SChand CLASS 9 Chapter 11 Rectilinear Figures Exercise 11(B)

 

 Exercise 11 B

Question 1

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(i) ABCD is a parallelogram in which diagonal AC = Diagonal 

BD and AC and BD intersect each other at O at right angle 

To prove: ABCD is square 

Proof: In $\triangle A B C$ and $\triangle A D B$ 

AB= AB 

AC = BD 

BC = AD 

So $\triangle A B C \cong \triangle A D C$

So $\angle A B C=\angle B A D$

But $\angle A B C+\angle B A D=180^{\circ}$

So $\angle A B C=\angle B A D=90^{\circ}$

Again in $\triangle A O B$ and $\triangle B O C$

$\begin{aligned}&\angle A O B=\angle B O C \\&B O=B O \\&A O=O C\end{aligned}$

So $\triangle A O B \cong \triangle B O C$

So $A B=B C$

But $A B=C D$ and $B C=A D$

So $A B=B C=C D=D A$

And each angle is  $90^{\circ} .$

Hence ABCD is a square 

Hence proved

(ii) (IMAGE TO BE ADDED)

ABCD is a rectangle and AC and BD arc its diagonals 

To prove: AC=BD

Proof: In$\triangle A B C$ and $\triangle A D B, A B=A B$

$B C=A D$

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

So $\triangle A B C \cong \triangle A D B$

So $A C=B D$ Hence proved

Question 2

Sol: In parallelogram ABCD , AC is its diagonal X and Y are the mid points of AB and CD respectively 

M is mid point of AC and  AX = CY

To prove: 

(i) $\triangle A X M \cong \triangle  C Y M$

(ii) XMY is a straight line 

Construction : Join XM and YM 

Proof : In $\triangle$ A M X and $\triangle C YM$

$A X=C Y$

$A M=C M$

And $\angle 4 \mathrm{~cm}=\angle M A X$

So 

$\begin{aligned} \triangle A M X & \cong \triangle  CMY \\ \text { So } A M X &=\angle CMY \end{aligned}$

But these are vertically opposite angles 

So XMY is a straight line 

Hence proved '

Question 3

Sol:(i) In a square ABCD , BD is its diagonal 

To Prove: $\angle A B D=45^{\circ}$

Proof : In $\triangle A B D$ and $\triangle B C D$ 

AB = DC 

AD= BC 

BD =BD 

So $\triangle A B D \cong \triangle B C D$

So $\angle A B D=\angle C B D$

But $\angle A B D+\angle C B D=90^{\circ}$

$\angle A B D=\frac{90^{\circ}}{2}=45^{\circ}$

Hence diagonal BD makes an angle of $45^{\circ}$ with the side of the square 

(ii) (IMAGE TO BE ADDED)

In rhombus ABCD its diagonals AC and BD bisects each other at O 

To Prove: $\angle A O B=\angle B O C=\angle C O D$

$\angle D O A=90^{\circ}$

Proof: In $\triangle A O B$ and $\triangle C O B$

$A O=O C$

$B O=B O$

$A B=B C$

So $\triangle A O B \cong \triangle C O B$

So $\angle A O B=\angle C O B$

But $\angle A O B+\angle C O B=180^{\circ}$

$\angle A O B=\angle C O B=90^{\circ}$

Similarly we can prove that 

$\angle C O D=\angle D O A=90^{\circ}$

Hence diagonals of a rhombus bisect each other at Right angles 

(iii) In rhombus ABCD, AC is its Diagonals 

To prove:  AC bisect $\angle A$ and $\angle C$

Proof: In $\angle A B C$ and $\triangle A D C$ 

$A B=C D$

$B C=A D$

$A C=A C$

So $\triangle A B C \cong \triangle A D C$

So $\angle C A B=\angle C A D$

and $\angle A C B=\angle A C D$

Hence diagonal of a rhombus bisects the angles at the vertices

Question 4

PQRS is a parallelogram in which PR is its diagonal ST 

$\perp P R$ and $Q U \perp P R$

To prove : 

