Here we are providing Quadrilaterals Class 9 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

## Extra Questions for Class 9 Maths Quadrilaterals with Answers Solutions

**Extra Questions for Class 9 Maths Chapter 8 Quadrilaterals with Solutions Answers**

### Quadrilaterals Class 9 Extra Questions Very Short Answer Type

Question 1.

If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram.

Solution:

Let the two adjacent angles be x and 2x.

In a parallelogram, sum of the adjacent angles are 180°

∴ x + 2x = 180°

⇒ 3x = 180°

⇒ x = 60°

Thus, the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Question 2.

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

quadrilateral.

Solution:

Rhombus.

Question 3.

Three angles of a quadrilateral are equal and the fourth angle is equal to 144o. Find each of the equal angles of the quadrilateral.

Solution:

Let each equal angle of given quadrilateral be x.

We know that, sum of all interior angles of a quadrilateral is 360°

∴ x + x + x + 144° = 360°

3x = 360° – 144°

3x = 216°

x = 72°

Hence, each equal angle of the quadrilateral is of 72o measures.

Question 4.

If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?

Solution:

∠A – ∠C = 0° (opposite angles of parallelogram are equal]

Question 5.

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram.

Solution:

Here, PQ = SR = 12 cm

Let PS = x and PS = QR

∴ x + 12 + x + 12 = Perimeter

2x + 24 = 40

2x = 16

x= 8

Hence, length of each side of the parallelogram is 12 cm, 8 cm, 12 cm and 8 cm.

Question 6.

Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram?

Solution:

We know that consecutive interior angles of a parallelogram are supplementary.

∴ (x + 60° + (2x + 30)° = 180°

⇒ 3x° + 90° = 180°

⇒ 3x° = 90°

⇒ x° = 30°

Thus, two consecutive angles are (30 + 60)°, 12 x 30 + 30)”. i.e., 90° and 90°.

Hence, the special name of the given parallelogram is rectangle.

Question 7.

ONKA is a square with ∠KON = 45°. Determine ∠KOA.

Solution:

Since ONKA is a square

∴ ∠AON = 90°

We know that diagonal of a square bisects its ∠s

⇒ ∠AOK = ∠KON = 45°

Hence, ∠KOA = 45°

Question 8.

In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then find the measure of ∠S.

Solution:

Let ∠Q = 2x, ∠R = 3x and ∠S = 7x

Now, ∠P + ∠Q + ∠R + ∠S = 360°

⇒ 60° + 2x + 3x + 7x = 360°

⇒ 12x = 300°

x = \\(\\frac{300^{circ}}{12}) = 25°

∠S = 7x = 7 x 25° = 175°

### Quadrilaterals Class 9 Extra Questions Short Answer Type 1

Question 1.

ABCD is a parallelogram in which ∠ADC = 75° and side AB is produced to point E as shown in the figure. Find x + y.

Solution:

Here, ∠C and ∠D are adjacent angles of the parallelogram.

∴ ∠C + ∠D = 180°

⇒ x + 75° = 180°

⇒ x = 105°

Also, y = x = 105° [alt. int. angles]

Thus, x + y = 105° + 105° = 210°

Question 2.

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

Solution:

Given: A parallelogram ABCD, in which AC = BD.

To Prove: ΔBCD is a rectangle.

Proof : In ΔABC and ΔBAD

AB = AB (common]

AC = BD (given]

BC = AD(opp. sides of a ||^{gm}]

⇒ ΔABC ≅ ΔBAD

[by SSS congruence axiom]

⇒ ∠ABC = ∠BAD (c.p.c.t.)

Also, ∠ABC + ∠BAD = 180° (co-interior angles)

∠ABC + ∠ABC = 180° [ ∵ ∠ABC = ∠BAD ]

2∠ABC = 180°

∠ABC = 1/2 x 180° = 90°

Hence, parallelogram ABCD is a rectangle.

Question 3.

In the figure, ABCD is a rhombus, whose diagonals meet at O. Find the values of x and y.

Solution:

Since diagonals of a rhombus bisect each other at right angle.

In ∴ ΔAOB, we have

∠OAB + ∠x + 90° = 180°

∠x = 180° – 90° – 35°

= 55°

Also,

∠DAO = ∠BAO = 35°

∠y + ∠DAO + ∠BAO + ∠x = 180°

⇒ ∠y + 35° + 35° + 55° = 180°

⇒ ∠y = 180° – 125o = 55°

Hence, the values of x amd y are x = 55°, y = 55°.

