NCERT Solutions Class 9 Maths Exercise 8.1 of chapter 8-Quadrilateral - Future Study Point

NCERT Solutions Class 9 Maths Exercise 8.1 of chapter 8-Quadrilateral

Exercise 8.1 class 9 maths

NCERT Solutions Class 9 Maths Exercise 8.1-Quadrilateral

Exercise 8.1 class 9 maths

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.

Exercise 8.2-Quadrilateral

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 = 360°

30x = 360°

x = 12

Therefore the required angles of quadrilateral are

3x = 3 × 12 = 36°

5x = 5× 12 = 60°

9x = 9 × 12 = 108°

13x = 13 × 12 = 156°

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

Ans.

Quadrilateral Q2 class 9 ex.1

GIVEN: ABCD is a perallogram

Both diagonals are equal i.e BD =AC

TO PROVE: ABCD is a rectangle i.e ∠A= ∠B = ∠C = ∠D = 90°

PROOF: In ΔADC and ΔBDC

AD = BC (opposite sides of parallogram)

DC = DC (common)

BD = AC (given)

ΔADC ≅ BCD (SSS rule congruency of the triangles)

∠D = ∠C (by CPCT)

∠D + ∠C  = 180° (Adjacent angles of parallogram)

∠D + ∠D = 180°

2 ∠D = 180°

∠D = 90°

Since  we have proved ∠ D =∠C

∴∠C = 90°

∠A = ∠C  (opposite angles of paralellogram)

∴∠A = 90°

∠B = ∠D  (opposite angles of pallogram)

∴B = 90°

∠A = ∠B=∠C=∠D = 90°

Therefore ABCD is a rectangle, Hence proved

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

Ans.

EX.8.1 Q3 CLASS 9 MATHS

 

GIVEN: Diagonal of quadrilateral bisect each other

∴AO =  CO and DO = BO

Diagonal of quadrilateral bisect each other at right angle

∠DOC = ∠AOD = ∠BOC = ∠AOB = 90°

TO PROVE:ABCD is a Rhombus

AB = BC = DC = AD

PROOF: In ΔAOD and ΔCOD

∠DOC = ∠AOD (given)

AO =  CO (given)

DO = DO (common)

ΔAOD ≅ΔCOD (SSS rule congruency of Δ’s)

DC = AD (by CPCT)

Similarly we can prove

ΔAOD ≅ ΔAOB,  ΔAOB  ≅ ΔBOC, ΔBOC ≅ΔDOC

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.

Q4. Ex 8.1 class 9 maths

 

 

GIVEN: ABCD is a square

TO PROVE: AC = BD

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

PROOF: In ΔABD  and ΔABC

AB = AB (common)

AD = BC (sides of square)

∠A = ∠B = 90° (angles of square)

ΔABD ≅ ΔABC (SSS rule of congruency)

AC = BD (by CPCT)

Hence diaogonals of square are equal

In ΔAOD and ΔBOC

∠AOD = ∠BOC (vertically opposite angles)

∠DAO = ∠BCO (alternate angles)

AD = BC (sides of square)

ΔAOD ≅ ΔCOB ( AAS rule)

AO = CO ( by CPCT)

DO = BO (by CPCT)

Hence diagonals of square bisect each other

Now, considering the ΔDOC and ΔDOA

AD = DC (sides of square)

DO = DO (common)

AO = CO ( proved above )

ΔDOC ≅ΔDOA (SSS rule)

∠DOC =∠DOA (by CPCT)

∠DOC +∠DOA = 180°

∠DOC +∠DOC= 180°

2∠DOC  = 180°

∠DOC  = 90°

∠DOA  = 90°

∠BOC = ∠DOA  = 90° (vertically opposite angle)

∠DOC  = ∠AOB = 90°(vertically opposite angle)

∠DOC = ∠DOA = ∠AOB = ∠BOC = 90°

Hence diagonal of square bisect each other at 90°

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

Ans.

Q5. exercise 8.1 class 9 maths

 

GIVEN:AC = BD

∠DOC = ∠DOA = ∠AOB = ∠BOC = 90°

AO = CO

DO = BO

TO PROVE:ABCD is a sqare

PROOF:In ΔAOD and ΔCOD

DO = DO (common)

AO = CO (given)

∠DOA =  ∠COD = 90°(given)

ΔAOD ≅ΔCOD (SAS rule)

AD = DC (by CPCT )

Similarly, ΔAOD ≅ΔAOB

AD = AB (by CPCT)

and ΔAOB≅ΔCOB

AB = BC

∴AB = BC= AD=DC

Therefore all sides of quadrilateral are equal

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

In ΔABC and ΔABD

AD =BC (proved above)

AB = AB (common)

BD = AC (given)

ΔABC≅ΔBAC

∠A = ∠B (by CPCT)

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

∴∠A + ∠B = 180°

Putting ∠A = ∠B

∠B + ∠B = 180°

2∠B= 180°⇒ ∠B = 90°and ∠A  = 90°

∠A = ∠C = 90°, ∠B = ∠D =90° (opposite angles of parallogram)

Each angle of the given quadrilateral are of 90°

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

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 ∠A (see the given figure). Show that 

(i) It bisects ∠C also

(ii)ABCD is a rhombus

Q6. Ex. 8.1 class 9 maths

 

Ans.

