# NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

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Exercise 8.2 Solutions : 7 Solved Questions

### Access Answers of Maths NCERT class 9 Chapter 8 Quadrilateral Exercise 8.1

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

Solution:

Let the common ratio between the angles be = x.

We know that the sum of the interior angles of the quadrilateral = 360Â°

Now,

3x+5x+9x+13x = 360Â°

â‡’ 30x = 360Â°

â‡’ x = 12Â°

, Angles of the quadrilateral are:

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.

Solution:

Given that,

AC = BD

To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal

To show ABCD is a rectangle we have to prove that one of its interior angles 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 [SSS congruency]

âˆ A = âˆ B [Corresponding parts of Congruent Triangles]

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

, ABCD is a rectangle.

Hence Proved.

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 whose diagonals bisect each other at right angles.

Given that,

OA = OC

OB = OD

and âˆ AOB = âˆ BOC = âˆ OCD = âˆ ODA = 90Â°

To show that,

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

i.e., we have to prove that 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 [SAS congruency]

Thus, AB = BC [CPCT]

Similarly we can prove,

BC = CD

, AB = BC = CD = AD

Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Hence Proved.

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

Solution:

Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To show that,

AC = BD

AO = OC

and âˆ AOB = 90Â°

Proof,

BC = BA (Common)

âˆ ABC = âˆ BAD = 90Â°

Thus,

AC = BD [CPCT]

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)

, Î”AOB â‰… Î”COD [AAS congruency]

Thus,

AO = CO [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)

, Î”AOB â‰… Î”COB [SSS congruency]

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.

Solution:

Given that,

Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O.

To prove that,

The Quadrilateral ABCD is a square.

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)

, Î”AOB â‰… Î”COD [SAS congruency]

Thus,

AB = CD [CPCT] â€” (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)

, Î”AOD â‰… Î”COD [SAS congruency]

Thus,

AD = CD [CPCT] â€” (ii)

also,

â‡’ AD = BC = CD = AB â€” (ii)

also,Â  âˆ ADC = âˆ BCDÂ  [CPCT]

and âˆ ADC+âˆ BCD = 180Â° (co-interior angles)

â‡’âˆ ADC = 90Â° â€” (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

Hence Proved.

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.

Solution:

AD = CB (Opposite sides of a parallelogram)

DC = BA (Opposite sides of a parallelogram)

AC = CA (Common Side)

, Î”ADC â‰… Î”CBA [SSS congruency]

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 above)

â‡’ AD = CD (Opposite sides of equal angles of a triangle are equal)

Also, AB = BC = CD = DA (Opposite sides of a parallelogram)

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.

Solution:

Given that,

ABCD is a rhombus.

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

, AC bisects âˆ C.

Similarly,

We can prove that diagonal AC bisects âˆ A.

Following the same method,

We can prove that the diagonal BD bisects âˆ B and 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.

Solution:

(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)

,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

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

Solution:

(i) In Î”APD and Î”CQB,

DP = BQ (Given)

âˆ ADP = âˆ CBQ (Alternate interior angles)

AD = BC (Opposite sides of a parallelogram)

Thus, Î”APD â‰… Î”CQB [SAS congruency]

(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 = CD (Opposite sides of a parallelogram)

Thus, Î”AQB â‰… Î”CPD [SAS congruency]

(iv) As Î”AQB â‰… Î”CPD

AQ = CP [CPCT]

(v) From the questions (ii) and (iv), it is clear that APCQ has equal opposite sides and also has equal and opposite angles. , APCQ is a parallelogram.

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

Solution:

(i) In Î”APB and Î”CQD,

âˆ ABP = âˆ CDQ (Alternate interior angles)

âˆ APB = âˆ CQD (= 90o as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

, Î”APB â‰… Î”CQD [AAS congruency]

(ii) As Î”APB â‰… Î”CQD.

, AP = CQ [CPCT]

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.

Solution:

(i) AB = DE and AB || DE (Given)

Two opposite sides of a quadrilateral are equal and parallel to each other.

Thus, quadrilateral ABED is a parallelogram

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii) 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) Since ACFD is a parallelogram

AC || DF and AC = DF

(vi) In Î”ABC and Î”DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

, Î”ABC â‰… Î”DEF [SSS congruency]

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

diagonal AC = diagonal BD

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

Solution:

To Construct: Draw a line through C parallel to DA intersecting AB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)

, BC = CE

â‡’âˆ CBE = âˆ CEB

also,

âˆ A+âˆ CBE = 180Â° (Angles on the same side of transversal and âˆ CBE = âˆ CEB)

âˆ B +âˆ CBE = 180Â° ( As Linear pair)

â‡’âˆ A = âˆ B

(ii) âˆ A+âˆ D = âˆ B+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

, Î”ABC â‰… Î”BAD [SAS congruency]

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

A figure formed by joining four points in an order is called a quadrilateral. This exercise questions are based on the different types of quadrilaterals, their properties and especially those of parallelograms. This exercise is based on the some very special properties of quadrilateral such as:

1. A diagonal of a parallelogram divides it into two congruent triangles.
2. In a parallelogram, opposite sides are equal, and opposite angles are equal.
3. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
4. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
5. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
6. A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

1. cherisha

thank you