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

 Exercise 11 D

Question 1

Sol: (i) Draw a line segment BC=5.2cm

 (ii) At B draw a ray BX making an angle of 45 and cut off BA=6.5cm

(iii) With center C and radius 6.5cm and with center A and radius 5.2 cm, draw arcs intersecting each other at D.

 (iv) Join AD and CD 

ABCD is the required ||gm

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Question 2

Sol:(i)Draw diagonal BD =6.8 cm and bisect it at O.

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(ii) From O, Draw a ray OX making an angle of 60 and  product 

(iii) Cut of OA = OC =4.4\2= 2.2cm

(iv) Join AB, BC , CD and DA 

ABCD is the required ||gm

Question 3

 Sol:(i) Draw a line segment AB=6cm

(ii) At A , draw a perpendicular AX and Cut off AP =2.9cm

(iii) From P, and EPF ||AB

(iv) With center B and A and radius 4cm draw arcs cutting EF at D and C.

(v) Join AD and BC.

Then ABCD is the required ||gm

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Question 4

Sol:(i) Draw a line segment AB=4.3cm

(ii) At B draw a ray BY. 

(iii) At A, draw perpendicular AX and Cut off AE =3cm

(iv) From E draw a parallel line to AB which Intersects By at C.

(v)With center A and radius BC, draw an arc cutting EF at D.

(vi) Join AD.

Then ABCD is the required ||gm

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Question 5

Sol: (i)Draw a line segment $A B=6 \mathrm{~cm}$

(ii) At A draw a ray AX making an angle of $45^{\circ}$ and cut off $A D=3 \mathrm{~cm}$.

(iii)With center B and Radius 3cm and with center D and Radius 6cm, Draw arcs intersecting each other at C.

(iv) Join BC and DC 

ABCD is the required ||gm

(v) Draw the angle bisector of A meeting CD at P and join BP

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To prove: $\angle A P B=90^{\circ}$

proof: In $\triangle A D B$

$\angle D P A=\angle P A B$

But $\angle D A P=\angle P A B$

$\angle D A P=\angle B P A=\frac{45^{\circ}}{2}=22 \frac{1}{2}^{\circ}$

$A D=D O=3 \mathrm{~cm}$

Similarly in $\triangle B C P$

${ }_{P C}=D C-D P=6-3=3 \mathrm{~cm}$

So $P C=C B=3 \mathrm{~cm}$

So $\angle C P B=\angle C B P$

But $\angle C P B+\angle C B P=180^{\circ}-\angle B C P$

$=180^{\circ}-45^{\circ}=135^{\circ}$

$\angle C P B=\angle C B P=\frac{139^{\circ}}{2}=67 \frac{1}{2}^{\circ}$

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

$\Rightarrow 22 \frac{1}{2}+67 \frac{1}{2}^{\circ}+\angle A P B=180^{\circ}$

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

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

Hence $\angle A P Q=90^{\circ}$

Question 6

Sol: The diagonal AC and BD of a ||gm ABCD bisect each other 

$A O=O C=\frac{10}{2}=5 \mathrm{~cm}$

and $B O=O D=\frac{6}{2}=3 \mathrm{~cm}$

(i) Draw a line segment AB =4cm 

(ii) With center a and radius 5cm and with center B and radius 3cm draw which intersect each other at O. 

(iii) Join OA and OC.

(iv) Produce AO to C such that Ao=OC  and produce BO to B such that BO=OD 

(v) Join AD, DC ,CB

Then ABCD is the required ||gm

Question 7

Sol: (i) Draw a line segment AC =10cm and bisect it at O.

(ii) AT O, Draw a ray OX making an angle of 60 and produce XO to Y. 

(iii) From O, cut off OD = OB = $\frac{6}{2}=3 \mathrm{~cm}$

(iv) Join AB,BC , CD and DA. 

ABCD is the required ||gm

Whose longer side is BC or AD on measuring it , it is 6cm long. 

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Question 8

Sol;(i) Draw a line AX.

(ii) AT A, Draw a perpendicular and cut off AL = 3cm 

(iii) Through L draw LY parallel to AX.

(iv) From A, Draw a ray making an angle of 60 which meets LY at D. 

