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Page No: 146

Exercise 8.1

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

Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360Â°
Now,
3x + 5x + 9x + 13x = 360Â°
â‡’ 30x = 360Â°
â‡’ x = 12Â°
3x = 3Ã—12Â° = 36Â°
5x = 5Ã—12Â° = 60Â°
9x = 9Ã—12Â° = 108Â°
13x = 13Ã—12Â° = 156Â°

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

Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, Î”ABC â‰… Î”BAD by SSS congruence condition.
âˆ A = âˆ B (by CPCT)
also,
âˆ A + âˆ B = 180Â° (Sum of the angles on the same side of the transversal)
â‡’ 2âˆ A = 180Â°
â‡’ âˆ A = 90Â° = âˆ B
Thus ABCD is a rectangle.

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

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and âˆ AOB = âˆ BOC = âˆ OCD = âˆ ODA = 90Â°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In Î”AOB and Î”COB,
OA = OC (Given)
âˆ AOB = âˆ COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, Î”AOB â‰… Î”COB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

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

Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
Â AC = BD, AO = OC and âˆ AOB = 90Â°
Proof,
BC = BA (Common)
âˆ ABC = âˆ BAD = 90Â°
Therefore, Î”ABC â‰… Î”BAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In Î”AOB and Î”COD,
âˆ BAO = âˆ DCO (Alternate interior angles)
âˆ AOB = âˆ COD (Vertically opposite)
AB = CD (Given)
Therefore, Î”AOB â‰… Î”COD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In Î”AOB and Î”COB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, Î”AOB â‰… Î”COB by SSS congruence condition.
also, âˆ AOB = âˆ COB
âˆ AOB + âˆ COB = 180Â° (Linear pair)
Thus, âˆ AOB = âˆ COB = 90Â° (Diagonals bisect each other at right angles)

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

Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Proof,
In Î”AOB and Î”COD,
AO = CO (Diagonals bisect each other)
âˆ AOB = âˆ COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, Î”AOB â‰… Î”COD by SAS congruence condition.
Thus, AB = CD by CPCT. --- (i)
also,
âˆ OAB = âˆ OCD (Alternate interior angles)
â‡’ AB || CD
Now,
In Î”AOD and Î”COD,
AO = CO (Diagonals bisect each other)
âˆ AOD = âˆ COD (Vertically opposite)
OD = OD (Common)
Therefore, Î”AOD â‰… Î”COD by SAS congruence condition.
Thus, AD = CD by CPCT. --- (ii)
also,
â‡’ AD = BC = CD = AB --- (ii)
also,Â  âˆ ADC = âˆ BCDÂ  by CPCT.
and âˆ ADC + âˆ BCD = 180Â° (co-interior angles)
â‡’ âˆ ADC = 90Â° --- (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects âˆ A (see Fig. 8.19). Show that
(i) it bisects âˆ C also,
(ii) ABCD is a rhombus.

(i)
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, Î”ADC â‰… Î”CBA by SSS congruence condition.
Thus,
âˆ ACD = âˆ CAB by CPCT
and âˆ CAB = âˆ CAD (Given)
â‡’ âˆ ACD = âˆ BCA
Thus, AC bisects âˆ C also.

(ii) âˆ ACD = âˆ CAD (Proved)
â‡’ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects âˆ A as well as âˆ C and diagonal BD bisects âˆ B as well as âˆ D.

Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
âˆ DAC = âˆ DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
â‡’ âˆ DAC = âˆ BCA (Alternate interior angles)
â‡’ âˆ DCA = âˆ BCA
Therefore, AC bisects âˆ C.
Similarly, we can prove that diagonal AC bisects âˆ A.

Also, by preceding above method we can prove that diagonal BD bisects âˆ B as well as âˆ D.

8. ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects âˆ B as well as âˆ D.

(i)âˆ DAC = âˆ DCA (AC bisects âˆ A as well as âˆ C)
â‡’ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.

(ii) In Î”BCD,
BC = CD
â‡’ âˆ CDB = âˆ CBD (Angles opposite to equal sides are equal)
also, âˆ CDB = âˆ ABD (Alternate interior angles)
â‡’ âˆ CBD = âˆ ABD
Thus, BD bisects âˆ B
Now,
â‡’ âˆ CDB = âˆ ADB
Thus, BD bisects âˆ D

Page No: 147

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) Î”APD â‰… Î”CQB
(ii) AP = CQ
(iii) Î”AQB â‰… Î”CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

(i) In Î”APD and Î”CQB,
DP = BQ (Given)
âˆ ADP = âˆ CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, Î”APD â‰… Î”CQB by SAS congruence condition.

