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Page No 222:

Exercise 21A

Question 1:

Find the perimeter of a rectangle in which:
(i) length = 16.8 cm and breadth = 6.2 cm
(ii) length = 2 m 25 cm and breadth = 1 m 50 cm
(iii) length = 8 m 5 dm and breadth = 6 m 8 dm

ANSWER:

We know: Perimeter of a rectangle = 2×(Length+Breadth)2×(Length+Breadth)

(i) Length = 16.8 cm
    Breadth = 6.2 cm
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                   = 2×(16.8+6.2) =46 cm2×(16.8+6.2) =46 cm
(ii) Length = 2 m 25 cm
                  =(200+25) cm       (1 m = 100 cm )
                  = 225 cm  
    Breadth =1 m 50 cm
                  = (100+50) cm      (1 m = 100 cm )
                  = 150 cm    
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                     = 2×(225+150) =750 cm2×(225+150) =750 cm
(iii) Length = 8 m 5 dm
                   = (80+5) dm   (1 m = 10 dm )
                   = 85 dm     
       Breadth = 6 m 8 dm
                     = (60+8) dm   (1 m = 10 dm )
                     = 68 dm  
    Perimeter = 2×(Length+Breadth)2×(Length+Breadth)
                   = 2×(85+68) =306 dm2×(85+68) =306 dm

Page No 222:

Question 2:

Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs 16 per metre.

ANSWER:

Length of the field = 62 m
Breadth of the field = 33 m
Perimeter of the field = 2(l + b) units
                                = 2(62 + 33) m =190 m
Cost of fencing per metre = Rs 16
Total cost of fencing = Rs (16××190) = Rs 3040

Page No 222:

Question 3:

The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field.

ANSWER:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 3x m
Perimeter of the rectangle = 2(l + b)
                                    = 2(5x + 3x) m
                                    = (16x) m
It is given that the perimeter of the field is 128 m.

∴16x=128⇒x=12816=8∴Length =(5×8)=40mBreadth =(3×8)=24m∴16x=128⇒x=12816=8∴Length =(5×8)=40mBreadth =(3×8)=24m

Page No 222:

Question 4:

The cost of fencing a rectangular field at Rs 18 per metre is Rs 1980. If the width of the field is 23 m, find its length.

ANSWER:

Total cost of fencing = Rs 1980
Rate of fencing = Rs 18 per metre

Perimeter of the field = Total costRate=Rs 1980Rs 18/m=(198018) m=110 mTotal costRate=Rs 1980Rs 18/m=(198018) m=110 m

Let the length of the field be x metre.
Perimeter of the field = 2(x + 23) m

∴2(x+23)=110⇒(x+23)=55x=(55−23)=32∴2(x+23)=110⇒(x+23)=55x=(55-23)=32
Hence, the length of the field is 32 m.

Page No 222:

Question 5:

The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs 25 per metre is Rs 3300. Find the dimensions of the field.

ANSWER:

Total cost of fencing = Rs 3300
Rate of fencing = Rs 25/m
Perimeter of the field = Total cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 mTotal cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 m

Let the length and the breadth of the rectangular field be 7x and 4x, respectively.
Perimeter of the field = 2(7x + 4x) = 22x

It is given that the perimeter of the field is 132 m.

∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m

Page No 222:

Question 6:

Find the perimeter of a square, each of whose sides measures:
(i) 3.8 cm
(ii) 4.6 cm
(iii) 2 m 5 dm

ANSWER:

(i) Side of the square = 3.8 cm
    Perimeter of the square = (4××side)
                                    = (4××3.8) = 15.2 cm

(ii) Side of the square = 4.6 cm
     Perimeter of the square = (4××side)
                                     = (4××4.6) = 18.4 cm

(iii) Side of the square = 2 m 5 dm
                                     = (20+5) dm   (1 m = 10 dm)
                                     = 25 dm
      Perimeter of the square = (4××side)
                                       = (4××25) = 100 dm

Page No 222:

Question 7:

The cost of putting a fence around a square field at Rs 35 per metre is Rs 4480. Find the length of each side of the field.

