)
DF || AC, DE || AB
DE = ½ AB and FD = ½ AC [by mid-point theorem]… (ii)
since ΔABC is an equilateral triangle
AB = BC = CA
=> ½ AB = ½ BC = ½ CA  [dividing by 2]
=> ∴ DE = EF = FD   [from Eqs. (i) and (ii)]
Thus, all sides of ADEF are equal.
Hence, ΔDEF is an equilateral triangle.
Hence proved.

Question 9:
Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

Solution:

Question 10:
In figure, P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2 CD.

Thinking Process
Firstly, use the property that sum of cointerior angles is 180°. Secondly, use the property that sum of all angles in a triangle is 180° and then prove the required result.
Solution:

Exercise 8.4: Long Answer Type Questions

Question 1:
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 bisects the hypotenuse.
Solution:
Given In isosceles triangle ABC, a square ΔDEF is inscribed.
To prove CE = BE
Proof In an isosceles ΔABC, ∠A = 90°
and AB=AC …(i)
Since, ΔDEF is a square.
AD = AF [all sides of square are equal] … (ii)
On subtracting Eq. (ii) from Eq. (i), we get
AB – AD = AC- AF
BD = CF ….(iii)

Question 2:
In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
Solution:
Given, a parallelogram ABCD in which AB = 10 cm and AD = 6 cm.
Now, draw a bisector of ∠A meets DC in E and produce it to F and produce BC to meet at F.

Question 3:
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus,
Thinking Process
Firstly, use the mid-point theorem in various triangles of a quadrilateral. Further show that the line segments formed by joining the mid-points are equal, which prove the required quadrilateral.
Solution:
Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

Question 4:
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC ⊥ BD. Prove that PQRS is a rectangle.
Solution:
Given In quadrilateral ABCD, P, O, S and S are the mid-points of the sides AB, BC, CD and DA, respectively.
Also, AC ⊥ BD
To prove PQRS is a rectangle.
Proof Since, AC ⊥ BD .
∠COD = ∠AOD= ∠AOB= ∠COB = 90°

Question 5:
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.
Solution:
Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.
Also, AC = BD and AC ⊥ BD.
To prove PQRS is a square.
Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

Question 6:
A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.
Solution:

Question 7:
P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.
Solution:
Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively.
To show PRQS is a parallelogram.
Proof Since, ABCD is a parallelogram.
AB||CD
=> AP || QC

Question 8:
ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D.
Solution:

Question 9:
In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF.

Solution:
Given In figure AB || DE and AC || DF, also AB = DE and AC = DF
To prove BC ||EF and BC = EF
Proof In quadrilateral ABED, AB||DE and AB = DE
So, ABED is a parallelogram. AD || BE and AD = BE
Now, in quadrilateral ACFD, AC || FD and AC = FD …..(i)
Thus, ACFD is a parallelogram.
AD || CF and AD = CF …(ii)
From Eqs. (i) and (ii), AD = BE = CF and CF || BE …(iii)
Now, in quadrilateral BCFE, BE = CF
and BE||CF [from Eq. (iii)]
So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

Question 10:
E is the mid-point of a median AD of ΔABC and BE is produced to meet AC at F. Show that AF = 1/3 AC.
Solution:

Question 11:
Show that the quadrilateral formed by joining the consecutive sides of a square is also a square.
Solution:

Question 12:
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = ½ (AB + CD).
Solution:

Question 13:
Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Solution:
Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively.
To prove Quadrilateral PQRS is a rectangle.
Proof Since, ABCD is a parallelogram, then DC || AB and DA is a transversal.
∠A+∠D= 180°
[sum of cointerior angles of a parallelogram is 180°]
=> ½ ∠A+  ½ ∠D = 90° [dividing both sides by 2]
∠PAD + ∠PDA = 90°
∠APD = 90°    [since,sum of all angles of a triangle is 180°]
∴ ∠SPQ = 90°     [vertically opposite angles]
∠PQR = 90°
∠QRS = 90°
and ∠PSR = 90°
Thus, PQRS is a quadrilateral whose each angle is 90°.
Hence, PQRS is a rectangle.

Question 14:
P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Thinking Process
Firstly, prove that ΔODP and ΔOBQ are congruent by ASA rule. Further show the required result by CPCT rule.
Solution:

Question 15:
ABCD is a rectangle in which diagonal BD bisects ∠B. Show that ABCD is a square.
Solution:

Question 16:
D, E and F are respectively the mid-points of the sides AB, BC and CA of a ΔABC. Prove that by joining these mid-points D, E and F, the ΔABC is divided into four congruent triangles.
Solution:
Given In a ΔABC, D, E and F are respectively the mid-points of the sides AB, BC and CA. To prove ΔABC is divided into four congruent triangles.
Proof Since, ABC is a triangle and D, E and F are the mid-points of sides AB, BC and CA, respectively.

Question 17:
Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.
Solution:
Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

Question 18:
P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.
Solution: