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

 Exercise 11(A)

Question 1

Sol :

(i) In a rectangle diagonals are equal and bisect each other at O

So, AO=BO

So, ∠OAB=∠OBA=18°

But ∠ABC=90°

So, x=90°-18°=72°

(ii) In rectangle RQRS its diagonals bisect each other at T

∠PTQ=120°

So, ∠TPQ+∠TQP=180°-120°=60°

But PT=QT

So, ∠TPQ=∠TQP$=\frac{60}{2}$=30°

Now, ∠SRT=∠TPQ

So, b=30°

a+30°=90°

a=90°-30°=60°

So, a=60° ,b=30°

(iii) In the rhombus EFGH

∠EFG=140°

But ∠EFG+∠FGH=180°

⇒140°+∠FGH=180°

⇒∠FGH=180°-140°=40°

If diagonals of a rhombus bisects the angles

So ,$x=\frac{1}{2} \times 40^{\circ}$=20°

(iv) If the diagonals of a rhombus bisect each other at right angles

So, ∠PLQ=90°

Now in ΔPLQ

So, ∠LPQ+∠LQP=90°

34°+x=90°

x=90°-34°

x=56°

If diagonals QS bisects ∠Q and ∠S

So, x=y 

y=56°

Hence, x=56 , y=56°

(v) In the Square ABCD

Diagonals AC intersects a line DY such that DY such that

∠DCX=112°

If diagonals AC bisects the angles A and C 

So, ∠ACB on ∠XCY=45°

But ∠CXY+∠CXD=180°

So, ∠CXY+112°=180°

∠CXY=180°-112°=68°

Now in ΔCXY,

∠CXY+∠XYC+∠XCY=180°

68°+d+45°=180°

113°+d=180°

d=180°-113°=67°

(vi) In square CDEF

DF is its diagonal

En is a line segment which intersect DF at M such that ∠EMF=38°, ∠MNC=x

If ∠EMF=∠DMN

So, DMN=38°

But diagonal DF bisects the ∠D and ∠F

So, ∠MND=45°

In ΔMND

Ext.∠MNC=∠DMN+∠MDN

=x+38°+45°=83°

Hence , x=83°

(vii) In the figure,

AB||CD and AD=BC

∠A=x, ∠C=75°

If ABCD is an isosceles trapezium

So,  ∠A+∠C=180° and ∠B+∠D=180°

Now, ∠A+∠C=180°

⇒x+75°=180°

⇒x=180°-75°

So, x=105°

(viii) The figure is of a kite whose diagonals AC and BD are perpendicular to each other

AB=AD and CB=CD

Now, in right ∠OCD, ∠DOC=90°

So, a+39°=90°

a=90°-39°=51°

In ΔABD, AB=AD

So, ∠ADO=∠ADO=90°

and in right ΔAOD, ∠AOD=90°

So, b+∠ADO=90°

b+74°=90°

b=90°-74°=16°

(ix) ABCD is an isosceles trapezium in which AB||DC 

AD=BC and Diagonals

AC and BD intersects each other at O

∠ODC=∠OCD=34°

But In ΔOCD

a+∠ODC+∠CD=180°

a+34°+34°=180°

a+68°=180°

a=180°-68°=112°

and ∠ADB=∠ACB=b

So, ΔOBC

ext. a=72°+b

Hence, ∠a=112° and ∠b=40°

Question 2

Sol :

In parallelogram ABCD,

∠A : ∠B=1:5

Let ∠A=x , then ∠B=5x

But ∠A+∠B=180°

So, x+5x=180°

6x=180°

$x=\frac{180}{6}=30^{\circ}$

So, ∠A=30°, ∠B=5x=5×30°=150°

and ∠C=∠A=30° and ∠D=∠B=150°

Question 3

Sol :

Let the smallest angle=x

Then other greatest angles=2x-20

But x+2x=180°

3x-20°=180°

3x=180°+20°=200°

So, $x=\frac{200}{3}=66 \frac{2}{3}$

and greatest angle=2x-20°

$=2 \times \frac{200}{3}-20$

$=\frac{400}{3}-20=\frac{400-60}{3}$

$=\frac{340}{3}=113 \frac{1}{3}$

So, The angles of parallelogram are

$66 \frac{2^{\circ}}{3}, 113 \frac{1}{3}, 66 \frac{2^{\circ}}{3}$ and $113 \frac{1^{\circ}}{3}$

Question 4

Sol :

ABCD is a rhombus in which ∠A=50° and its diagonals AC and BD bisect each other at O right angles.