ANSWER:

Total cost of fencing = Rs 4480
Rate of fencing = Rs 35/m
Perimeter of the field = Total costRate=Rs 4480Rs 35/m=448035 m=128 mTotal costRate=Rs 4480Rs 35/m=448035 m=128 m

Let the length of each side of the field be x metres.
Perimeter = (4x) metres
∴4x=128⇒x=1284=32 ∴4x=128⇒x=1284=32 

Hence, the length of each side of the field is 32 m.

Page No 222:

Question 8:

Each side of a square field measures 21 m. Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both the fields are equal, find the dimensions of the rectangular field.

ANSWER:

Side of the square field = 21m
Perimeter of the square field = (4××21) m
                                       = 84 m  

Let the length and the breadth of the rectangular field be 4x and 3x, respectively.
 Perimeter of the rectangular field = 2(4x + 3x) = 14x

 Perimeter of the rectangular field = Perimeter of the square field

∴14x=84 ⇒x=8414=6∴14x=84 ⇒x=8414=6
∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m

Page No 222:

Question 9:

Find the perimeter of
(i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm,
(ii) an equilateral triangle of side 9.4 cm,
(iii) an isosceles triangle with equal sides 8.5 cm each and third side 7 cm.

ANSWER:

(i) Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side) cm
                                      = (7.8 + 6.5 + 5.9) cm
                                      = 20.2 cm

(ii) In an equilateral triangle, all sides are equal.
     Length of each side of the triangle = 9.4 cm
     ∴∴ Perimeter of the triangle = (3 ×× Side) cm
                                            = (3 ×× 9.4) cm
                                            = 28.2 cm

(iii) Length of two equal sides = 8.5 cm
       Length of the third side = 7 cm
    ∴∴ Perimeter of the triangle = {(2 ×× Equal sides) + Third side} cm
                                           = {(2 ×× 8.5) + 7} cm
                                           = 24 cm

Page No 222:

Question 10:

Find the perimeter of
(i) a regular pentagon of side 8 cm,
(ii) a regular octagon of side 4.5 cm,
(iii) a regular decagon of side 3.6 cm,

ANSWER:

(i) Length of each side of the given pentagon = 8 cm
   ∴∴ Perimeter of the pentagon = (5××8) cm
                                                  = 40 cm

(ii) Length of each side of the given octagon = 4.5 cm
   ∴∴ Perimeter of the octagon = (8××4.5) cm
                                                = 36 cm

(iii) Length of each side of the given decagon = 3.6 cm
   ∴∴ Perimeter of the decagon = (10××3.6) cm
                                                 = 36 cm

Page No 222:

Question 11:

Find the perimeter of each of the following figures:
Figure

ANSWER:

(i) Perimeter of the figure = Sum of all the sides
                                         =(27 + 35 + 35 + 45) cm
                                         = 142 cm
(ii) Perimeter of the figure = Sum of all the sides
                                         =(18 + 18 + 18 + 18) cm
                                         = 72 cm
(iii) Perimeter of the figure = Sum of all the sides
                                         =(8 + 16 + 4 + 12 + 12 + 16 + 4) cm
                                         = 72 cm

Page No 224:

Exercise 21B

Question 1:

Find the circumference of a circle whose radius is
(i) 28 cm
(ii) 10.5 cm
(iii) 3.5 m

ANSWER:

(i) Radius, r = 28 cm

∴ Circumference of the circle, C=2πr                                                    =(2×227×28)                                =176 cmHence, the circumference of the given circle is 176 cm.∴ Circumference of the circle, C=2πr                                                    =(2×227×28)                                =176 cmHence, the circumference of the given circle is 176 cm.

(ii) Radius, r = 10.5 cm
      ∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.

(iii) Radius, r = 3.5 m
     ∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.

