NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Ex 8.1 Class 9 Maths** **Question 1.

The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

[Angle sum property of a quadrilateral]

⇒ 30x = 360°

⇒ x = 360∘/30 = 12°

∴ 3x = 3 x 12° = 36°

5x = 5 x 12° = 60°

9x = 9 x 12° = 108°

13a = 13 x 12° = 156°

⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.

Ex 8.1 Class 9 Maths** **Question 2.

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

Solution:

Let ABCD is a parallelogram such that AC = BD.

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

Ex 8.1 Class 9 Maths** **Question 3.

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

OB = OD [O is the mid-point of BD]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

Ex 8.1 Class 9 Maths** **Question 4.

Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

Let ABCD be a square such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, we need to prove AC = BD.

In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

AC = BD [By C.P.C.T.] …(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]

∴ ∠1 = ∠3

[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved]

∴ ∆OAD ≅ ∆OCB [By ASA congruency]

⇒ OA = OC and OD = OB [By C.P.C.T.]

i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii) In ∆OBA and ∆ODA, we have

OB = OD [Proved]

BA = DA [Sides of a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency]

⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)

∵ ∠AOB and ∠AOD form a linear pair.

∴∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD …(4)

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.

Ex 8.1 Class 9 Maths** **Question 5.

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.

Now, in ∆AOD and ∆AOB, We have

∠AOD = ∠AOB [Each 90°]

AO = AO [Common]

OD = OB [ ∵ O is the midpoint of BD]

∴ ∆AOD ≅ ∆AOB [By SAS congruency]

⇒ AD = AB [By C.P.C.T.] …(1)

Similarly, we have

AB = BC … (2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4), we have

AB = BC = CD = DA

∴ Quadrilateral ABCD have all sides equal.

In ∆AOD and ∆COB, we have

AO = CO [Given]

OD = OB [Given]

∠AOD = ∠COB [Vertically opposite angles]

So, ∆AOD ≅ ∆COB [By SAS congruency]

∴∠1 = ∠2 [By C.P.C.T.]

But, they form a pair of alternate interior angles.

∴ AD || BC

Similarly, AB || DC

∴ ABCD is a parallelogram.

∴ Parallelogram having all its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ∆ABC and ∆BAD, we have

AC = BD [Given]

BC = AD [Proved]

AB = BA [Common]

∴ ∆ABC ≅ ∆BAD [By SSS congruency]

∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5)

Since, AD || BC and AB is a transversal.

∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles]

⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]

So, rhombus ABCD is having one angle equal to 90°.

Thus, ABCD is a square.

Ex 8.1 Class 9 Maths** **Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

We have a parallelogram ABCD in which diagonal AC bisects ∠A

⇒ ∠DAC = ∠BAC

(i) Since, ABCD is a parallelogram.

∴ AB || DC and AC is a transversal.

∴ ∠1 = ∠3 …(1)

[ ∵ Alternate interior angles are equal]

Also, BC || AD and AC is a transversal.

∴ ∠2 = ∠4 …(2)

[ v Alternate interior angles are equal]

Also, ∠1 = ∠2 …(3)

[ ∵ AC bisects ∠A]

From (1), (2) and (3), we have

∠3 = ∠4

⇒ AC bisects ∠C.

(ii) In ∆ABC, we have

∠1 = ∠4 [From (2) and (3)]

⇒ BC = AB …(4)

[ ∵ Sides opposite to equal angles of a ∆ are equal]

Similarly, AD = DC ……..(5)

But, ABCD is a parallelogram. [Given]

∴ AB = DC …(6)

From (4), (5) and (6), we have

AB = BC = CD = DA

Thus, ABCD is a rhombus.

Ex 8.1 Class 9 Maths** **Question 7.

ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.

Solution:

Since, ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || CD and AD || BC

Now, CD = AD ⇒ ∠1 = ∠2 …….(1)

[ ∵ Angles opposite to equal sides of a triangle are equal]

Also, AD || BC and AC is the transversal.

