SELINA Solution Class 9 Chapter 17 Circle Exercise 17B

Question 1

The figure shows two concentric circles and AD is a chord of a larger circle.
Prove that: AB = CD.

Sol:

Drop OP ⊥ AD
∴ OP bisects AD.     ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ AP = PD               .....(i)

Now, BC is a chord for the inner circle and OP ⊥ BC.
∴ OP bisects BC      ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ BP = PC               .....(ii)

Subtracting (ii) from (i),
AP - BP = PD - PC 
⇒ AB = CD.

Question 2

A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centers O and O'. Prove that the chords AB and CD, which are intercepted by the two circles, are equal.

Sol:

Given: A straight line AD intersects two circles of equal radii at A, B, C and D.
The line joining the centers OO' intersect AD at M and M is the midpoint of OO'.

To Prove: AB = CD.

Construction: From O, draw OP ⊥ AB and from O', draw O'Q ⊥ CD.

Proof:  
In ΔOMP and ΔO'MQ,
∠OMP = ∠O'MQ            ...( Vertically Opposite angles )
∠OPM = ∠O'QM            ...( each = 90° ) 
OM = O'M                      ...( Given )

By Angle-Angle-Side criterion of congruence,
∴ ΔOMP ≅ ΔO'MQ,       ...( by AAS )

The corresponding parts of the congruent triangles are congruent.
∴ OP = O'Q                     ...( c.p.c.t. )

We know that two chords of a circle or equal circles which are equidistant from the center are equal.
∴ AB = CD.

Question 3

M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O.
Prove that: (i) ∠BMN = ∠DNM
                  (ii) ∠AMN = ∠CNM

Sol:

Drop OM ⊥ AB and ON ⊥ CD.
∴ OM bisects AB and ON bisects CD.    ...( Perpendicular drawn from the centre of a circle to a chord bisects it. )

⇒ BM = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{CD}}$ = DN  ....(1)

Applying Pythagoras theorem,
OM² = OB² - BM²
         = OD² - DN²                         ....( By 1 )
         = ON²
∴ OM = ON
⇒  ∠OMN = ∠ONM                   ....(2)
( Angles opp to equal sides are equal. )  

(i) ∠OMB = ∠OND                  .....( both 90° )
Subtracting (2) from above,
∠BMN = ∠DNM

(ii) ∠OMA = ∠ONC                  .....( both 90° )
Adding (2) to above,
∠AMN = ∠CNM.

Question 4

In the following figure; P and Q are the points of intersection of two circles with centers O and O'. If straight lines APB and CQD are parallel to OO';
prove that: (i) OO' = ${\displaystyle \frac{1}{2}}$AB ; (ii) AB = CD

Sol:

Drop OM and O'N perpendicular on AB and OM' and O'N' perpendicular on CD.

∴ OM, O'N, OM' and O'N' bisect AP, PB, CQ and QD respectively.
( Perpendicular is drawn from the center of a circle to a chord bisects it. )

∴ MP = ${\displaystyle \frac{1}{2}\text{AP},\text{PN}=\frac{1}{2}\text{BP},\text{M'Q}=\frac{1}{2}\text{CQ},\text{QN}\prime =\frac{1}{2}\text{QD}}$ 

Now, OO' = MN = MP + PN = ${\displaystyle \frac{1}{2}\left(\text{AP + BP}\right)=\frac{1}{2}\text{AB}}$   ...(i)

and OO' = M'N' = M'Q + QN' = ${\displaystyle \frac{1}{2}\left(\text{CQ + QD}\right)=\frac{1}{2}\text{CD}}$  ...(ii)

By (i) and (ii),
AB = CD.

Question 5

Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle.
Prove that: (i) AP = CP ; (ii) BP = DP

Sol:

Drop OM and ON perpendicular on AB and CD.
Join OP, OB, and OD.