Question 4.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :

(i) ΔAPB = ΔCQD

(ii) AP = CQ

Solution:

Given : In ||gm ABCD, AP and CQ are perpendiculars from the

vertices A and C on the diagonal BD.

To Prove: (i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Proof : (i) In ΔAPB and ΔCQD

AB = DC (opp. sides of a ||gm ABCD]

∠APB = ∠DQC (each = 90°)

∠ABP = ∠CDQ (alt. int. ∠s]

⇒ ΔAPB ≅ ΔCQD[by AAS congruence axiom]

(ii) ⇒ AP = CQ [c.p.c.t.]

### Quadrilaterals Class 9 Extra Questions Short Answer Type 2

Question 1.

The diagonals of a quadrilateral ABCD are perpendicular to each other. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Solution:

Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P, Q, R and S are mid-points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS.

To Prove: PQRS is a rectangle.

Proof : In ∆ABC, P and Q are mid-points of AB and BC respectively.

∴ PQ || AC and PQ = \\(\\frac{1}{2}) AC … (i) (mid-point theorem]

Further, in SACD, R and S are mid-points of CD and DA respectively.

SR || AC and SR = \\(\\frac{1}{2}) AC … (ii) (mid-point theorem]

From (i) and (ii), we have PQ || SR and PQ = SR

Thus, one pair of opposite sides of quadrilateral PQRS are parallel and equal.

∴ PQRS is a parallelogram.

Since PQ|| AC PM || NO

In ∆ABD, P and S are mid-points of AB and AD respectively.

PS || BD (mid-point theorem]

⇒ PN || MO

∴ Opposite sides of quadrilateral PMON are parallel.

∴ PMON is a parallelogram.

∠ MPN = ∠ MON (opposite angles of ||gm are equal]

But ∠MON = 90° [given]

∴ ∠MPN = 90° ⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram whose one angle is 90°

∴ PQRS is a rectangle.

Question 2.

In the fig., D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.

Solution:

Since line segment joining the mid-points of two sides of a triangle is half of the third side.

Therefore, D and E are mid-points of BC and AC respectively.

⇒ DE = \\(\\frac{1}{2})AB …(i)

E and F are the mid-points of AC and AB respectively.

∴ EF = \\(\\frac{1}{2})BC … (ii)

F and D are the mid-points of AB and BC respectively.

∴ FD = \\(\\frac{1}{2}) AC … (iii)

Now, SABC is an equilateral triangle.

⇒ AB = BC = CA

⇒ \\(\\frac{1}{2})AB = \\(\\frac{1}{2})BC = \\(\\frac{1}{2})CA

⇒ DE = EF = FD (using (i), (ii) and (iii)]

Hence, DEF is an equilateral triangle.

Question 3.

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Solution:

Since BXDY is a parallelogram.

XO = YO

DO = BO

[∵ diagonals of a parallelogram bisect each other]

But AX = CY …. (iii) (given]

Adding (i) and (iii), we have

XO + AX = YO + CY

⇒ AO = CO …. (iv)

From (ii) and (iv), we have

AO = CO and DO = BO

Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Question 4.

ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that

∠X +∠Y = \\(\\frac{1}{2})(∠A + ∠C)

Solution:

Here, ∠1 = ∠2 and ∠3 = ∠4

In ΔXBC, we have

∠X + ∠B + ∠4 = 180°

∠X + ∠B + \\(\\frac{1}{2})∠C = 180

In ΔADY, we have

∠2 + ∠D + ∠Y= 180°

\\(\\frac{1}{2}) ∠A + ∠D + ∠Y = 180°

Adding (i) and (ii), we have

∠X + ∠Y + ∠B + ∠D + \\(\\frac{1}{2}) ∠C + \\(\\frac{1}{2}) ∠A = 360°

Also, in quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

∠X + ∠Y + ∠B + ∠D + \\(\\frac{1}{2}) ∠C + \\(\\frac{1}{2}) ∠A = ∠A + ∠B + ∠C + ∠D

∠X + ∠Y = ∠A – \\(\\frac{1}{2}) ∠A + \\(\\frac{1}{2}) ∠C – \\(\\frac{1}{2}) ∠C

∠X+ ∠Y = \\(\\frac{1}{2}) (∠A + ∠C)

### Quadrilaterals Class 9 Extra Questions Long Answer Type

Question 1.

In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = \\(\\frac{1}{4}) BC.

Solution:

Here, in ΔABC, R and Q are the mid-points of AB and AC respectively.