Page No 224:

Question 2:

Find the circumference of a circle whose diameter is
(i) 14 cm
(ii) 35 cm
(iii) 10.5 m

ANSWER:

(i)

Circumference=2πr                          =π(2r)                            =π× Diameter of the circle (d)       (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.Circumference=2πr                          =π(2r)                            =π× Diameter of the circle (d)       (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.

(ii)
Circumference=2πr                          =π(2r)                        =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.Circumference=2πr                          =π(2r)                        =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.

(iii)
Circumference=2πr                         =π(2r)                         =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.Circumference=2πr                         =π(2r)                         =π×Diameter of the circle(d)       (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.

Page No 224:

Question 3:

Find the radius of a circle whose circumference is 176 cm.

ANSWER:

Let the radius of the given circle be r cm.
Circumference of the circle = 176 cm
Circumference = 2πr2πr
∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.

Page No 224:

Question 4:

Find the diameter of a wheel whose circumference is 264 cm.

ANSWER:

Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.

Page No 224:

Question 5:

Find the distance covered by the wheel of a car in 500 revolutions if the diameter of the wheel is 77 cm.

ANSWER:

Radius of the wheel =Diameter of the wheel2Diameter of the wheel2
⇒r=772cm⇒r=772cm
Circumference of the wheel =2π r2π r
=(2×227×772)=242 cm=(2×227×772)=242 cm

In 1 revolution the wheel covers a distance equal to its circumference.

∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm                                                                =121000 cm                                                               =1210 m   (100 cm= 1m )                                                               =1.21 km   (1000 m=1 km  )∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm                                                                =121000 cm                                                               =1210 m   (100 cm= 1m )                                                               =1.21 km   (1000 m=1 km  )

Page No 224:

Question 6:

The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel 1.65 km?

ANSWER:

Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35)                                   =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution    (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions                                        = 750   revolutions                 Thus,the wheel will make 750 revolutions to travel 1.65 km.Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35)                                   =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution    (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions                                        = 750   revolutions                 Thus,the wheel will make 750 revolutions to travel 1.65 km.

Page No 226:

Exercise 21C

Question 1:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 12 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares ×× Area of the square
                                         =(12×1) sq cm(12×1) sq cm
                                         =12 sq cm

Page No 226:

Question 2:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 18 complete squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of complete squares ×× Area of the square
                                         =(18×1) sq cm(18×1) sq cm
                                         =18 sq cm

Page No 226:

Question 3:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 14 complete squares and 1 half square.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(14 × 1) + (1 × 12)]sq cm(14 × 1) + (1 × 12)sq cm
                                         =14121412 sq cm

Page No 226:

Question 4:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 6 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
      ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(6×1)+(4×12)] sq cm(6×1)+(4×12) sq cm
                                         =8 sq cm

Page No 226:

Question 5:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 9 complete squares and 6 half squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares ×× Area of the square
                                         =[(9 × 1) + (6 × 12)] sq cm(9 × 1) + (6 × 12) sq cm
                                         =12 sq cm

Page No 226:

Question 6:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 16 complete squares.
    Area of 1 small square = 1 sq cm
     ∴ Area of the figure = Number of squares ×× Area of a square
                                         =(16×1) sq cm(16×1) sq cm
                                         =16 sq cm

Page No 226:

Question 7:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

In the given figure, there are 4 complete squares, 8 more than half parts of squares and 4 less than half parts of squares.
We neglect the less than half parts and consider each more than half part of the square as a complete square.

                  ∴ Area = (4 + 8) sq cm
                            = 12 sq cm

Page No 226:

Question 8:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

In the given figure, there are 9 complete squares, 5 more than half parts of squares and 7 less than half parts of squares.
We neglect the less than half parts of squares and consider the more than half squares as complete squares.
∴ Area of the figure = (9 + 5) sq cm
                                      = 14 sq cm

Page No 226:

Question 9:

Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm².
Figure

ANSWER:

The figure contains 14 complete squares and 4 half squares.
    Area of 1 small square = 1 sq cm
     Area of the figure = Number of squares ×× Area of one square
                                         =[(14×1)+(4×12) ]sq cm(14×1)+(4×12) sq cm
                                         =16 sq cm