(i) $\Delta S T R \cong \Delta Q U P \quad$ (ii) $S T=Q U$

Proof : In $\triangle$ STR and $\triangle Q U P$

$P S=S R$

$\angle T=\angle U$

$\angle S P T=\angle U R Q$

(i) So $\triangle$ STR $\cong \triangle Q U P$

(ii) so $S T=Q U$

Question 5

Sol: In ||gm ABCD , PO and QO are the angle bisectors of $\triangle P$ and $\triangle Q$ respectively meeting Each other at O .  $\triangle QM$ is drawn P parallel to PQ through O 

To prove: 

(i) $P L=Q P \quad$ (iii L O=O M$

 Proof : If PO is the bisects of $\angle P$ 

So $\angle 1=\angle 2$

Similarly QO is the bisector of  $\angle Q$

so $\angle 5=\angle 6$

if $\angle M \| P Q$

 $\angle PQM is a parallelogram 

And $\angle 3=\angle 2=\angle 1$

And $\angle 4=\angle 5=\angle 6$ 

(i) LP = QM

(ii) PL = LO 

and OM = OM 

But PL = OM 

OL =OM 

Hence proved

Question 6

(i) Sol: In ||gm ABCD, BD is its diagonal L and M are points on AB and CD respectively such that AL =CM

To prove: LM and BD bisect each other 

i.e  BO =OD and LO = OM

Proof : In || gm ABCD, 

AB=CD

But AL = CM 

So AB-AL = CD - CM

=LM =MD 

Now in $\triangle L O B$ and $\triangle D O M$

$\angle B=M D$

$\angle L B O=\angle M D O$

$\angle L O B=\angle D O M$

So $\triangle L O B \cong \triangle D O M$

SO $O B=O D$ and $O L=O M$

Hence BD and LM bisect each other 

(ii)  In a parallelogram ABCD, AB is produced to E such that BE = AB 

To prove: ED bisects BC 

Proof: IF AB= DC 

and AB= BE 

DC = BE 

Now in $\triangle B O E$ and $\triangle C O D$

BE=DC

$\begin{aligned} \angle B O E &=\angle C O D \\ \angle \angle B E O &=\angle C D O \\ \text { SO } \triangle B O E & \cong \triangle C O D \\ \text { SO } B O &=O C \end{aligned}$

Hence DE bisects BC

Question 7

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(i) In parallelogram ABCD diagonals AC and BD intersect each other at O. 

A line XOY is drawn which meets AB in X and CD in Y 

To prove: OX =OY

Proof: In $\triangle X O B$ and $\triangle Y O D$

$O B=O D$

$\angle B O X=\angle D O Y$

$\angle O B X=\angle O D Y$

So $\triangle X O B \cong \triangle Y O D$

So $O X=O Y$

 Hence proved

 

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(ii) In  $\triangle A B C, A D$ is the median which is produced to X such that DX =AB BX and XC are joined 

To Prove: ABXC is a parallelogram 

Proof: In $\triangle A D B$ and $\triangle X D C$ 

$A D=D X$

$B D=D C$

$\angle A D B=\angle X D C$

So $\triangle A D B \cong \triangle \X D C$

SO $A B=C X$

$\angle B A D=\angle C X D$

But these are alternate angles 

So AB||XC

If AB =XC and AB||XC 

So ABXC is a parallelogram (HENCE PROVED)

Question 8

Sol: (IMAGE TO BE ADDED)

(i) In parallelogram ABCD, AE is the bisector of $\angle A$ which meets CD in E. AB = 2AD 

ED is joined 

To prove: 

