**NCERT Solutions Class 9 Maths Exercise 8.1-Quadrilateral**

**NCERT Solutions Class 9 Maths Exercise 8.1 of chapter 8-Quadrilateral** are based on the questions on different type of **quadrilaterals** like **square,rectangle, rhombus and parallelogram**. These **NCERT solitions** are the **solutions** of unsolved problems of **class 9 maths NCERT text** book **exercise 8.1 of chapter 8-Quadrilaterals**.All questions are solved by the subject expert.You can study here** maths** and science **NCERT solutions,** sample papers ,**solutions** of previous years question papers,important science and **maths** notes, articles on different type of competitice entrance exams and carrier in online jobs.

**NCERT Solutions Class 9 Maths Exercise 8.1-Quadrilateral**

**Q1.The angles of quadrilateral are in the ratio 3 : 5 : 9: 13. Find all the angles of the quadrilateral.**

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

According to angle sum property of quadrilateral

3x + 5x + 9x + 13x = 360ยฐ

30x = 360ยฐ

x = 12

Therefore the required angles of quadrilateral are

3x = 3 ร 12 = 36ยฐ

5x = 5ร 12 = 60ยฐ

9x = 9 ร 12 = 108ยฐ

13x = 13 ร 12 = 156ยฐ

**Q2.If diagonals of parallelogram are equal, then show that it is a rectangle.**

Ans.

**GIVEN:** ABCD is a perallogram

Both diagonals are equal i.e BD =AC

**TO PROVE:** ABCD is a rectangle i.e โ A= โ B = โ C = โ D = 90ยฐ

**PROOF:ย **In ฮADC and ฮBDC

AD = BC (opposite sides of parallogram)

DC = DC (common)

BD = AC (given)

ฮADC โ BCD (SSS rule congruency of the triangles)

โ D = โ C (by CPCT)

โ D + โ Cย = 180ยฐ (Adjacent angles of parallogram)

โ D + โ D = 180ยฐ

2 โ D = 180ยฐ

โ D = 90ยฐ

Sinceย we have proved โ D =โ C

โดโ C = 90ยฐ

โ A = โ Cย (opposite angles of paralellogram)

โดโ A = 90ยฐ

โ B = โ Dย (opposite angles of pallogram)

โดB = 90ยฐ

โ A = โ B=โ C=โ D = 90ยฐ

Therefore ABCD is a rectangle, Hence proved

**Q3. Show that diagonal of a quadrilateralย bisect each other at right angles, then it is a rhombus.**

Ans.

**GIVEN:** Diagonal of quadrilateral bisect each other

โดAO =ย CO and DO = BO

Diagonal of quadrilateral bisect each other at right angle

โ DOC = โ AOD = โ BOC = โ AOB = 90ยฐ

**TO PROVE:**ABCD is a Rhombus

AB = BC = DC = AD

**PROOF:ย **In ฮAOD and ฮCOD

โ DOC = โ AOD (given)

AO =ย CO (given)

DO = DO (common)

ฮAOD โ ฮCOD (SSS rule congruency of ฮ’s)

DC = AD (by CPCT)

Similarly we can prove

ฮAOD โ ฮAOB,ย ฮAOBย โ ฮBOC, ฮBOC โ ฮDOC

So,AD = AB, AB = BC,BC =DC (by the rule of CPCT)

Therefore ABCD is a rhombus

Hence proved

**Q4. Show that diagonals of a square are equal and bisect each other at right angle.**

Ans.

**GIVEN: **ABCD is a square

**TO PROVE:ย **AC = BD

AC and BD bisect each other at right angle i.e โ DOC = โ DOA = โ AOB = โ BOC = 90ยฐ

**PROOF:ย **In ฮABDย and ฮABC

AB = AB (common)

AD = BC (sides of square)

โ A = โ B = 90ยฐ (angles of square)

ฮABD โ ฮABC (SSS rule of congruency)

**AC = BD (by CPCT)**

Hence diaogonals of square are equal

In ฮAOD and ฮBOC

โ AOD = โ BOC (vertically opposite angles)

โ DAO = โ BCO (alternate angles)

AD = BC (sides of square)

ฮAOD โ ฮCOB ( AAS rule)

