**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 = 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.

**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.

**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.

**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.

**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**

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.

**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.

**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**

(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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