Page No 229:

Exercise 21D

Question 1:

Find the area of a rectangle whose
(i) length = 46 cm and breadth = 25 cm
(ii) length = 9 m and breadth = 6 m
(iii) length = 14.5 m and breadth = 6.8 m
(iv) length = 12 m 5 cm and breadth = 60 cm
(v) length = 3.5 km and breadth = 2 km

ANSWER:

 (i) Length = 46 cm
Breadth = 25 cm
  Area of the rectangle = (Length ××Breadth) sq units
                              = (46××25) cm² = 1150 cm²
 
(ii) Length = 9 m
Breadth = 6 m
  Area of the rectangle = (Length ××Breadth) sq units
                             = (9××6) m² = 54 m²

 (iii) Length = 14.5 m
Breadth = 6.8 m
  Area of the rectangle = (Length ××Breadth) sq units
                             = (14510×681014510×6810) m² = 98601009860100 m² =98.60 m²

 (iv) Length = 2 m 5 cm
                   = (200+5) cm   (1 m = 100 cm )
                   =205cm
       Breadth = 60 cm
       Area of the rectangle = (Length ××Breadth) sq units
                                    = (205××60) cm² = 12300 cm²

 (v) Length = 3.5 km
Breadth = 2 km
  Area of the rectangle = (Length ××Breadth) sq units
                             = (3.5××2) km² = (3510×2)(3510×2) km² =7 km² 

Page No 230:

Question 2:

Find the area of a square plot of side 14 m.

ANSWER:

Side of the square plot = 14 m
Area of the square plot = (Side)² sq units
                                     = (14)² m²
                                     = 196  m²

Page No 230:

Question 3:

The top of a table measures 2 m 25 cm by 1 m 20 cm. Find its area in square metres.

ANSWER:

Length of the table = 2 m 25 cm
                         = (2 + 0.25) m     (100 cm = 1 m)
                         = 2.25 m
Breadth of the table = 1 m 20 cm
                                 = (1 + 0.20) m    (100 cm = 1 m)
                                 =1.20 m
Area of the table = (Length × Breadth) sq units
                             = (2.25 × 1.20) m²
       
                              = (225100×120100)(225100×120100) m²
                              = 2.7 m²

Page No 230:

Question 4:

A carpet is 30 m 75 cm long and 80 cm wide. Find its cost at Rs 150 per square metre.

ANSWER:

Length of the carpet = 30 m 75 cm
                           =(30 + 0.75) cm        (100 cm = 1 m)
                           = 30.75 m
Breadth of the carpet = 80 cm
                            = 0.80 m                (100 cm = 1 m)

Area of carpet = ( Length ×× breadth ) sq units

                            =(30.75 × 0.80) m2= (3075100×80100) m2=24.6 m2=(30.75 × 0.80) m2= (3075100×80100) m2=24.6 m2

Cost of 1 m² carpet= Rs 150
Cost of 24.6 m² carpet = Rs (24.6××150)
                                      =  Rs 3690

                      

Page No 230:

Question 5:

How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72 cm. If each envelope requires a piece of paper of size 18 cm by 12 cm?

ANSWER:

Length of the sheet of paper = 3 m 24 cm = 324 cm
Breadth of the sheet of paper = 1 m 72 cm = 172 cm
Area of the sheet = (Length ×× Breadth)
                           =(324×172) cm2 =55728 cm2=(324×172) cm2 =55728 cm2

Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
Area of the piece of paper required to make 1 envelope = (18××12) cm²
                                                                                   = 216 cm²

No. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopesNo. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopes

Page No 230:

Question 6:

A room is 12.5 m long and 8 m wide. A square carpet of side 8 m is laid on its floor. Find the area of the floor which is not carpeted.