∴ By using mid-point theorem, we have

RQ || BC and RQ = \\(\\frac{1}{2}) BC

∴ RQ = BP = PC [∵ P is the mid-point of BC]

∴ RQ || BP and RQ || PC

In quadrilateral BPQR

RQ || BP, RQ = BP (proved above]

∴ BPQR is a parallelogram. [∵ one pair of opp. sides is parallel as well as equal]

∴ X is the mid-point of PR. [∵ diagonals of a ||gm bisect each other]

Now, in quadrilateral PCQR

RQ || PC and RQ = PC [proved above)

∴ PCQR is a parallelogram [∵ one pair of opp. sides is parallel as well as equal]

∴ Y is the mid-point of PQ [∵ diagonals of a ||gm bisect each other]

In ΔPQR

∴ X and Y are mid-points of PR and PQ respectively.

Question 2.

In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.

Solution:

Since AE = DE

∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]

Again, BC || AD

∠EBC = ∠A …. (ii) (corresponding ∠s]

From (i) and (ii), we have

∠D = ∠EBC …. (iii)

But ∠EBC + ∠ABC = 180° (a linear pair]

∠D + ∠ABC = 180° (using (iii)]

Now, a pair of opposite angles of quadrilateral ABCD is supplementary

Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA

∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]

∠BAD = ∠CDA [using (i)]

AD = AD (common]

So, by using AAS congruence axiom, we have

ΔABD ≅ ΔDCA

Hence, BD = CA [c.p.c.t.]

Question 3.

In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.

Solution:

Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.

In ΔAOB, X and Y are the mid-points of AO and BO.

∴ By using mid-point theorem, we have

XY = \\(\\frac{1}{2}) AB = \\(\\frac{1}{2}) x 8 cm = 4 cm

Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.

∴ By using mid-point theorem, we have

YZ = \\(\\frac{1}{2}) BC = \\(\\frac{1}{2}) x 9cm = 4.5 cm

And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.

∴ ZX = \\(\\frac{1}{2}) AC = \\(\\frac{1}{2}) x 10 cm = 5 cm

Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Question 4.

PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°

Solution:

Since PQRS is a square.

∴ PQ = QR … (I) [∵ sides of a square are equal]

Also, BQ = CR … (ii) [given]

Subtracting (ii) from (i), we obtain

PQ – BQ = QR – CR

⇒ PB = QC … (iii)

In Δ𝜏APB and Δ𝜏BQC

AP = BQ[given

∠APB = ∠BQC = 90°](each angle of a square is 90°)

PB = QC (using (iii)]

So, by using SAS congruence axiom, we have

ΔAPB ≅ ΔBQC

∴ AB = BC [c.p.c.t.]

Now, in ΔABC

AB = BC [proved above]

∴ ∠ACB = ∠BAC = x° (say) [∠s opp. to equal sides]

Also, ∠B + ∠ACB + ∠BAC = 180°

⇒ 90° + x + x = 180°

⇒ 2x° = 90°

x° = 45°

Hence, ∠BAC = 45°

Question 5.

ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.

Solution:

Since DP and CP are angle bisectors of ∠D and ∠C respectively.

: ∠1 = ∠2 and ∠3 = ∠4

Now, AB || DC and CP is a transversal

∴ ∠5 = ∠1 [alt. int. ∠s]

But ∠1 = ∠2 [given]

∴ ∠5 = ∠2

Now, in ABCP, ∠5 = ∠2

⇒ BC = BP … (I) [sides opp. to equal ∠s of a A]

Again, AB || DC and DP is a transversal.

∴ ∠6= ∠3 (alt. int. Δs]

But ∠4 = ∠3 [given]

∴ ∠6 = ∠4

Now, in ΔADP, ∠6 = ∠4

⇒ DA = AP …. (ii) (sides opp. to equal ∠s of a A]

Also, BC = DA… (iii) (opp. sides of parallelogram)

From (i), (ii) and (iii), we have

BP = AP

Hence, P is the mid-point of side AB.

Question 6.

In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that :

(i) EO || AB

(ii) AO = CO

Solution:

Here, E and F are the mid-points of AD and BC respectively.

In ΔBFG and ΔCFD

BF = CF [given]

∠BFG = ∠CFD (vert. opp. ∠s]

∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)

So, by using AAS congruence axiom, we have

ΔBFG ≅ ΔCFD

⇒ DF = FG [c.p.c.t.)

Now, in ΔAGD, E and F are the mid-points of AD and GD.

∴ By mid-point theorem, we have

EF || AG

or EO || AB

Also, in ΔADC, EO || DC

∴ EO is a line segment from mid-point of one side parallel to another side.

Thus, it bisects the third side.

Hence, AO = CO