∴ OM and ON bisect AB and CD respectively.     ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
∴ MB = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{CD}=\text{ND}}$....(i)

In right ΔOMB,
OM² = OB² - MB²                            ....(ii)
In right ΔOND,
ON² = OD² - ND²                           ....(iii)

From (i), (ii), and (iii),
OM = ON

In ΔOPM and ΔOPN,
∠OMP = ∠ONP          ....( both 90° )
OP = OP                     ....( common )
OM = ON                   ....( proved above )
By Right Angle-Hypotenuse-Side criterion of congruence,
∴ ΔOPM ≅ ΔOPN      ....( by RHS )

The corresponding parts of the congruent triangles are congruent.
∴ PM = PN                ....( c.p.c.t. )

Adding (i) to both sides,
MB + PM = ND + PN
⇒ BP = DP
Now, AB = CD
∴  AB - BP = CD - DP     ...( ∵ BP = DP )
⇒ AP = CP.

Question 6

In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q.
Prove that:
( i ) ΔOPA ≅ ΔOQC 
( ii ) ΔBPC ≅ ΔBQA

Sol:

(i) In ΔOPA and ΔOQC,
 OP = OQ                          ....[ radii of same circle ]
∠AOP = ∠COQ                  ... [ both 90° ] 
OA = OC                           ... [ sides of the square ]

By Side- Angle - Side criterion of congruence.
∴ ΔOPA ≅ ΔOQC             ...[ by SAS ]

(ii) Now, OP = OQ           ...[ radii ]
 and  OC = OA                 ...[ sides of the square ]
∴ OC - OP = OA - OQ  
⇒ CP = AQ                      ....(i)

In ΔBPC and ΔBQA,
BC = BA                           ...[ sides of the square ]
∠PCB = ∠QAB                 ...[ both 90° ]
 PC = QA                         ...[ by ( i ) ]

By Side- Angle-Side criterion of congruence,
∴ ΔBPC ≅ ΔBQA                ...[ by SAS ]

Question 7

The length of the common chord of two intersecting circles is 30 cm. If the diameters of these two circles are 50 cm and 34 cm, calculate the distance between their centers.

Sol:

OA = 25 cm and AB = 30 cm

∴    AD = ${\displaystyle \frac{1}{2}\times \text{AB} =(\frac{1}{2}\times 30)}$ cm = 15 cm 

Now in right angled ADO
OA² + AD² + OD²  
⇒  OD² = OA² - OD²  = 25² - 15² 
             = 625 - 225 = 400
∴ OD = ${\displaystyle \sqrt{400}}$ = 20 cm
Again, we have  O'A = 17 cm.

In right-angle ADO'
O'A² = A'D² + O'D²  
⇒  O'D² = O'A² - AD² 
= 17² - 15² 
= 289 - 225 = 64

∴ O'D = 8 cm
∴ OO' = ( OD + O'D )
          = ( 20 + 8 ) = 28 cm

∴ the distance between their centres is 28 cm.

Question 8

The line joining the midpoints of two chords of a circle passes through its center.
Prove that the chords are parallel.

Sol:

Given: AB and CD are the two chords of a circle with center O.
L and M are the mid-points of AB and CD and O lies in the line joining ML.

To prove : AB || CD.

Proof:
AB and CD are two chords of a circle with center O.
Line LOM bisects them at L and M.
Then,                OL ⊥ AB
and,                OM ⊥ CD
∴               ∠ALM = ∠LMO = 90°
But they are alternate angles
∴                 AB || CD.

Question 9

In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles.
Show that:
(i) AB = CD ;
(ii) AC = BD.

Sol:

In the circle with center Q, QO ⊥ AD
∴ OA = OD                            ....(i)   ...[ perpendicular drawn the center of a circle to a chord bisects it ]

In circle with center P, PO ⊥ BC
∴ OB  = OC                          ....(ii)  ....[  perpendicular drawn the center of a circle to a chord bisects it ]

(i) (i) - (ii) gives,
AB = CD                              ....(iii)

(ii) Adding BC to both sides of equation (iii) 
 AB + BC + CD + BC
⇒ AC = BC

Question 10

AB and CD are two equal chords of a circle with center O which intersect each other at a right angle at point P.
If OM ⊥ AB and ON ⊥ CD;
show that OMPN is a square.

Sol:

Clearly , all the angles of OMPN are 90°.
OM ⊥ AB and ON ⊥ CD

∴ BM = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$CD = CN      ....(i) ...[ perpendicular drawn from the center of a circle to a chord bisects it ]

As the two equal chords, AB and CD intersect at point P inside the circle,

∴ AP = DP and CP = BP                 .....(ii)
Now, CN - CP = BM - BP                ...[ by (i) and (ii) ]
⇒ PN = MP

∴ Quadrilateral OMPN is A square.

