NCERT Solutions for Class 9 Maths Chapter 8: Quadrilaterals - Exercise 8.2

EXERCISE NO: 8.1

1. The angles of quadrilateral are in the ration \[3:5:9:13\]. Find all the angles of the quadrilateral.

Ans: Given that the ratio of the angles of a quadrilateral are $3:5:9:13$.

Let us assume the common ratio between the angle of quadrilateral as $x$.

Now the angles of the quadrilateral are $3x$, $5x$, $9x$ and $13x$ respectively.

We know that the sum of the angle of the quadrilateral is $360{}^\circ $. Hence, we can write 

$3x+5x+9x+13x=360{}^\circ $

$\Rightarrow 30x=360{}^\circ $

$\Rightarrow x=\dfrac{360{}^\circ }{30}$

$\Rightarrow x=12{}^\circ $

Now the value of the angles by using the common ratio value are given by 

$3x=3\left( 12{}^\circ  \right)=36{}^\circ $

$5x=5\left( 12{}^\circ  \right)=60{}^\circ $

$9x=9\left( 12{}^\circ  \right)=108{}^\circ $

$13x=13\left( 12{}^\circ  \right)=156{}^\circ $

Hence the angles of the quadrilateral are $36{}^\circ $, $60{}^\circ $, $108{}^\circ $ and $156{}^\circ $.

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

Ans: Given that the diagonals of a parallelogram are equal. Let us assume a parallelogram $ABCD$ whose diagonals are equal as shown in below figure.

Here $ABCD$ is a parallelogram, so we can write that $AB=DC$ since opposite sides of a parallelogram are equal.

Consider the triangles $ABC$ and $DCB$. In these two triangles we have $AB=DC$, $BC=BC$ and as per given data $AC=DB$. So, we can write that $\Delta ABC\cong \Delta DCB$ by SSS Congruence rule.

$\Rightarrow \angle ABC=\angle DCB$

We know that the sum of the angles on the same side of the transversal is $180{}^\circ $.

$\therefore \angle ABC+\angle DCB=180{}^\circ $

Substitute the value $\angle ABC=\angle DCB$ in the above equation.

$\Rightarrow \angle ABC+\angle ABC=180{}^\circ $

$\Rightarrow 2\angle ABC=180{}^\circ $

$\Rightarrow \angle ABC=\dfrac{180{}^\circ }{2}$

$\Rightarrow \angle ABC=90{}^\circ $

Here we have the angle $ABC$ and $DCB$ as $90{}^\circ $. When the angles of the parallelogram are equal to $90{}^\circ $, then it is called as Rectangle.

Hence, if the diagonals of the parallelogram are equal then it is called as Rectangle.

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

Ans: Given that the diagonals of a quadrilateral bisect each other at right angles.

Let $ABCD$ will be the quadrilateral, whose diagonals are bisecting each other at right angle. The diagram of the quadrilateral $ABCD$ is given below.

In the above quadrilateral the diagonals bisect each other. So, we can write that $OA=OC$, $OB=OD$ and $\angle AOB=\angle BOC=\angle COD=\angle AOD=90{}^\circ $.

Consider $\Delta AOD$ and $\Delta COD$. In both the triangles we can write 

$OA=OC$, 

$\angle AOD=\angle COD$ since $ABCD$ is a quadrilateral.

$OD=OD$.

By using SAS congruence rule, we can write that $\Delta AOD\cong \Delta COD$. So, 

$AD=CD$

Similarly, we can prove that 

$AD=AB$ and $CD=BC$.

Finally, we can write that $AB=BC=CD=AD$.

Here we have opposite sides of the quadrilateral $ABCD$ are equal, so it will be a parallelogram. Again in $ABCD$ we have all sides are equal, so it can be a Rhombus also.

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

Ans: Consider a square $ABCD$ whose diagonals are $AC$ and $BD$. The intersecting point of the diagonals will be $O$ as shown in below figure.

According to the given statement we need to prove that $AC=BD$, $OA=OC$, $OB=OD$ and $\angle AOB=90{}^\circ $.

Consider $\Delta ABC$ and $\Delta DCB$.

In both the triangles we can write 

$AB=DC$ since the sides of a square are equal to each other, $\angle ABC=\angle DCB=90{}^\circ $ and 

$BC=CB$.

