- Quadrilateral
- |
- Parallelogram
- |
- Trapezium
- |
- Rhombus
- |
- Rectangle
- |
- Square
- |
- Mid-point Theorem for Triangles
- |
- How to solve the angle Problem in Quadilateral
- |
- Additional Topics

In this page we have *NCERT book Solutions for Class 9th Maths:Quadilaterals* for
EXERCISE 1 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

We must revise these notes before proceding with the questionsAAA( Angle Angle Angle) is not the right condition to Prove congurence.

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

Let the common ratio between the angles be y. Therefore, the angles will be 3y, 5y, 9y, and 13y respectively.

As the sum of all interior angles of a quadrilateral is 360

3y + 5y + 9y + 13y = 360

30y= 360

y = 12

Hence, the angles are

3y=36

5y = 5 × 12 = 60

9y = 9 × 12 = 108

13y = 13 × 12 = 156

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

Let ABCD is a parallelogram

AC=BD AD=BC and CD=AB (in parallelogram opposite side are equal)

Show that it is a rectangle

In ? ADC and ?BCD

AD=BC

CD =CD

AC=BD

So by

ΔADC≅ΔBCD

So ∠ ADC=∠ BCD by CPCT (corresponding parts of the two congruent triangles)

But, we also have ∠ ADC+∠ BCD=180

So 2∠ ADC=180

=> ∠ ADC=90

So ∠ BCD=90

So all the angles A, B, C, D are 90

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

Let ABCD be the quadrilateral and AC and BD are diagonal which bisect at right angles

OA = OC, OB = OD and AOB = BOC = COD = AOD = 90º

Now, in ?AOD and ? COD

OA = OC (Diagonal bisects each other)

∠ AOD = ∠ COD (given)

OD = OD (common)

So by

ΔAOD≅ ΔCOD AD = CD (by CPCT) (i)

Similarly we can prove that

AD = AB and CD = BC (ii)

From equations (i) and (ii), we can say that

AB = BC = CD = AD

Since all sides are equal, it is a rhombus

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

Let ABCD is a square whose diagonal BD and AC intersect art O

AB=BC=CD=AD

OA=OC

OD=OB

AC=BD

In Δ ABC and Δ DCB,

AB = DC ( From Sides of a square are equal to each other)

∠ ABC =∠ DCB (All interior angles are of 90)

BC = CB (Common side)

So by

ΔABC≅ΔDCB

AC = DB (By CPCT)

Hence, the diagonals of a square are equal in length.

In Δ AOB and Δ COD,

∠ AOB = ∠ COD (Vertically opposite angles)

∠ ABO = ∠ CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal)

;ΔAOB≅ΔCOD

∴ AO = CO and OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other.

In Δ AOB and Δ COB,

As we had proved that diagonals bisect each other, therefore,

AO = CO

AB = CB ( we know that Sides of a square are equal)

BO = BO (Common)

ΔAOB≅ΔCOB

∴ ∠ AOB = ∠ COB (By CPCT)

However, ∠ AOB + ∠ COB = 180

2∠ AOB = 180

∠ AOB = 90

Hence, the diagonals of a square bisect each other at right angles.

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

Let ABCD is a quadrilateral whose diagonal BD and AC bisect each other at right angle at O

AO=OC, BO=OD, AC=BD

∠ AOB=∠ COD

In Δ AOB and Δ COD

AO=CO (Given)

∠ AOB=∠ COD (Each given equal to 90

BO=DO (Given)

So by

ΔAOB≅ ΔCOD.