(v) AT AD. draw a perpendicular and cut off AM = 4cm

(vi) Through M, draw a line parallel to AD which Meets LY at C. 

Then ABCD is the required parallelogram 

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Question 9

Sol: If diagonal of a ||gm bisect each other 

AO=OC and OB =OD = 2.5cm

Steps of construction :

(i) Draw a line segment AC=7cm and bisect it at O. 

(ii) AT O. draw a line making an angle of 75 and produce it to both sides

(iii) From O, Cut off OB =OD $=\frac{5}{2}=2.5 \mathrm{~cm} .$ 

(iv) Join AB, BC, CD and DA 

ABCD is the required ||gm

Measuring its shorter side AD=4cm 

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Question 10

Sol: (i) Draw a line segment AB = 3.6cm

(ii) AT, A draw a ray AX making an angle of 45 and cut off AB=3.6cm

(iii)With center B and D and radius 3.6cm draw arcs intersecting each other at C.

(iv) Join BC and CD 

Then ABCD is the required rhombus 

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Question 11

Sol: (i) Draw a line segment AB =3.2cm

(ii) AT A and B draw perpendiculars AX and BY and Cut off AD = BC =3.2cm

(iii) Join DC 

ABCD is the required Square 

(iv) Join its diagonals AC and BD 

On Measuring each =4.5 cm and each angles between these diagonals is 90 $\angle A O B=90^{\circ}$

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Question 12

Sol: We know that diagonals of a square bisect each other at right angles and diagonals are also equal 

Steps of construction: 

(i) Draw a line segment PR =5cm and draw its perpendicular bisects XY intersecting it at O.

(ii)Cut off OQ = OS =$\frac{5}{2}=2.5 \mathrm{~cm}$

(iii) Join PQ, QR, RS and SP respectively. 

Then PQRS is the required square on measuring its side PQ which is = 3.6cm

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Question 13

Sol: Steps of construction: 

(i) Draw a line segment AB = 4.2 cm 

(ii)AT A and B, Draw perpendicular AX and BY 

(iii) Join CD

Then ABCD is the required rectangle 

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Question 14

Sol: Diagonals of a rectangle bisect each other and arc equal in length 

Steps of Construction: 

(i) Draw a line segment AC = 6cm

(ii) Bisect AC and O. 

(iii) From O, draw a line XY 

Making an angle of 37 and produce it to both side of O.

(iv) From XY Cut off OC =OD = 3cm

 (v) Join AB, BC , CD and AD,

Then ABCD is the required rectangle.

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Question 15

Sol: The diagonals of a square bisect each other at right angles and arc also equal in length. 

Steps of Construction: 

(i) Draw a line segment AC =5cm

(ii) Draw its perpendicular bisector XY intersecting at O. 

(iii) Cut off OB =OD =2.5CM

(iv) Join AB, BC, CD and DA respectively 

Then ABCD is the required square

On measuring its side it is 3.6 cm each 

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Question 16

Sol: Diagonals of rectangle bisect each other and are equal to each other in length.

Steps of constructions:

(i) draw a line segment $A C=6 \mathrm{~cm}$ and bisect it at O .

(ii)At O , draw a line XY making an angle of $45^{\circ}$ and produce it to both of O .

(iii) cut off $O B=O D=\frac{6}{2}=3 \mathrm{~cm}$.

(iv)Join AB, BC, CD and DA respectively

Then ABCD is the required rectangle. 

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Question 17

SoL: Diagonals of a rhombus bisect each other at right angles.

Steps of constructions:

(i) Draw a line segment $P R=8 \mathrm{~cm}$

(ii) Draw its perpendicular bisector XY which intersects $P R$ at 0 .

(iii) from XY act off $O Q=O S=\frac{6}{2}=3 \mathrm{~cm}$.

(iv) Join $P Q, Q R, R S$ and SP

Then PQRS is the required rhombus 

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(v) if the required point X is equidistant rom $p$ and PS.