So, by using SAS congruency law we can say that $\Delta ABC\cong \Delta DCB$.

$\therefore AC=DB$ by CPCT.

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

Now consider $\Delta AOB$ and $\Delta COD$.

In both the triangles we can write 

$\angle AOB=\angle COD$ since vertically opposite angles are equal, 

$\angle ABO=\angle CDO$ since alternate interior angles are also equal and 

$AB=CD$ since the sides of a square are always equal.

So, by using AAS congruence rule we can say that $\Delta AOB\cong \Delta COD$.

$\therefore AO=CO$ and $OB=OD$ by CPCT.

Now consider $\Delta AOB$ and $\Delta COB$,

As we have proved earlier $AO=CO$,

$AB=CB$ since sides of a square are equal,

$BO=BO$.

By using SSS congruency rule we can write that $\Delta AOB\cong \Delta COB$.

$\therefore \angle AOB=\angle COB$ by CPCT.

But the angles $AOB$ and $COB$ are linear pair.

$\therefore \angle AOB+\angle COB=180{}^\circ $

$2\angle AOB=180{}^\circ $

$\angle AOB=90{}^\circ $

Hence, the diagonals of a square 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.

Ans: - Consider an equilateral $ABCD$ whose diagonals are $AC$ and $BD$. The intersecting point of the diagonals will be $O$ as shown in below figure.

According to the given statement we can write that $AC=BD$, $OA=OC$, $OB=OD$ and $\angle AOB=\angle BOC=\angle COD=\angle AOD=90{}^\circ $.

Consider $\Delta AOB$ and $\Delta COD$. In these two triangles we can write that 

$AO=CO$ since diagonals bisect each other,

$OB=OD$ since diagonals bisect each other,

$\angle AOB=\angle COD$ since vertically opposite angles.

By using SAS congruence rule, we can say that $\Delta AOB\cong \Delta COD$.

$\therefore AB=CD$ and $\angle OAB=\angle OCD$ by CPCT.

However, these are alternative interior angles for line $AB$ and $CD$ However alternative interior angles are equal to each other only when the two lines are parallel. 

$AB\parallel CD$

From all the above proofs we can say that $ABCD$ is a parallelogram.

Consider $\Delta AOD$ and $\Delta COD$, in these two triangles we can write that 

$AO=CO$ since diagonals bisect each other,

$\angle AOD=\angle COD=90{}^\circ $,

$OD=OD$.

By using SAS congruence rule, we can say that $\Delta AOD\cong \Delta COD$.

$\therefore AD=DC$

But $ABCD$ is a parallelogram so we can write that $AB=BC=CD=DA$.

That means all sides of the quadrilateral $ABCD$ are equal.

Consider $\Delta ADC$ and $\Delta BDC$, in these two triangles we can write that 

$AD=BC$, $AC=BD$ and $DC=CD$. 

By using SSS congruence rule, we can say that $\Delta ADC\cong \Delta BCD$.

$\therefore \angle ADC=\angle BCD$ by CPCT.

But we know that the angles $ADC$ and $BCD$ are co interior angles.

$\therefore \angle ADC+\angle BCD=180{}^\circ $

$\Rightarrow 2\angle ADC=180{}^\circ $

$\Rightarrow \angle ADC=90{}^\circ $

Here the interior angles of the quadrilateral are right angles.

So, for a $ABCD$ parallelogram the sides are equal to each other and all angles are right angles.

Hence $ABCD$ is a square.

6. Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ as shown in figure. 

i. Show that it bisects $\angle C$.

Ans: Given that $ABCD$ is a parallelogram.

$\therefore \angle DAC=\angle BCA$ and $\angle BAC=\angle DCA$ since alternative interior angles

 But in the problem, we have given that $AC$ bisect $\angle A$.

$\therefore \angle DAC=\angle BAC$

From all the above equations we can write that 

$\angle DAC=\angle BCA=\angle BAC=\angle DCA$

$\angle DCA=\angle BAC$

Hence, $AC$ bisects $\angle C$.

ii. Show that $ABCD$ is a rhombus.

Ans: From the equation $\angle DAC=\angle BCA=\angle BAC=\angle DCA$, we can write that $\angle DAC=\angle DCA$.