S.chand class 6 Mathematics Chapter 17 Exercise 17B

Exercise 17 B

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Question 1

In the given fig. Q1 , which setting will the pointer face after it turns 

(i) 45º clockwise from normal ?

Sol :

A circle consist of total angle of 360º  . Now lets find the angle between two options .

And to do so , we take $=\dfrac{360}{\text{total options}}=\dfrac{360^º}{8}$

= 45º

According to question as we move from normal in clockwise direction 45º , then pointer face normal .

 

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(ii) 135º anti-clockwise from off ?

Sol :

As we already know angle between two options is $=\dfrac{360º}{8}= 45º$ .

So , if we divide 135º by 45º $=\dfrac{135^º}{45^º}=3$

Which means from off position pointer have to move 3 steps in anti-clockwise direction .

First step = off→very hot

Second step = very hot→hot

Third step = hot→warm

So , at last pointer stop at warm position .

 

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(iii) 360º anti-clockwise from hot ?

Sol :

We know angle between two options is $=\dfrac{360º}{8}= 45º$

So , if we divide 360º by 45º $=\dfrac{360º}{45º}=8$

Which means from hot position pointer have to move 8 steps in anti-clockwise direction .

First step = hot→warm

Second step = warm→normal

Third step = normal→cool

Fourth step = cool→cold

Fifth step = cold →very cold

Sixth step = very cold →off

Seventh step = off →very hot

Eighth step = very hot →hot

So , at last pointer stop at hot position .

 

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(iv) 270º clockwise from cold ?

Sol :

We know angle between two options is $=\dfrac{360º}{8}= 45º$

So , if we divide 270º by 45º $=\dfrac{270º}{45º}=6$

Which means from cold position pointer have to move 8 steps in anti-clockwise direction .

First step = cold→cool

Second step = cool→normal

Third step = normal→warm

Fourth step = warm→hot

Fifth step = hot →very hot

Sixth step = very hot → off

So , at last pointer stop at off position .

 

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Question 2

There are markings as shown on a domestic appliance . Through how many degrees does the pointer turn when it move in the fig.Q1

(i) clockwise from off to cold

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from off to cold option in clockwise direction there are total 2 steps i.e.

First step = off→very cold

Second step = very cold→cold

And single step covers 45º , then 2 steps cover 2×45º = 90º

 

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(ii) anti-clockwise from hot to warm

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from hot to warm option in anti-clockwise direction there are total 1 steps i.e.

hot→warm

And single step covers 45º

 

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(iii) anti-clockwise from very cold to very hot

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from very cold to very hot option in anti-clockwise direction there are total 2 steps i.e.

First step = very cold →off

Second step = off→ very hot

And single step covers 45º , then 2 steps cover 2×45º = 90º

 

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(iv) clockwise from warm to cool

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from warm to cool option in clockwise direction there are total 6 steps i.e.

First step = warm→hot

Second step = hot→very hot

Third step = very hot→off

Fourth step = off→very cold

Fifth step = very cold →cold

Sixth step = cold → cool

And single step covers 45º , then 2 steps cover 6×45º = 270º

 

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Question 3

Answer the following questions related to  a clock in the fig. Q3.

(i) When the seconds hand has moved from 12 to 6 , how many degrees has it turned through ?

Sol :

As we know $=\dfrac{360^º}{12}=30º$ which means 30º per hours  .

And from 12 to 6 we have 6 hours ,= 30º×6

= 180º

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(ii) When the seconds hand has moved from 6 to 8 , how many degrees has it turned through ?

Sol :

As we know $=\dfrac{360^º}{12}=30º$ which means 30º per hours  .

And from 6 to 8 we have 2 hours ,= 30º×2

= 60º

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(iii) What is the time on the clock when hour hand moves clockwise ?