So it lies on the angle bisectors of LQPS which is the diagonal PR

it is equidistant from $R$ and S

 Then it will be on the perpendicular bisector of ks Now draw perpendicular bisector of RS which intersects PR at X

Then X is the required point 

Construction the trapezium PQRS in which PQ is parallel to RS from the given measurement in problem 18 to 20 without using set square and protractor as far as possible

Question 18

Sol:  Steps of constructions:

(i) Draw a line segment $P Q=7 \mathrm{~cm}$.

(ii) Take a point E such that AE =SR = 2.5 cm and then EQ = 7 -2.5 =4.5cm

(iii) With center E and radius 3cm and with center Q and radius 3.5cm Draw arcs which intersect each other at R 

(iv)Join $E R$ and $Q R$.

(v)with conter $R$ and radius $2.5 \mathrm{~cm}$ and with center $P$ and radius $3 \mathrm{~cm}$ draw arcs intersecting each other at S .

 (vi) Join RS and PS.

Then PQRS is the required trapezium 

(IMAGE TO BE ADDED)

Question 19

Sol: Steps of constructions:

(i) draw a line segment $P Q=8 \mathrm{~cm}$.

(ii) cut off $P E=R S=3 \mathrm{~cm}$.

(iii) With center E and Radius 2cm and with center    Q and radius 4cm draw intersecting each other at R. 

(IV) Join $E R$ and $Q R$

(v) with center $P$ and radius $2 \mathrm{~cm}$ and with conter $R$ and radius $3 \mathrm{~cm}$ draw arcs intersecting each other at S.

(vi) Join PS and SR.

Then PQRS is the required trapezium On measuring the angle QPS it is 50

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Question 20

Sol:if $\angle A B C=120^{\circ}$ and $A B \| D C$

 So $\angle B C D=180^{\circ}-120^{\circ}=60^{\circ}$

(i) draw a line segment $A B=4 \mathrm{~cm}$

(ii) AT B draw a ray BX making an angle of 120 and cut off BC = 2.5cm

(iii) At c draw and the ray CY making an angle of $60^{\circ}$

(iv) with centre $A$ and radius $2.4 \mathrm{~cm}$ draw an are which intersects $\mathrm{CY}$ at $\mathrm{D}^{\prime}$ and O

If $\angle A$ is obtuse and by joining AD then angle 

So Join AD 

Then ABCD is the required trapezium on measuring AC it is 5.5cm

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Question 21

(i) draw a line segment $A B=6.5 \mathrm{~cm}$

(ii)produce it to $P$ such that- BP $=3 \mathrm{~cm}$

(iii)with conter p and radius $5 \mathrm{~cm}$ and with center $A$ and radius $7 \mathrm{~cm}$ draw ares which intersect each other at C

(iv) Join $A C$ and $P C$ and $B C$.

(v)With conter B and recdius $5 \mathrm{~cm}$ and with conter c and radius $3 \mathrm{~cm}$, draw ares intersecting each other at $D$.

(vi) Join AD , BD and CD 

Then $A B C D$ is the required trapezium.

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Question 22

Sol: The radius of the circumcircle al a regular hexagon is the length of the side of the regular hexagon.

(i) Draw a circle with center O and radius 4cm

(ii) Take a point $A$ on it and from $A$, draw ars of $4 \mathrm{~cm}$ cutting the circle at $B, C, D$, F and $F$.

(iii) Join $A D, B C, C D, D E, E F$ and $F A$.

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Then $A B C D E F$ is the required regular $h$ exagon The area ag regular hexagon $=6 \times \frac{\sqrt{3}}{4}(\text { side })^{2}$

$=\frac{6 \sqrt{3}}{4}(4)^{2}$

$=\frac{6 \times \sqrt{3} \times 16}{4}=24 \sqrt{3} \mathrm{~cm}^{2}$

$=24(1.732) \mathrm{cm}^{2}$

$=41.568 \mathrm{~cm}^{2}$

Question 23

Sol; The radius of the circumcircle of a regular hexagon is the length of the side of this hexagon. 

(i) Draw a circle with center O and radius 3.2cm

(ii)Take a point A on the circle 

(iii) Start from A and Radius 3.2cm the arcs on the circle at B,C,D E and F

(IV) Join AB, BC , CD , DE , EF and FA 

Then ABCDEF is required regular hexagon 

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