So, ∠AOB=90°

If diagonals bisect the opposite angles

So, ∠OAB$=\frac{50^{\circ}}{2}=25^{\circ}$

So, ∠OBA=180°=(90°+25°)

=180°-115°=65°

Hence angles are 25°, 65° and 90°

Question 5

Sol :

In rectangle ABCD , diagonals AC and BD bisect each other at P.

∠ABD=50°

In ΔAPB

AP=BP

So, ∠PAB=∠PBA or ∠ABD=50°

So, ∠APB=180°-(50°+50°)

=180°-100°=80°

But ∠CPD=∠APD (vertically opposite angles)

So, ∠CPD=80°

Question 6

Sol :

In parallelogram ABCD

Let ∠B=x , then

∠A$=\frac{2}{3}x$

But ∠A+∠B=180°

$\frac{2}{3} x+x$ or $\frac{5}{3} x=180^{\circ}$

$x=\frac{180 \times 3}{5}$=36×3=108°

So, ∠B=108° and 

∠A$=\frac{2}{3} \times 108^{\circ}$=2×36=72°

Hence, ∠A=72° and ∠B=108°

Question 7

Sol :

In parallelogram ABCD

BD is its diagonal

∠DAB=70° , ∠DBC=80°

∠CDB=x and ∠ADB=y

If AD||BC and BD is its transversal

So, ∠ADB=∠DBC

=y=80°

In ΔABD,

∠DAB+∠ABD+∠ADB=180°

=70°+∠ABD+80°=180°

150°+∠ABD=180°

∠ABD=180°-150°=30°

But ∠CDB=∠ABD

x=30°

Hence ∠ABD=30° and ∠ADB=80°

Question 8

Sol :

In a rhombus ABCD diagonals AC=24 cm, and BD=18cm

If diagonals of a rhombus bisect each other

So, AO=OC$=\frac{24}{2}$=12 cm

BO=OD$=\frac{18}{2}$=9 cm

and ∠AOB=90°

So, In right ΔAOB

$A B^{2}=A O^{2}+B O^{2}$

$=12^{2}+9^{2}$

=144+81=225

$A B^{2}={15}^2$

So, AB=15

So, Each sides of rhombus ABCD=15 cm

Question 9

Sol :

In rhombus ABCD

AC and BD are its diagonals AB=5 cm and AC=8 cm

If the diagonals of a rhombus bisect each other at right angles

So, AO=OC$=\frac{8}{2}$=4 cm and BO=OD and ∠AOB=90°

Now, In right ΔAOB

$A B^{2}=A O^{2}+B O^{2}$

$(5)^{2}=(4)^{2}+B 0^{2}$

$25=16+B 0^{2}$

$B O^{2}=25-16=9$

$BO^{2}=9=(3)^{2}$

So, BO=3

So, BD=2BO=2×3=6 cm

Now area of rhombus

$=\frac{\text{Product of diagonals}}{2}$

$\frac{A C \times B D}{2}=\frac{8 \times 6}{2}=24 \mathrm{~cm}^{2}$

Question 10

Sol :

In rhombus ABCD

Side PQ=3 cm

Height RL=2.5cm

and diagonals AC and BD cut each ther at O

To find:

(i) Perimeter of PQRS

=4×3cm=12cm

(ii) Area of PQRS

Base PQ=3cm and Height RL=2.5cm

Area of PQRS=Base×Height

=3×2.5=7.5 cm²

(iii) Measure of ∠POQ

If the diagonals of a rhombus bisect each other at right angles

So, ∠POQ=90°

Question 11

Sol :