(a) 60º from 6 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 60º by 30º we get $=\dfrac{60^º}{30^º}=2$ which means 2 hours

Which means 6 + 2 = 8 O' clock is the answer

 

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(b) 180º from 10 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 180º by 30º we get $=\dfrac{180^º}{30^º}=2$ which means 6 hours

Which means 10 + 6 = 16 = 12 + 4  i.e at first hour hand moves from 10→12 then 12→4

4 O' clock is the answer

 

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(c) 270º from 12 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 270º by 30º we get $=\dfrac{270^º}{30^º}=9$ which means 9 hours

Which means hour hand moves from 12 → 9

9 O' clock is the answer

 

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(iv) Through how many degrees does the minute hand of a clock turn in 

(a) 1 minute

Sol :

As we know $=\dfrac{360^º}{60^º}=6º$ which means minute hand moves 6º per minutes .

So , in 1 minute it covers 6º

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(b) 8 minutes 

Sol :

As we know minute hand moves 6º per minutes . Then , in 8 minutes it covers 6×8 = 48º

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(c) $\dfrac{3}{4}\text{ hours}$ 

Sol :

Firstly , lets convert hours to minutes . We know 1 hour = 60 minutes

on multiplying both sides by 3/4 we get

$\dfrac{3}{4}\text{ hour}=\dfrac{3}{4}\times 60 \text{ minutes}$

$\dfrac{3}{4}\text{ hour}=\dfrac{3\times 15}{4}\text{ minutes}$

$\dfrac{3}{4}\text{ hour}=45 \text{ minutes}$

and as we know minute hand moves 6º per minutes . Then , in 45 minutes it covers 6×45 = 270º

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(d) $1\dfrac{1}{2}\text{ hours}$

Sol :

Firstly , we convert mixed fraction to improper fraction

$=1\dfrac{1}{2}=\dfrac{3}{2}$

Lets convert hours to minutes . We know 1 hour = 60 minutes

On multiplying both sides by 3/2 we get

$\dfrac{3}{2}\text{ hour}=\dfrac{3}{2}\times 60 \text{ minutes}$

$\dfrac{3}{2}\text{ hour}= 90 \text{ minutes}$

and as we know minute hand moves 6º per minutes . Then , in 90 minutes it covers 6×90 = 540º

 

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(v) Through how many degrees does the hour hand of a clock turn in 

(a) 1 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 1 minute it turns only 0.5º or $\dfrac{1}{2}$

 

(b) 10 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 10 minute it turns 10×0.5º = 5º

 

(c) 30 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 30 minute it turns 30×0.5º = 15º

 

(d) 2 hours 

Sol :

$=\dfrac{360^º}{12}=30^º$ which means hour hand moves 30º per hour

So in 2 hour , the hour hand moves 2×30 = 60º

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Question 4

In the given Fig. Q.4 , find the number of degrees , when we turn in a clockwise directions

(i) From NW to SW

Sol :

 

(ii) From N to S

Sol :

 

(iii) From N to E

Sol :

 

(iv) From NE to W

Sol :

 

 

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Question 5

Look at Fig. Q.5 . The circle is divided into 16 equal parts . Find

(i)  ∠A₁OA₂

(ii) ∠A₁OA3

(iii) ∠A₁OA5

(iv) How many times is the angle ∠A₁OA12 of the angle ∠A₁OA₂ ?

(v) How many times is the angle ∠A₁OA4 of the angle ∠A₁OA2 ?

(vi) What fraction of the whole revolution is the angle ∠A₅OA₇ ?

Sol :

 

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Question 6

Add the following .

(i) 36º 45' and 56º 87'

(ii) 47º 28' 55'' and 27º 35' 49''

 

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Question 7

Subtracting the following

(i) 39º 39' from 50º 15'

(ii) 26º 31'  from 50º 8'

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Multiple Choice Questions (MCQs)

Tick (✔) the correct option.

8. A boat is sailing North East. A little later it is found sailing south. Through what angle has it turned?

(a) $150^{\circ}$

(b) $180^{\circ}$

(c) $135^{\circ}$

(d) $45^{\circ}$

9. The minute hand when it moves $330^{\circ}$ from $110^{\prime}$ Clock is now at

(a) 9 O clock $^{\prime}$

(b) $110^{\prime}$ Clock

(c) 12 O' Clock

(d) 10 O' Clock

RS Aggarwal solution class 8 chapter 17 Construction of Quadrilaterals Exercise 17B

Exercise 17B

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Q1 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 1:

Construct a parallelogram ABCD in which AB = 5.2 cm, BC = 4.7 cm and AC = 7.6 cm.