? ∠ OBA=∠ ODC (by CPCT)

But, these are alternate interior angles which means that AB? CD. (i)

Similarly, we can prove that BC? AD (ii)

From (1) and (2), we can say that quadrilateral ABCD is a parallelogram. Hence, we have AB=CD and BC=AD because opposite sides of a parallelogram are equal. (iii)

Now, in Δ AOB and Δ AOD

AO=AO (Common)

∠ AOB=∠ AOD (Each given equal to 90

OB=OD (Given)

So by

ΔAOB≅ΔAOD

? AB=AD (by CPCT) (iv)

In Δ ACD and Δ BDC

AC=BD (Given)

AD=BC (Proved above in (1) )

CD=DC (Common)

So by SSS congruence rule

ΔACD≅ΔBDC

∠ ADC=∠ BCD (by CPCT) (v)

But, we also have ∠ ADC+∠ BCD=180

From (5) and (6), we can say that

∠ ADC+∠ ADC=180

2∠ ADC=180

∠ ADC=180/2=90

From (iii), (iv) -> we can say that ABCD is a parallelogram having all the sides equal.

(vii) -> we have showed that it's one angle is equal to 90

Diagonal AC of a parallelogram ABCD bisects ∠ A (see the given figure). Show that

(i) It bisects ∠ C also,

(ii) ABCD is a rhombus.

i)

ABCD is a parallelogram and AC bisect angle A

∠DAC = ∠ BAC

**To Prove:** AC bisects ∠C

**Proof:**

As ABCD is a parallelogram

∠ DAC = ∠ BCA (Alternate interior angles) ... (i)

And ∠ BAC = ∠ DCA (Alternate interior angles) ... (ii)

But it is given that AC bisects ∠ A.

∠ DAC = ∠ BAC ... (iii)

From equations (i), (ii) and (iii), we have

∠ DAC = ∠ BCA =∠ BAC = ∠ DCA ... (iv)

=> ∠ DCA = ∠ BCA

Hence, AC bisects ∠C.

(ii)

**To Prove**: ABCD is a rhombus

**Proof :**

From equation (iv), we have

∠ DAC = ∠ DCA
DA = DC (we know that side opposite to equal angles are equal)

But DA = BC and AB = CD (opposite sides of parallelogram are equal)

AB = BC = CD = DA

Hence, ABCD is rhombus

**Question 7:**

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

**Solution:**

**Given:** ABCD is a rhombus

**To Prove**: Diagonal BD bisect angle B and D

Diagonal AC bisect angle A and C

**Proof:**

Let us join AC.

In Δ ABC,

BC = AB (Sides of a rhombus are equal to each other)

∴ ∠ BAC = ∠ BCA (Angles opposite to equal sides of a triangle are equal)

However, ∠ BAC = ∠ DCA (Alternate interior angles for parallel lines AB and CD)

∠ BCA= ∠ DCA

Therefore, AC bisects ∠ C.

Also, ∠ BCA= ∠ DAC (Alternate interior angles for || lines BC and DA)

∠ BAC = ∠ DAC

Therefore, AC bisects ∠ A.

Similarly, it can be proved that BD bisects ∠ B and ∠ D as well.

**Question 8:**

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

**Solution**

**Given**

ABCD is a rectangle and ∠DAC=∠BAC and ∠DCA=∠BCA

**To Prove**: ABCD is a square

**Proof:**

In Δ ADC and Δ ABC

∠ CAD=∠ CAB (Given)

AC=AC (Common)

∠ DCA=∠ BCA (Given)

Therefore, by ASA congruence rule

ΔADC≅ ΔABC

AD=AB (by CPCT) (2)

From (1) and (2), we can say that ABCD is a rectangle having all the sides equal. It means that ABCD is a square.

**To prove **diagonal BD bisects ∠ B as well as ∠ D.

In solution (i), we have showed that ABCD is a square.

Now in Δ CBD and Δ ABD

BC=BA (Sides of square are equal)

BD=BD (Common)

CD=AD (Sides of square are equal)

So by **SSS congruence** rule

ΔCBD≅ΔABD

∠ CBD=∠ ABD (by CPCT) (3)

And, ∠ CDB=∠ ADB (by CPCT) (4)

From (3) and (4), we can say that BD bisects ∠ B as well as ∠ D.

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