[ ∵ Every rhombus is a parallelogram]

⇒ ∠1 = ∠3 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠2 = ∠3 …(3)

Since, AB || DC and AC is transversal.

∴ ∠2 = ∠4 …(4)

[ ∵ Alternate interior angles are equal] From (1) and (4),

we have ∠1 = ∠4

∴ AC bisects ∠C as well as ∠A.

Similarly, we can prove that BD bisects ∠B as well as ∠D.

Ex 8.1 Class 9 Maths** **Question 8.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

(i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

Solution:

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

(i) Since, every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || CD and AC is a transversal.

∴∠2 = ∠4 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠3 = ∠4

In ∆ABC, ∠3 = ∠4

⇒ AB = BC

[ ∵ Sides opposite to equal angles of a A are equal]

Similarly, CD = DA

So, ABCD is a rectangle having adjacent sides equal.

⇒ ABCD is a square.

(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.

So, BD bisects ∠B as well as ∠D.

Ex 8.1 Class 9 Maths** **Question 9.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

Solution:

We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

(i) Since, AD || BC and BD is a transversal.

∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite sides of a parallelogram ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB || CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

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

∴ ∆AQB = ∆CPD [By SAS congruency]

(iv) Since, ∆AQB = ∆CPD [Proved]

⇒ AQ = CP [By C.P.C.T.]

(v) In a quadrilateral ∆PCQ,

Opposite sides are equal. [Proved]

∴ ∆PCQ is a parallelogram.

Ex 8.1 Class 9 Maths** **Question 10.

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

Solution:

(i) In ∆APB and ∆CQD, we have

∠APB = ∠CQD [Each 90°]

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

∠ABP = ∠CDQ

[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]

∴ ∆APB = ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [By C.P.C.T.]

Ex 8.1 Class 9 Maths** **Question 11.

In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ABC ≅ ∆DEF

Solution:

(i) We have AB = DE [Given]

and AB || DE [Given]

i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.

∴ ABED is a parallelogram.

(ii) BC = EF [Given]

and BC || EF [Given]

i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.

∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram [Proved]

∴ AD || BE and AD = BE …(1)

[ ∵ Opposite sides of a parallelogram are equal and parallel] Also, BEFC is a parallelogram. [Proved]

BE || CF and BE = CF …(2)

[ ∵ Opposite sides of a parallelogram are equal and parallel]

From (1) and (2), we have

AD || CF and AD = CF

(iv) Since, AD || CF and AD = CF [Proved]

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

∴Quadrilateral ACFD is a parallelogram.

(v) Since, ACFD is a parallelogram. [Proved]

So, AC =DF [∵ Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have

AB = DE [Given]

BC = EF [Given]

AC = DE [Proved in (v) part]

∆ABC ≅ ∆DFF [By SSS congruency]

Ex 8.1 Class 9 Maths** **Question 12.

ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i )∠A=∠B

(ii )∠C=∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

Solution:

We have given a trapezium ABCD in which AB || CD and AD = BC.

(i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC

⇒ AE || DC Also AD || CF

∴ AECD is a parallelogram.

⇒ AD = CE …(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …(2) [Given]

By (1) and (2), BC = CF

Now, in ∆BCF, we have BC = CF

⇒ ∠CEB = ∠CBE …(3)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° … (4)

[Linear pair]

and ∠A + ∠CEB = 180° …(5)

[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (3)]

⇒ ∠B = ∠A …(6)

(ii) AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180° …(7) [Co-interior angles]

Similarly, ∠B + ∠C = 180° … (8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]

### NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

## Quadrilterals Class 9 Extra Questions Maths Chapter 8

**Extra Questions for Class 9 Maths Chapter 8 Quadrilterals**

**IMPORTANT QUESTIONS**

**VERY SHORT ANSWER TYPE QUESTIONS**

**Question.1 Three angles of a quadrilateral are equal and the fourth angle is equal to 144°. Find each of the equal angles of the quadrilateral.**