(i) BE bisects LE

(ii)$\angle A E B$ is a right angle 

Construction: Take F , a mid point of AB 

So AF =AD 

If $\angle 1=\angle 2$

So AFED is a rhombus 

$A D\|E F\| C D$ and E is mid point of CD 

So FBCE is also a rhombus 

Whose BE is diagonal 

But diagonal of a rhombus Bisects the angles at the vertices 

So BE bisects LE 

(ii) SO $\angle 1=\angle 2$ and $\angle 3=\angle 4$

But $\angle A+\angle B=180^{\circ}$

So $2 \angle 2+2 \angle 3=180^{\circ}$

$\Rightarrow \angle 2+\angle 3=90^{\circ}$

so In $\triangle A E B$,

if $\begin{aligned} & \angle 2+\angle 3+\angle A E B=180^{\circ} \\ \Rightarrow & 90^{\circ}+\angle A E B=180^{\circ} \\ \Rightarrow & \angle A E B=180^{\circ}-90^{\circ} \end{aligned}$

$\Rightarrow \angle A E B=90^{\circ}$

Hence $\angle A E B$ is a right angle 

Question 9

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Sol: (i) In ||gm ABCD, 

X and Y are mid points of the side AD and BC 

respectively . BD is its diagonals AY and CX are joined 

To prove: 

(i) ATCX is a ||gm

(ii) XY and BD bisect each other 

Proof : In AB||BC 

= AX ||CY

But X and Y are mid point of AD and BC respectively and AD = BC 

AX=CY

so AXCY is a ||gm

(ii) In $\triangle X O D$ and $\triangle B O Y$

$X D=B Y$

$\angle X O D=\angle B O Y$

$\angle A D O=O B Y$

 So $\triangle X O D \cong \triangle B O Y$

XO =OY 

and DO= OB 

Hence XY and BD Bisect Each other (Hence proved)

(ii) In ||gm ABCD , 

BM is the bisector of $\angle B$ and DN is the bisector of $\angle D$ 

To prove: 

(i) BNDM is a ||gm

(ii) BM=ON 

Proof: If bisector of $\angle A B C$ Meets AD at M 

So M is mid point of AD

Similarly DN is the bisector of $\angle A D C$

So N is mid point of BC 

So MD||BN 

and DM||BN 

so BNDM is a ||gm

So BM =DM

 Hence proved

(iii) In ||gm ABCD, BM is the bisector of $\angle B$ and AN is the bisector of $\angle A$

To prove: 

(i) $M N=C D$

(ii) ABNM is a rhombus

Proof: $\angle A+\angle B=180^{\circ}$

If AN and BM are the bisector of $\angle A$ and $\angle B$ respectively 

So $\begin{aligned} \frac{1}{2} \angle A+\frac{1}{2} \angle B &=90^{\circ} \\ \Rightarrow \angle O A B+\angle D B A &=90^{\circ} \end{aligned}$

So In $\triangle A O B$,

$\angle A O B=180^{\circ}-90^{\circ}=90^{\circ}$

So AN and BM are perpendicular to each other Now in $\triangle A O B$ and $\triangle M O N$

$A B=M N$

$\angle A O B=\angle M O N$

$\angle B A N=\angle A N M$

SO $\triangle A O B \cong \triangle M O N$

SO $A O=O N$ and $B O=0 M$

If diagonals BM and AN bisects each other at right angles. 

So ABNM is a rhombus. 

Question 10

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Sol:  Two parallel line PQ and RS and  a transversal LM intersect them at A and B respectively 

This bisectors of interior angles at A and B, Meet each other at C and D as shown in the figure forming a quadrilateral ABCD

To prove: 

(i) ABCD is a rectangles 

(ii) CD is parallel to the original parallel lines PQ and RS 

Construction: Join CD 

Proof: 