(ii) AP = CQ by CPCT as Î”APD â‰… Î”CQB.

(iii) In Î”AQB and Î”CPD,
BQ = DP (Given)
âˆ ABQ = âˆ CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, Î”AQB â‰… Î”CPD by SAS congruence condition.

(iv) AQ = CP by CPCT as Î”AQB â‰… Î”CPD.

(v) From (ii)Â  and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) Î”APB â‰… Î”CQD
(ii) AP = CQ

(i) In Î”APB and Î”CQD,
âˆ ABP = âˆ CDQ (Alternate interior angles)
âˆ APB = âˆ CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, Î”APB â‰… Î”CQD by AAS congruence condition.

(ii) AP = CQ by CPCT as Î”APB â‰… Î”CQD.

11. In Î”ABC and Î”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 Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) Î”ABC â‰… Î”DEF.

(i) AB = DE and AB || DE (Given)
Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.
(ii) Again BC = EF and BC || EF.
Thus, quadrilateral BEFC is a parallelogram.
(iii)Â  Since ABED and BEFC are parallelograms.
â‡’ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)
Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.
(vi) In Î”ABC and Î”DEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (Opposite sides of a parallelogram)
Thus, Î”ABC â‰… Î”DEF by SSS congruence condition.

12. ABCD is a trapezium in which AB || CD and
AD = BC (see Fig. 8.23). Show that
(i) âˆ A = âˆ B
(ii) âˆ C = âˆ D
(iv) diagonal AC = diagonal BD
[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
Therefor, BC = CE
â‡’ âˆ CBE = âˆ CEB
also,
âˆ AÂ + âˆ CBE = 180Â° (Angles on the same side of transversal and âˆ CBE = âˆ CEB)
âˆ B + âˆ CBE = 180Â° (Linear pair)
â‡’ âˆ A = âˆ B

(ii) âˆ A + âˆ D = âˆ BÂ + âˆ C = 180Â° (Angles on the same side of transversal)
â‡’ âˆ A + âˆ D = âˆ A + âˆ C (âˆ A = âˆ B)
â‡’ âˆ D = âˆ C

AB = AB (Common)
âˆ DBA = âˆ CBA
Thus, Î”ABC â‰… Î”BAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as Î”ABC â‰… Î”BA.

Page No: 150

Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that :
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

(i) In Î”DAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SR || AC and SR = 1/2 AC

(ii) In Î”BAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQ || AC and PQ = 1/2 AC
also, SR = 1/2 AC
Thus, PQ = SR

(iii) SR || AC - from (i)
and, PQ || AC - from (ii)
â‡’ SR || PQ - from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Given,
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
AC and BD are joined.
Proof,
In Î”DRS and Î”BPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
âˆ SDR = âˆ QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, Î”DRS â‰… Î”BPQ by SAS congruence condition.
RS = PQ by CPCT --- (i)
In Î”QCR and Î”SAP,
RC = PA (Halves of the opposite sides of the rhombus)
âˆ RCQ = âˆ PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, Î”QCR â‰… Î”SAP by SAS congruence condition.
RQ = SP by CPCT --- (ii)
Now,
In Î”CDB,
R and Q are the mid points of CD and BC respectively.
â‡’ QR || BDÂ
also,
P and S are the mid points of AD and AB respectively.
â‡’ PS || BD
â‡’ QR || PS
Thus, PQRS is a parallelogram.
also, âˆ PQR = 90Â°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
âˆ Q = 90Â°
Thus, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Given,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In Î”ABC
P and Q are the mid-points of AB and BC respectively
Thus, PQ || AC and PQ = 1/2 AC (Mid point theorem) --- (i)
SR || AC and SR = 1/2 AC (Mid point theorem) --- (ii)
So, PQ || SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PS || QR and PS = QR (Opposite sides of parallelogram) --- (iii)
Now,
In Î”BCD,
Q and R are mid points of side BC and CD respectively.
Thus, QR || BD and QR = 1/2 BD (Mid point theorem) --- (iv)
AC = BD (Diagonals of a rectangle are equal) --- (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.