**Solution.**

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

**Solution.**

**Question.3 If one angle of a parallelogram is 30° less than twice the smallest angle, then find the measure of each angle.**

**Solution.**

**Question.4 If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram. [CBSE-15-6DWMW5A]**

**Solution.**

**Question.5**

**Solution.**

**Question.6.If the diagonals of a quadrilateral bisect each other at right angles, then name the quadrilateral.**

**Solution.** Rhombus.

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

**Solution.**

**Question.8 If an angle of a parallelogram is two-third of its adjacent angle, then find the smallest angle of the parallelogram.**

**Solution.**

**Question.9 In the given figure, ABCD is a parallelogram. If ∠B = 100°, then find the value of ∠A +∠C.**

**Solution.**

**Question.10 If the diagonals of a parallelogram are equal, then state its name.**

**Solution.** Rectangle

**Question.11 ONKA is a square with ∠KON = 45°. Determine ∠KOA.**

**Solution.**

**Question.12 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.**

**Question.13**

**Solution.**

**Question.14**

**Solution.**

**Question. 15.If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?**

**Solution.** **∠**A –**∠**C = 0° [opposite angles of parallelogram are equal]

**SHORT ANSWER QUESTIONS TYPE-I**

**Question.16 Prove that a diagonal of a parallelogram divide it into two congruent triangles. ****[CBSE March 2012]**

**Solution.** **Given :** A parallelogram ABCD and AC is its diagonal.

**Question.17 ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and ****C on diagonal BD (see fig.). Show that :**

**(i) AAPB ≅ ACQD (ii) AP = CQ [CBSE March 2012]**

**Solution.**

**Question.18**

**Solution.**

**Question.19**

**Solution.**

**Question.20**

**Solution.**

**Question.21 If the diagonals of a parallelogram are equal, then show that it is a rectangle. [CBSE March 2012]**

**Solution.**

**Question.22 ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. Show that AX \\CY. D x c**

**Solution.**

**SHORT ANSWER QUESTIONS TYPE-II**

**Question.23**

**Solution**.

**Question.24 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 : [CBSE-15-6DWMW5A]**

**Solution.**

**Question.25 In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. [CBSE March 2012]**

**Solution.**

**Question.26 D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles. [NCERT Exemplar Problem]**

**Solution.**

**Question.27**

**Solution.**

**Question.28**

**Solution.**

**Question.29**

**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.

**LONG ANSWER TYPE QUESTIONS**

**Question.30 ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. Show that:**

**(i) D is the mid-point of AC**

**(ii) MD ⊥ AC**

**(iii) CM = MA = 1/2 AB. [CBSE March 2012]**

**Solution.**

**Question.31**

**Solution.**

**Question.32 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.**

**Solution.**

**Question.33**

**Solution.**

**Question.34**

**Solution.**

**Question.35 ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC at D. Show that:**

**(i) D is the mid-point of AC**

**(ii) MD⊥ AC**

**(iii) CM = MA =1/2 AB. [CBSE March 2012]**

**Solution.**

**Question.36**

**Solution.**

**Question.37 ABCD is a rhombus. Show that diagonals AC bisects ∠A as well as ∠C and diagonal BD bisects∠B as well as ∠D**

**Solution.**

**Question.38**

**Solution.**

**Question.39**

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

**Question.40**

**Solution.**

**Question.41**

**Solution.**

**Question. 42 ABCD is a parallelogram in which diagonal AC bisects∠A as well as ∠C. Show that ABCD is a rhombus. [CBSE-14-17DIG1U]**

**Solution.**

**Question. 43**

**Solution.**

**Question.44 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. ****[CBSE-15-NS72LP7]**

**Solution.**

**Value Based Questions (Solved)**

**Question.1**

**Solution.**

**Question.2**

**Solution.**

**Question.3**

**Solution.**