(i) AC and BD are the bisector of $\angle P A B$ and $\angle A B S$ and AC||BD 

Similarly AD||BC 

ACBD is a parallelogram

Now $\angle P A B+\angle A B R=180^{\circ}$

So $\frac{1}{2} \angle P A B+\frac{1}{2} \angle A B R=90^{\circ}$

$\Rightarrow \angle C A B+\angle C B A=90^{\circ}$

So In $\triangle A B C$

$\angle A C B=90^{\circ}$

So A parallelogram with each angle a right angle is a rectangle 

Hence ABCD is a rectangle

(ii) If diagonals of  a rectangle are equal and bisect each other 

Diagonal AB and CD of rectangle ABCD Bisects each other  At O 

So AB = OC

So $\angle O A C=\angle A C O \Rightarrow C C A P=\angle A C O$ 

But these are alternate angels

So $C D \| P Q$ or $R S$

Question 11

Sol: (i) In $\triangle A B C, B Y$ is the bisector of  Meeting AC at Y 

Through Y, YX ||AB and YZ ||BC are drawn 

TO prove: BXYX is a rhombus 

Proof: if XY||BA OR BZ and YZ||BC or BX 

(IMAGE TO BE ADDED)

So BZYX is parallelogram 

But BY is the diagonal which bisects $\angle B$.

So $B Z Y X$ is a rhombus

(ii)HJKL is a square in which diagonals HK and JL Intersect each other at O 

X is a point on HJ such that HX =HO 

XO is joined 

To prove: $\angle H O X=3 \angle X O J$

Proof : If diagonals of square HJKL Bisects each other at right angle at O 

So $\angle H O J=90^{\circ}$ and $\angle K H J$ or $\angle O H X=45^{\circ}$

If In $\Delta O H X$

OH=HX 

so $\angle H O X=\angle H XO$

But $\angle O H X+\angle H O X+\angle H X O=180^{\circ}$

$\Rightarrow 45^{\circ}+\angle H O X+\angle H O X=180^{\circ}$

$\Rightarrow 2 \angle H O X=180^{\circ}-45^{\circ}=135^{\circ}$

$\Rightarrow \angle H O X=\frac{135}{2}=67 \frac{1}{2}^{\circ}$.........(i)

$\begin{aligned} \text { But } \angle X O T &=\angle H O J=\angle H O X \\ 90^{\circ}-67 \frac{1^{\circ}}{2} &=22 \frac{1^{\circ}}{2} \end{aligned}$...........(ii)

From (i) and (ii)

$\angle H O X=67 \frac{1}{2}^{\circ}=3 \times 22 \frac{1}{2}$

3 $\angle XOJ$

(iii) In the figure 

ABCD is ||gm 

ABL and ADXY are square 

To prove : $\triangle$ CXM  is an isosceles triangle

Construction: Join MC, MX and CX 

Proof : In parallelogram ABCD 

AB=CD and BC =AD

and $\angle A B C=\angle C D A$ and $\angle B C D=\angle B A D$

So $\angle M B C=\angle M B A+\angle A B C=90^{\circ}+\angle A B C$.............(i)

$\angle C D X=\angle C D A+\angle A D X$..............(ii)

$=\angle A B C+90^{\circ}$

So From (i) and (ii)

$\angle M B C=\angle C D X$

Now in $\triangle B M C$ and $\triangle D C X$

BM =BA =CD 

BC =AD = DX

and $\angle M B X=\angle C D X$

So $\triangle B A C \cong \triangle D C X$

So $CM=CX$

Hence $\triangle CMX$ is an isosceles triangle

(iv) ABCD and ALMN are two square as shown in the figure 

(IMAGE TO BE ADDED)

To prove: BL=DN

Construction : Join DM and BL 

Proof: In $\triangle A D N$ and $\triangle A L B$ 

$A D=A B$

$A N=A L$

$\angle D A N=\angle L A B$

SO $\triangle A D N \approx \triangle A L B$

SO $D N=B L$

Hence proved 

Question 12

Sol: (i) In ||gm ABCD, AE and BE are the bisector of adjacent $\angle A$ and $\angle B$ which mect at $E$