**AO = CO ( by CPCT)**

**DO = BO (by CPCT)**

Hence diagonals of square bisect each other

Now, considering the ฮDOC and ฮDOA

AD = DC (sides of square)

DO = DO (common)

AO = CO ( proved above )

ฮDOC โ ฮDOA (SSS rule)

โ DOC =โ DOA (by CPCT)

โ DOC +โ DOA = 180ยฐ

โ DOC +โ DOC= 180ยฐ

2โ DOCย = 180ยฐ

โ DOCย = 90ยฐ

โ DOAย = 90ยฐ

โ BOC = โ DOAย = 90ยฐ (vertically opposite angle)

โ DOCย = โ AOB = 90ยฐ(vertically opposite angle)

**โ DOC = โ DOA = โ AOB = โ BOC = 90ยฐ**

Hence diagonal of square bisect each other at 90ยฐ

**Q5. Show that diagonal of a quadrilateral are equal and bisect each other at right angle,then it is a square.**

Ans.

**GIVEN:**AC = BD

โ DOC = โ DOA = โ AOB = โ BOC = 90ยฐ

AO = CO

DO = BO

**TO PROVE:**ABCD is a sqare

**PROOF:**In ฮAOD and ฮCOD

DO = DO (common)

AO = CO (given)

โ DOA =ย โ COD = 90ยฐ(given)

ฮAOD โ ฮCOD (SAS rule)

AD = DC (by CPCT )

Similarly, ฮAOD โ ฮAOB

AD = AB (by CPCT)

and ฮAOBโ ฮCOB

AB = BC

โดAB = BC= AD=DC

Therefore all sides of quadrilateral are equal

Now, we have to prove all angles of quadrilateral are of 90ยฐ

In ฮABC and ฮABD

AD =BC (proved above)

AB = AB (common)

BD = AC (given)

ฮABCโ ฮBAC

โ A = โ B (by CPCT)

Since we have proved all sides are equal so ABCD is a parallogram and โ A , โ B are adjacent angles.

โดโ A + โ B = 180ยฐ

Putting โ A = โ B

โ B + โ B = 180ยฐ

2โ B= 180ยฐโ โ B = 90ยฐand โ Aย = 90ยฐ

โ A = โ C = 90ยฐ, โ B = โ D =90ยฐ (opposite angles of parallogram)

Each angle of the given quadrilateral are of 90ยฐ

In given quadrilateral all sides are equal and each of four angles are of 90ยฐ

Hence ABCD is square.

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**NCERT Solutions Class 9 Maths Exercise 8.1-Quadrilateral**

**Q6. Diagonal AC of a parallelogram ABCD bisects โ A (see the given figure). Show thatย **

**(i) It bisects โ C also**

**(ii)ABCD is a rhombus**

Ans.

**GIVEN: **ABCD is a parallelogram

in which diagonal AC bisects โ A

**TO PROVE:**Diagonal AC bisects โ C

ABCD is a rhombus i.e AB = BC = CD = AD

**PROOF :**In ฮADC and ฮABC

โ DAC = โ CAB…..(i) (given)

โ DAC = โ ACB…..(ii) (alternate angle )

โ ACD = โ CAB…..(iii)(alternate angle )

From equation (i) and equation (ii)

โ DAC = โ ACB….(iv)

From (i) and (iii)

โ DAC = โ ACD ….(v)

From equation (iv) and (v)

โ ACD = โ ACB

**Therefore AC bisects โ C also**

Now, we have to prove ABCD is rhombus

โ A = โ C (opposite angles of parallogram)

โ A /2= โ C/2

โ DAC = โ ACD (AC bisects โ A is given AC bisects โ C, proved above)

DC = AD (opposite sides of equal engle in a triangle)

DC = AB (opposite sides of parallelogram)

AD =BC (opposite sides of parallelogram)

โดDC = AD =AB = BC

Hence ABCD is a rhombus

**Q7. ABCD is a rhombus . Show that diagonal AC bisects โ A as well as โ C and diagonal BD bisects โ B as well as โ D.**

Ans.