GIVEN: ABCD is a parallelogram

in which diagonal AC bisects ∠A

TO PROVE:Diagonal AC bisects ∠C

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

PROOF :In ΔADC and ΔABC

∠DAC = ∠CAB…..(i) (given)

∠DAC = ∠ACB…..(ii) (alternate angle )

∠ACD = ∠CAB…..(iii)(alternate angle )

From equation (i) and equation (ii)

∠DAC = ∠ACB….(iv)

From (i) and (iii)

∠DAC = ∠ACD ….(v)

From equation (iv) and (v)

∠ACD = ∠ACB

Therefore AC bisects ∠C also

Now, we have to prove ABCD is rhombus

∠A = ∠C (opposite angles of parallogram)

∠A /2= ∠C/2

∠DAC = ∠ACD (AC bisects ∠A is given AC bisects ∠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)

∴DC = AD =AB = BC

Hence ABCD is a rhombus

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

Ans.

EX. 8.1 Q7 class 9 maths

 

GIVEN: ABCD is a rhombus

TO PROVE: ∠ODA= ∠CDO, ∠OBC = ∠OBA

∠DCO = ∠BCO, ∠OAB = ∠DAO

PROOF: In ΔADB

AD = AB (sides of rhombus)

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

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

From equation (i) and equation (ii)

∠OBA =∠OBC …..(iii)

In ΔBDC

DC = BC (sides of rhombus)

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

∠OBC = ∠ODA….(v)(alternative angle)

From equation (iv) and equation (v)

∠CDO = ∠ODA…..(vi)

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

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

Q8.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

Ans.

Q8 EX 8.1 Class 9 maths

 

GIVEN: ABCD is a rectangle

∠DAC = ∠BAC

∠DCA = ∠BCA

TO PROVE :

(i) ABCD is a square

(ii) Diagonal BD bisects ∠B as well as ∠D

∠ADB=∠BDC, ∠ADB = ∠DBC

PROOF: In ΔBCA

∠DAC = ∠BAC (given)

∠DCA = ∠BCA (given)

∠A = ∠C  = 90° (angles of rectangle)

∠DAC = ∠BAC = ∠A/2 = 90°/2 = 45°

∠DCA = ∠BCA = ∠C/2 = 90°/2 = 45°

∠BAC = ∠BCA

AB = BC

∠DAC = ∠DCA

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

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

⇒ AB = BC = DC = AD

Hence ABCD is a square

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

TO PROVE: Diagonal BD bisects ∠B as well as ∠D

PROOF: In ΔAOD and ΔCOD

BC = AB(proved above)

DO = DO (common)

AO = CO (diagonal of rectangle bisect each other)

ΔAOD ≅ COD (SSS rule)

∠ADO =∠CDO (by CPCT)…..(i)

In ΔAOB and ΔCOB

AD = DC (proved above)

BO = BO (common)

AO = CO (diagonal of rectangle bisect each other)

ΔAOB≅ COB (SSS rule)

∠OBA=∠OBC (by CPCT)……(ii)

Equation (i) and equation (ii) shows that BD bisects ∠B as well as ∠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

Q9. EX 8.1 CLASS 9 MATHS

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

Ans.

GIVEN: ABCD is a parallelogram

P and Q are the points on diagonal BD

DP = BQ

TO PROVE:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

PROOF: (i) In ΔAPD and ΔCQB

AD = BC (opposite sides of parallelogram)

DP = BQ (given)

∠ADP = ∠CBQ (alterative angle)

ΔAPD ≅ ΔCQB (SAS rule)

PROOF: (ii) AP = CQ

Since,we have already proved above in (i)

ΔAPD ≅ ΔCQB

Therefore AP = CQ (by CPCT)

PROOF: (iii) ΔAQB ≅ ΔCPD

In ΔAQB and ΔCPD

AB = CD (opposite sides of parallelogram)

DP = BQ (given)

∠ABQ = ∠CDP (alterative angles)

ΔAQB ≅ ΔCPD( SAS rule)

PROOF:(iv) AQ = CP

ΔAQB ≅ ΔCPD(proved above)

AQ = CP (by CPCT)

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