$DA=DC$ since side opposite to equal angles are equal.

But we have $DA=BC$ and $AB=CD$ since opposite sides of the parallelogram are equal.

$\therefore AB=BC=CD=DA$

7. A rhombus $ABCD$ is given. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Ans: Let the diagram of the given rhombus will be 

Consider $\Delta ABC$, in this triangle we can write that 

$BC=AB$ since sides of a rhombus are equal.

$\therefore \angle 1=\angle 2$ since angles opposite to equal sides of a triangle are equal.

But we have $\angle 1=\angle 3$ because they are alternate interior angles for parallel lines $AB$ and $CD$.

$\therefore \angle 2=\angle 3$

Hence the diagonal $AC$ bisects $\angle A$.

Also, $\angle 2=\angle 4$ since these two angles are also alternate interior angles for the parallel lines $BC$ and $DA$.

$\therefore \angle 1=\angle 4$

Hence the diagonal $AC$ bisects $\angle C$.

Similarly, we can prove the same for the diagonal $BD$.

8. $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. 

i. $ABCD$ is a square.

Ans: - Let the diagram of the given $ABCD$ rectangle is given by 

We have given that $ABCD$ is a rectangle. So, we can write that 

$\angle A=\angle C$

$\Rightarrow \dfrac{1}{2}\angle A=\dfrac{1}{2}\angle C$

But they have mentioned that $AC$ bisects $\angle A$ as well as $\angle C$.

$\Rightarrow \angle DAC=\dfrac{1}{2}\angle DCA$

$\therefore CD=DA$ since sides opposite to equal angles are also equal.

But $DA=BC$ and $AB=CD$ since opposite sides of a rectangle are equal.

$\therefore AB=BC=CD=DA$

Here $ABCD$ is given to be a rectangle but all the sides are equal.

Hence $ABCD$ is a square.

ii. Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Ans: Now consider $\Delta BCD$, in this triangle we can write that 

$BC=CD$ since sides of a square are equal,

$\angle CDB=\angle CBD$ since angles opposite to equal sides are equal.

But we have $\angle CDB=\angle ADB$ since alternate interior angles for parallel lines $AB$ and $CD$.

$\therefore \angle CBD=\angle ABD$

Hence $BD$ bisects $\angle B$.

Also, $\angle CBD=\angle ADB$ which are alternate interior angles for parallel lines $BC$ and $AD$.

$\therefore \angle CDB=\angle ADB$

Finally, $BD$ bisects $\angle D$ and $\angle B$.

9. In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ$ as shown in below figure. 

(Image Will Be Updated Soon)

i. Show that $\Delta APD\cong \Delta CQB$

Ans: Consider $\Delta APD$ and $\Delta CQB$, in these two triangles we can write 

$\angle ADP=\angle CBQ$ since alternate interior angles for the parallel lines $BC$ and $AD$,

$AD=CB$ since opposite sides of parallelogram are equal.

$DP=BQ$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta APD\cong \Delta CQB$

ii. Show that $AP=CQ$

Ans: As we have $\Delta APD\cong \Delta CQB$. By using CPCT we can write that $AP=CQ$.

iii. Show that $\Delta AQB\cong \Delta CPD$

Ans: Consider $\Delta AQB$ and $\Delta CPD$, in these two triangles we can write 

$\angle ABQ=\angle CDP$ since alternate interior angles for the parallel lines $AB$ and $CD$,

$AB=CD$ since opposite sides of parallelogram are equal.

$BQ=DP$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta AQB\cong \Delta CPD$

iv. Show that $AQ=CP$

Ans: As we have $\Delta AQB\cong \Delta CPD$. By using CPCT we can write that $AQ=CP$.

v. Show that $APCQ$ is a parallelogram.

Ans: From the results obtained in (ii) and (iv), which are $AQ=CP$ and $AP=CQ$. That means the opposite sides of the quadrilateral $APCQ$ are equal, so $APCQ$ is a parallelogram.

10. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ as shown in figure. 

(Image Will Be Updated Soon)

i. Show that  $\Delta APB\cong \Delta CQD$

Ans: Consider $\Delta APB$ and $\Delta CQD$, in these two triangles we can write that 

$\angle APB=\angle CQD=90{}^\circ $,

$AB=CD$ since opposite sides of parallelogram,

$\angle ABP=\angle CDQ$ since alternate interior angles for parallel lines $AB$ and $CD$.