ABCD is a trapezium in which AB=10cm  AD=4cm, ∠DAB=∠CBA=60°

Draw CL⟂AB and DM⟂AB

If ∠DAB=∠CBA

So, ABCD is an isosceles trapezium

So, AD=BC=4cm

In right ΔADM,

Sin 60°$=\frac{D M}{AD}$  

$\frac{\sqrt{3}}{2}=\frac{DM}{4}$

$DM=\frac{4 \sqrt{3}}{2}=2 \sqrt{3}$cm

and Cos60°$=\frac{AM}{AD}$ 

$\frac{1}{2}=\frac{A B}{4}$

$AB=\frac{4}{2}$=2 cm

So, LB=AM=2 cm

Now, CD=ML=AB-(Am+LB)

=10-(2+2)=10-4=6 cm

Hence

(i) CD=6 cm

(ii) Distance between AB and CD$=2\sqrt{3}$ cm

Question 12

Sol :

EFGH is an isosceles trapezium

S, EH=FG

and ∠HEF=∠GFE

$2 y^{2}-25=y^{2}+24^{\circ}$

$2 y^{2}-y^{2}=24+25$

$y^{2}=49=(\pm 7)^{2}$

So, y=7 or -7

Question 13

Sol :

In rhombus PQRS, PR and QS are the diagonals which intersects each other at  

PR is produced to T

∠SRT=152°

If ∠PRS+∠SRT=180°

∠PRS=180°-152°=28°

So, ∠SPQ=∠SRQ=2∠PRS

=2×28°=56°

(∵Diagonals of  a rhombus bisect the angles)

But ∠PQR+∠SRQ=180°

∠PQR+56°=180°

∠PQR=180°-56°=124°

So, $x=\frac{1}{2}\angle PQR=\frac{1}{2} \times 124^{\circ}=62^{\circ}$

So, x=62°, y=90°

z=∠PRS=28°

Question 14

Sol :

ABCD is a rhombus and ΔABE is an equilateral triangle.

BD and DE are joined.

∠BCD=78°

∠BAD=∠BCD=78°

and ∠EAB=60°

So, ∠EAD=78°+60°=138°

But in ΔEAD

AE=AD

So, ∠AED=∠ADE

But ∠AED+∠ADE+∠EAB=180°

∠ADE+∠ADE+138°=180°

2∠ADE=180°-138°=42°

(i) So, ∠ADE$=\frac{42}{2}$=21°

(ii) ∠AED=60°

So, ∠BED=∠AEB-∠AED

=60°-21°=39°

(iii) ∠BCD+∠CDA=180°

So, ∠CDA=180°-∠BCD=180°-78°=102°

If BD is the diagonal of rhombus

So, $\angle B D A=\frac{1}{2} \angle C D A=\frac{1}{2} \times 102^{\circ}$=51°

But ∠ADE=21°

So, ∠BDE=51°-21°=30°

Hence, ∠ADE=21°, ∠BDE=30° and ∠BED=39°

Question 15

Sol :

ABCD is a square and EBC is an equilateral triangle on BC

ED is joined

In ΔECD

CD=CE

So, ∠CED=∠CDF

But ∠DCF=∠DCB+∠BCE

=90°+60°=150°

So, ∠CED+∠CDE=180°-150°=30°

∠CED+∠CED=30°

2∠CED=30°

∠CED$=\frac{30}{2}$=15°

But ∠BEC=60°

So, ∠BED=∠BEC-∠CED

=60°-15°=45°

Question 16

Sol :

ABCD is a square

So, Its each angle=90°

ABO is an equilateral triangle

So, its each angle=60°

Now in ΔOAD,

∠OAD=∠BAD-∠OAB=90°-60°=30°

So, ∠AOB+∠ADO=180°-30°=150°

But ∠AOD=∠ADO

So, ∠AOD=∠ADO$=\frac{150^{\circ}}{2}=75^{\circ}$

Similarly, In ΔOBC

∠BOC=75°

But ∠DOC+∠AOD+∠BOC+∠AOB=360°

∠DOC+75°+75°+60°=360°

∠DOC+210°=360°

∠DOC=360°-210°=150°

Hence, ∠DOC=150°