Answer 1:

Steps of construction:
Step 1: Draw AB = $5.2cm$
Step 2: With B as the centre, draw an arc of $4.7 cm$.
Step 3: With A as the centre, draw another arc of $7.6 cm$, cutting the previous arc at C.
Step 4: Join A and C.
Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with C as the centre, draw an arc of $5.2cm$.
Step 6: With A as the centre, draw another arc of $4.7 cm$, cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Q2 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 2:

Construct a parallelogram ABCD in which AB = 4.3 cm, AD = 4 cm and BD = 6.8 cm.

Answer 2:

Steps of construction:
Step 1: Draw AB= $4.3cm$
Step 2: With B as the centre, draw an arc of $6.8 cm$.
Step 3: With A as the centre, draw another arc of $4cm$, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: We know that the opposite sides of a parallelogram are equal.
            Thus, with D as the centre, draw an arc of $4.3cm$.
Step 6: With B as the centre, draw another arc of $4 cm$, cutting the previous arc at C.
Step 7: Join CD and BC.
​then, ABCD is the required parallelogram.

Q3 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 3:

Construct a parallelogram PQRS in which QR = 6 cm, PQ = 4 cm and ∠PQR = 60° cm.

Answer 3:

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $\angle PQR={60}^{\circ }$
Step 2: With Q as the centre, draw an arc of 6 cm and name that point as R.
Step 3: With R as the centre, draw an arc of 4 cm and name that point as S.
Step 4: Join SR and PS.
Then, PQRS is the required parallelogram.

Q4 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 4:

Construct a parallelogram ABCD in which BC = 5 cm, ∠BCD = 120° and CD = 4.8 cm.

Answer 4:

Steps of construction:
Step 1: Draw BC= $5cm$
Step 2: Make an $\angle BCD={120}^{\circ }$
Step 2: With C as centre draw an arc of $4.8 \mathrm{cm}$, name that point as D
Step 3: With D as centre draw an arc $5\mathrm{cm}$, name that point as A
Step 4: With B as centre draw another arc $4.8 \mathrm{cm}$ cutting the previous arc at A.
Step 5: Join AD and AB
​then, ABCD is a required parallelogram.

Q5 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 5:

Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side.

Answer 5:

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB= $4.4cm$
Step 2: With A as the centre and radius $2.8cm$, draw an arc.
Step 3: With B as the centre and radius $3.5cm$, draw another arc, cutting the previous arc at point O.
Step 4: Join OA and OB.
Step 5: Produce OA to C, such that OC= AO. Produce OB to D, such that OB=OD.
Step 5: Join AD, BC, and CD.
Thus, ABCD is the required parallelogram. The other side is 4.5 cm in length.

Q6 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 6:

Construct a parallelogram ABCD in which AB = 6.5 cm, AC = 3.4 cm and the altitude AL from A is 2.5 cm. Draw the altitude from C and measure it.

Answer 6:

Steps of construction:
Step 1: Draw AB= 6.5cm
Step 2: Draw a perpendicular at point A. Name that ray as AX. From point A, draw an arc of length 2.5 cm on the ray AX and name that point as L.
Step 3: On point L, make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX.
Step 4: Cut an arc of length 3.4 cm on the line YZ and name it as C.
Step 5: From point C, cut an arc of length 6.5 cm on the line YZ. Name that point as D.
 Step 6: Join BC and AD.

Therefore, quadrilateral ABCD is a parallelogram.

The altitude from C measures 2.5 cm in length.

Q7 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 7:

Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm, diagonal BD = 4.6 cm and the angle between AC and BD is 60°.

Answer 7:

 We know that the diagonals of a parallelogram bisect each other.

Steps of construction:
Step 1: Draw AC= $3.8\mathrm{cm}$
Step 2: Bisect AC at O.
Step 3: Make $\angle COX={60}^{\circ }$ 
Produce XO to Y.
Step 4:

Step 5: Join AB, BC, CD and AD.
​Thus, ABCD is the required parallelogram.

Q8 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 8:

Construct a rectangle ABCD whose adjacent sides are 11 cm and 8.5 cm.