To prove: $\angle E$ is a right angle 

Proof : If AE is the bisector of $\angle A$ 

So $\angle E A B=\frac{1}{2} \angle A$ 

Similarly BE is the bisector of $\angle B$

$\angle E B A=\frac{1}{2} \angle B$

But $\angle A+\angle B=180^{\circ}$

So $\frac{1}{2} \angle A+\frac{1}{2} \angle B=90^{\circ}$

$\angle E A B+\angle E B A=90^{\circ}$

But in $\triangle A E B$

$\angle E A B+\angle E B A+\angle A E B=180^{\circ}$

$\Rightarrow 90^{\circ}+\angle A E B=180^{\circ}$

$\Rightarrow \angle A E B=180^{\circ}-90^{\circ}=90^{\circ}$

So $\angle E$ is a right angle 

 

( ii) In ||gm ABCD PA, QD, RC and SB are the bisector of 

$\angle P, \angle Q, \angle R$ and $\angle S$ Respectively Which meet each others 

At A,C,D and B respectively forming a quad, ABCD 

To prove: ABCD is rectangle 

Proof: If PD and QD are the bisector of two adjacent angles $\angle P$ and $\angle Q$ Respectively 

So $\quad \angle D=90^{\circ}$

Similarly $\angle B=90^{\circ}$

And PA and SA are the bisector of $\angle P$ and $\angle S$ respectively 

So $\angle A=90^{\circ}$ 

Similarly $\angle C=90^{\circ}$ 

So the quad. ABCD has its each angle equal to 90

So ABCD is a rectangle   Hence proved 

Question 13

Sol:  ABCD is a ||gm and X is mid point of BC AX is joined and produced to meet BC produced at Q. The ||gm ABPQ is completed as shown in the figure 

To prove: 

(i)$\triangle A B X \cong \triangle Q C X$

(ii)DC =CQ = QP 

Proof: IN $\triangle A B X$ and $\triangle Q C X$

$B X=X C$

$\angle A \times B=\angle C \times Q$

$\angle A B X=\angle X C Q$

So $\triangle A D X \cong \triangle Q C X$ 

So AB=CQ 

But AB = DC and AB =QP 

So DC =CQ = QP 

Hence proved

Question 14

Sol: P,Q and R are the mid points of the sides AB,BC and CD of rhombus ABCD PQ and QR are joined 

To prove: $P Q \perp Q R$

Construction: Join diagonals AC and BD of the rhombus ABCD 

Proof: In $\triangle A B C$ 

P and Q are mid points of AB and BC respectively 

So 

PQ||AC ...................(i)

Similarly in $\triangle B C D$ 

Q and R the mid points of BC and CD respectively 

So $Q R \quad \| B D$ ................(ii)

But AC and BD bisects each other at right angles 

$P Q \perp Q R$

(IMAGE TO BE ADDED)

Question 15

Sol: In a rhombus ABCD 

RABS is straight line such that RA = AB = RS 

RD and SC are joined and produced to meets at T 

To prove: $\angle T=90^{\circ}$

Proof: In $\triangle R A D$ 

RA =AB = AD 

so $\angle D R A=\angle R D A$

and $Ext. \cdot \angle D A B=\angle D R A+\angle R D A$

$=\angle D R A+\angle D R A=2 \angle D R A$..........(i)

Similarly in $\triangle B S C$

BS = AB = BC 

So $\angle B C S=\angle B S C$

and Ext. $\angle A B C=\angle B C S+\angle B S C$

$=\angle B S C+\angle B S C=2 \angle B S C$.........(ii)

But in rhombus ABCD 

$\angle D O B+\angle A B C=2 \angle B S C$............(iii)

$\angle D O B+\angle A B C=180^{\circ}$

2 $\angle D R A+2 \angle B S C=180^{\circ}$

$\angle D R A+\angle B S C=90^{\circ}$

Now in $\triangle R T S$.

$\angle D R A+\angle B S C=90^{\circ}$

Or $\angle T R S+\angle R S T=90^{\circ}$

$\angle R T S=180^{\circ}-90^{\circ}=90^{\circ}$

Hence $\angle T=90^{\circ}$