**GIVEN:** ABCD is a rhombus

**TO PROVE:** โ ODA= โ CDO, โ OBC = โ OBA

โ DCO = โ BCO, โ OAB = โ DAO

**PROOF:** In ฮADB

AD = AB (sides of rhombus)

โ ADO = โ OBA ….(i)(opposite angles of equal sides)

โ ADO = โ OBC …..(ii)(alternative angle)

From equation (i) and equation (ii)

โ OBA =โ OBC …..(iii)

In ฮBDC

DC = BC (sides of rhombus)

โ OBC=โ CDO…..(iv)(opposite angles of equal sides)

โ OBC = โ ODA….(v)(alternative angle)

From equation (iv) and equation (v)

โ CDO = โ ODA…..(vi)

Equation (iii) and equation (vi) shows that BD bisects โ B as well as โ D

Similarly we can prove that AC bisects โ A as well as โ C

**Q8.ABCD is a rectangle in which diagonal AC bisects โ A as well as โ C. Show that**

**(i) ABCD is a square**

**(ii) Diagonal BD bisects โ B as well as โ D**

Ans.

**GIVEN:** ABCD is a rectangle

โ DAC = โ BAC

โ DCA = โ BCA

**TO PROVE :**

**(i) ABCD is a square**

**(ii) Diagonal BD bisects โ B as well as โ D**

โ ADB=โ BDC, โ ADB = โ DBC

**PROOF:ย **In ฮBCA

โ DAC = โ BAC (given)

โ DCA = โ BCA (given)

โ A = โ Cย = 90ยฐ (angles of rectangle)

โ DAC = โ BAC = โ A/2 = 90ยฐ/2 = 45ยฐ

โ DCA = โ BCA = โ C/2 = 90ยฐ/2 = 45ยฐ

โ BAC = โ BCA

AB = BC

โ DAC = โ DCA

AD = DC (opposite sides of equal angles in a triangle)

Therefore AB = DC, BC = AD (opposite sides of rectangle)

โย AB = BC = DC = AD

Hence ABCD is a square

We have to prove (ii) Diagonal BD bisects โ B as well as โ D

**TO PROVE:** Diagonal BD bisects โ B as well as โ D

**PROOF:** In ฮAOD and ฮCOD

BC = AB(proved above)

DO = DO (common)

AO = CO (diagonal of rectangle bisect each other)

ฮAOD โ COD (SSS rule)

โ ADO =โ CDO (by CPCT)…..(i)

In ฮAOB and ฮCOB

AD = DC (proved above)

BO = BO (common)

AO = CO (diagonal of rectangle bisect each other)

ฮAOBโ COB (SSS rule)

โ OBA=โ OBC (by CPCT)……(ii)

Equation (i) and equation (ii) shows that BD bisects โ B as well as โ D, Hence proved.

**NCERT Solutions Class 9 Maths Exercise 8.1-Quadrilateral**

**Q9. In parallelogram ABCD, two points P ad Q are taken on diagonal BD such that DP = BQ(see the given figure). Show that**

(i) ฮAPD โ ฮCQB

(ii) AP = CQ

(iii) ฮAQB โ ฮCPD

(iv) AQ = CP

Ans.

**GIVEN: **ABCD is a parallelogram

P and Q are the points on diagonal BD

DP = BQ

**TO PROVE:**

(i) ฮAPD โ ฮCQB

(ii) AP = CQ

(iii) ฮAQB โ ฮCPD

(iv) AQ = CP

**PROOF: (i)ย **In ฮAPD and ฮCQB

AD = BC (opposite sides of parallelogram)

DP = BQ (given)

โ ADP = โ CBQ (alterative angle)

ฮAPD โ ฮCQB (SAS rule)

**PROOF:** (ii) AP = CQ

Since,we have already proved above in (i)

ฮAPD โ ฮCQB

Therefore AP = CQ (by CPCT)

**PROOF:** (iii) ฮAQB โ
ฮCPD

In ฮAQB and ฮCPD

AB = CD (opposite sides of parallelogram)

DP = BQ (given)

โ ABQ = โ CDP (alterative angles)

ฮAQB โ ฮCPD( SAS rule)

**PROOF:**(iv) AQ = CP

ฮAQB โ ฮCPD(proved above)

AQ = CP (by CPCT)

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