Answer 8:

Steps of construction:
Step 1: Draw AB = $11cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
Step 3: Draw an arc of 8.5 cm from point A and name that point as D.
Step 4: Draw an arc of 8.5 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is the required rectangle.

Q9 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 9:

Construct a square, each of whose sides measures 6.4 cm.

Answer 9:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB = $6.4cm$ 
Step 2: Make $$\begin{gathered} \angle A={90}^{\circ } \\ \angle B={90}^{\circ } \end{gathered}$$
Step 3: Draw an arc of length 6.4 cm from point A and name that point as D.
Step 4: Draw an arc of length 6.4 cm from point B and name that point as C.
Step 5: Join C and D.
​Thus, ABCD is a required square.

Q10 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 10:

Construct a square, each of whose diagonals measures 5.8 cm.

Answer 10:

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC= $5.8 \mathrm{cm}$
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:
: $$\begin{gathered} From O: \\ OB=\frac{1}{2}(5.8) \mathrm{cm} = 2.9 \mathrm{cm} \\ OD=\frac{1}{2}(5.8) \mathrm{cm}= 2.9 \mathrm{cm} \end{gathered}$$
Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

Q11 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 11:

Construct a rectangle PQRS in which QR = 3.6 cm and diagonal PR = 6 cm. Measure the other side of the rectangle.

Answer 11:

Steps of construction:
Step 1: Draw QR = $3.6cm$ 
Step 2: Make  $$\begin{gathered} \angle Q={90}^{\circ } \\ \angle R={90}^{\circ } \end{gathered}$$
Step 3:
 

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
​Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
​Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
From point P, draw an arc of length 3.6 cm, cutting the previous arc. Name that point as S.
Step 7: Join P and S.
Thus, PQRS is the required rectangle. The other side is 4.8 cm in length.

Q12 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 12:

Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm.

Answer 12:

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6cm$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:

Draw an arc of length 4 cm on OX and name that point as B.
Draw an arc of length 4 cm on OY and name that point as D.
Step 4 : Join AB, BC, CD and AD.
​Thus, ABCD is the required rhombus, as shown in the figure.

Q13 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 13:

Construct a rhombus ABCD in which AB = 4 cm and diagonal AC is 6.5 cm.

Answer 13:

Steps of construction:
Step 1: Draw AB = $4cm$
Step 2: With B as the centre, draw an arc of $4 \mathrm{cm}$.
Step 3: With A as the centre, draw another arc of $6.5 \mathrm{cm}$, cutting the previous arc at C.
​Step 4: Join AC and BC.
Step 5: With C as the centre, draw an arc of 4 cm.
Step 6: ​With A as the centre, draw another arc of $4 \mathrm{cm}$, cutting the previous arc at D.
Step 7: Join AD and CD.
ABCD is the required rhombus.

Q14 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 14:

Draw a rhombus whose side is 7.2 cm and one angle is 60°.

Answer 14:

Steps of construction:
Step1: Draw AB = 
Step2: Draw
Sum of the adjacent angles is 180°.

Step 3:  

Step 4: Join C and D.
Then, ABCD is the required rhombus.

Q15 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 15:

Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, ∠B = 75° and DC||AB.

Answer 15:

Steps of construction:
Step 1: Draw AB=$6 \mathrm{cm}$
Step 2: Make  $\angle ABX={75}^{\circ }$
Step 3: With B as the centre, draw an arc at $4cm$. Name that point as C.
Step 4:  $AB\parallel CD$


Make 
At C, draw an arc of length $3.2 \mathrm{cm}$.
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Q16 | Ex-17B | Class 8 | RS AGGARWAL | chapter 17 | Construction of Quadrilaterals

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Question 16:

Draw a trapezium ABCD in which AB||DC, AB = 7 cm, BC = 5 cm, AD = 6.5 cm and ∠B = 60°.

Answer 16:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle,
Step3: With B as the centre, draw an arc of $5 \mathrm{cm}$. Name that point as C. Join B and C.
Step4:


Draw an angle,  $\angle BCY, \mathrm{equal} \mathrm{to} 120°.$

Step4: With A as the centre, draw an arc of length $6.5 \mathrm{cm}$, which cuts CY. Mark that point as D.

Step5: Join A and D.

​Thus, ABCD is the required trapezium.