Given,
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
E is the mid point of AD and also EG || AB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In Î”BDC,
G is the mid point of BD and also GF || AB || DC.
Thus, F is the mid point of BC (Converse of mid point theorem)

Page No: 151

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Given,
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,
ABCD is a parallelogram
Therefor, AB || CD
also, AE || FC
Now,
AB = CD (Opposite sides of parallelogram ABCD)
â‡’ 1/2 AB = 1/2 CD
â‡’ AE = FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AF || EC (Opposite sides of a parallelogram)
Now,
In Î”DQC,
F is mid point of side DC and FP || CQ (as AF || EC).
P is the mid-point of DQ (Converse of mid-point theorem)
â‡’ DP = PQ --- (i)
Similarly,
In APB,
E is mid point of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
â‡’ PQ = QB --- (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now,
In Î”ACD,
R and S are the mid points of CD and DA respectively.
Thus, SR || AC.
Similarly we can show that,
PQ || AC
PS || BD
QR || BD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC

(ii) MD âŠ¥ AC
(iii) CM = MA = 1/2 AB

(i) In Î”ACB,
M is the mid point of AB and MD || BC
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) âˆ ACB = âˆ ADM (Corresponding angles)
also, âˆ ACB = 90Â°
Thus, âˆ ADM = 90Â° and MD âŠ¥ AC

(iii)Â  In Î”AMD and Î”CMD,
AD = CD (D is the midpoint of side AC)
âˆ ADM = âˆ CDM (Each 90Â°)
DM = DM (common)
Thus, Î”AMD â‰… Î”CMD by SAS congruence condition.
AM = CM by CPCT
also, AM =Â  1/2 AB (M is mid point of AB)
Hence, CM = MA =Â  1/2 AB

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Chapter 8 Quadrilaterals NCERT Solutions is very important for the preparation of exams. A figure formed by joining four points in an order is called a quadrilateral. A quadrilateral has four sides, four angles and four vertices. In this chapter, we will be discussing different types of quadrilaterals, their properties and about parallelograms.

â€¢ Angle Sum Property of a Quadrilateral: The sum of the angles of a quadrilateral is 360Âº. This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

A square is a rectangle and also a rhombus.
A parallelogram is a trapezium.
A kite is not a parallelogram.
A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram).
A rectangle or a rhombus is not a square.

â€¢ Properties of a Parallelogram:
1. A diagonal of a parallelogram divides it into two congruent triangles.
2. In a parallelogram, opposite sides are equal.
3. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
4. In a parallelogram, opposite angles are equal.
5. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
6. The diagonals of a parallelogram bisect each other.
7. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

â€¢ Another Condition for a Quadrilateral to be a Parallelogram: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

â€¢ The Mid-point Theorem:
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.Â
The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

There are only two exercises in Chapter 8 Quadrilaterals NCERT Solutions which are provided below which can be helpful in completing your homework on time.

Studyrankers experts have taken every care while preparing these Class 9 Maths NCERT Solutions so you can easily clear your doubts. These NCERT Solutions are updated as per the latest marking scheme released by CBSE.

### NCERT Solutions for Class 9 Maths Chapters:

 Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations in Two Variables Chapter 5 Introduction to Euclidâ€™s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 9 Areas of Parallelograms and TrianglesÂ Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron's Formula Chapter 13 Surface Areas And Volumes Chapter 14 Statistics Chapter 15 Probability

#### What are the benefits of NCERT Solutions for Chapter 8 Quadrilaterals Class 9 NCERT Solutions?

Through the help of NCERT Solutions of Chapter 8 Quadrilaterals you will be able to solve difficult questions easily and revising the chapter properly. It will improve the learning behaviour of the students and learning diverse topics.

#### What is a Rhombus?

A parallelogram having all sides equal is called a rhombus.

#### What is mid-point theorem?

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.

#### In a quadrilateral, âˆ  A : âˆ  B : âˆ  C : âˆ  D = 1 : 2 : 3 : 4, then find the measure of each angle of the quadrilateral.

Since âˆ  A : âˆ  B : âˆ  C : âˆ  D = 1 : 2 : 3 : 4
âˆ´ If âˆ  A = x, then âˆ  B = 2x, âˆ  C = 3x and âˆ  D = 4x. âˆ´ âˆ  A + âˆ  B + âˆ  C + âˆ  D = 360Â°
â‡’ x + 2x + 3x + 4x = 360Â° â‡’ 10x = 36Â°
â‡’ x= (360Â°/10)= 36Â°
âˆ´ âˆ  A = x = 36Â° âˆ  B = 2x = 2 x 36Â° = 72Â° âˆ  C = 3x = 3 x 36Â° = 108Â° âˆ  D = 4x = 4 x 36Â° = 144Â°.
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