By using AAS congruency rule we can say that $\Delta APB\cong \Delta CQD$.

ii. Show that $AP=CQ$

Ans: From the statement $\Delta APB\cong \Delta CQD$, by using CPCT we can write that $AP=CQ$.

11. In $\Delta ABC$ and $\Delta DEF$, $AB=DE$, $AB\parallel DE$, $BC=EF$ and $BC\parallel EF$. Vertices $A$, $B$ and $C$ are joined to vertices $D$, $E$ and $F$ respectively as shown in figure.

(Image Will Be Updated Soon)

i. Show that quadrilateral $ABED$ is a parallelogram.

Ans: Given that $AB=DE$ and $AB\parallel DE$.

If two opposite sides of a quadrilateral are equal and parallel to each other, then it is called as parallelogram.

Hence quadrilateral $ABED$ is a parallelogram.

ii. Show that quadrilateral $BEFC$ is a parallelogram.

Ans: We have also given $BC=EF$ and $BC\parallel EF$.

Hence quadrilateral $BEFC$ is a parallelogram.

iii. Show that $AD\parallel CF$ and $AD=CF$

Ans: We have the parallelograms $ABED$ and $BEFC$. So, we can write that $AD=BE$, \[AD\parallel BE\] and $BE=CF$, $BE\parallel CF$ since opposite sides of a parallelogram are equal and parallel.

$\therefore AD=CF$ and $AD\parallel CF$

iv. Show that quadrilateral $ACFD$ is a parallelogram

Ans: As we had observed that one pair of opposite sides ($AD$ and $CF$) of quadrilateral $ACFD$ are equal and parallel to each other, hence it is a parallelogram.

v. Show that $AC=DF$

As $ACFD$ is a parallelogram, so the pair of opposite sides will be equal and parallel to each other. 

$\therefore AC=DF$ and $AC\parallel DF$.

vi. Show that $\Delta ABC\cong \Delta DEF$

Ans: Consider $\Delta ABC$ and $\Delta DEF$, in these two triangles we have 

$AB=DE$, $BC=EF$ and $AC=DF$.

By using SSS congruence rule we can say that $\Delta ABC\cong \Delta DEF$.

12. $ABCD$ is a trapezium on which $AB\parallel CD$ and $AD=BC$ as shown in figure. (Hint: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ at point $E$).

(Image Will Be Updated Soon)

i. Show that $\angle A=\angle B$

Ans: We have $AD=CE$ opposite sides of parallelogram and $AD=BC$.

$\therefore BC=CE$ and $\angle CEB=\angle CBE$ since angle opposite to equal sides are also equal.

Consider the parallel lines $AD$ and $CE$. $AE$ is the transversal line for them.

So, 

$\angle A+\angle CEB=180{}^\circ $ since angles on the same side of transversal.

$\Rightarrow \angle A+\angle CBE=180{}^\circ $

Also, $\angle B+\angle CBE=180{}^\circ $ as they are linear pair of angles. So, from these two equals we can write that 

$\angle A=\angle B$

ii. Show that $\angle C=\angle D$

Ans: Given $AB\parallel CD$, so 

$\angle A+\angle D=180{}^\circ $ , $\angle C+\angle B=180{}^\circ $since angles on the same side of the transversal.

$\therefore \angle A+\angle D=\angle C+\angle B$

But we have $\angle A=\angle B$, so we will have 

$\angle C=\angle D$

iii. Show that $\Delta ABC\cong \Delta BAD$

Ans: Consider $\Delta ABC$ and $\Delta BAD$, in these two triangles we can write that 

$AB=BA$, $BC=AD$ and $\angle B=\angle A$.

By using SAS congruence rule, we can say that $\Delta ABC\cong \Delta BAD$.

iv. Show that diagonal $AC=\text{ diagonal }BD$

Ans: We have $\Delta ABC\cong \Delta BAD$, so by using CPCT we can write that $AC=BD$

NCERT Solutions for Class 9 Maths  | Chapterwise PDF

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solution Class 9 Maths of Chapter 8 All Exercises

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

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