ANSWER:

Length of the room = 12.5 m
Breadth of the room = 8 m
Area of the room = (Length××Breadth)
                     =(12.5×8) m2 = 100 m2=(12.5×8) m2 = 100 m2
Side of the square carpet = 8 m
Area of the carpet = (Side)²
                              = 8² m²
                              = 64 m²

Area of the floor which is not carpeted = Area of the room − Area of the carpet
                                                               = (100 − 64) m²
                                                               = 36 m²

Page No 230:

Question 7:

A lane, 150 m long and 9 m wide, is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. Find the number of bricks required.

ANSWER:

Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length××Breadth)
                     =15000×900  cm2=13500000 cm2=15000×900  cm2=13500000 cm2
Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length××Breadth)
                             =(22.5×7.5)  cm2=168.75 cm2=(22.5×7.5)  cm2=168.75 cm2

Number of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricksNumber of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricks

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Question 8:

A room is 13 m long and 9 m broad. Find the cost of carpeting the room with a carpet 75 cm broad at the rate of Rs 65 per metre.

ANSWER:

Length of the room = 13 m
Breadth of the room = 9 m
Area of the room = (13××9) m² = 117 m²

Let length of required carpet be x m.
Breadth of the carpet = 75 cm
                            = 0.75 m         (100 cm = 1 m)
Area of the carpet = (0.75××x) m²
                       = 0.75x m²
For carpeting the room:
Area covered by the carpet = Area of the room
     ⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m

So, the length of the carpet is 156 m.
Cost of 1 m carpet = Rs 65
Cost 156 m carpet = Rs (156××65)
                              = Rs 10140

Page No 230:

Question 9:

The length and the breadth of a rectangular park are in the ratio 5 : 3 and its perimeter is 128 m. Find the area of the park.

ANSWER:

Let the length of the rectangular park be 5x.
∴ Breadth of the rectangular park = 3x
Perimeter of the rectangular field = 2(Length + Breadth)
                                                      =2(5x + 3x)
                                                      = 16x

It is given that the perimeter of rectangular park is 128 m.
⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m                                            =40 mBreadth of the park=(3×8) m                                     =24 m⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m                                            =40 mBreadth of the park=(3×8) m                                     =24 m

Area of the park = (Length ×× Breadth) sq units
                           
                          =(40×24) m2=960 m2=(40×24) m2=960 m2

Page No 230:

Question 10:

Two plots of land have the same perimeter. One is a square of side 64 m and the other is a rectangle of length 70 m. Find the breadth of the rectangular plot. Which plot has the greater area and by how much?

ANSWER:

Side of the square plot = 64 m
Perimeter of the square plot = (4×Side) m =(4×64) m=256 m(4×Side) m =(4×64) m=256 m
Area of the square plot = (Side)²
= 64² m²
= 4096 m²

Let the breadth of the rectangular plot be x m.
Perimeter of the rectangular plot = 2(l+b)  m
= 2(70+x) m

Perimeter of the rectangular plot = Perimeter of the square plot   (Given)
⇒2(70+x)=256⇒140+2x=256⇒2x=256−140⇒2x=116⇒x=1162=58⇒2(70+x)=256⇒140+2x=256⇒2x=256-140⇒2x=116⇒x=1162=58
So, the breadth of the rectangular plot is 58 m.
Area of the rectangular plot = (Length × Breadth)=(70 × 58) m2=4060 m2(Length × Breadth)=(70 × 58) m2=4060 m2
Area of the square plot − Area of the rectangular plot
= (4096 − 4060)
= 36 m²
Area of the square plot is 36 m² greater than the rectangular plot.

Page No 230:

Question 11:

The cost of cultivating a rectangular field at Rs 35 per square metre is Rs 71400. If the width of the field is 40 m, find its length. Also, find the cost of fencing the field at Rs 50 per metre.

ANSWER:

Total cost of cultivating the field = Rs 71400
Rate of cultivating the field = Rs 35/m²


Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2

Let the length of the field be x m.

Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m 

Perimeter of the field = 2(l+b)
= 2(51+40) m
= 182 m
Cost of fencing 1 m of the field = Rs 50
Cost of fencing 182 m of the field = Rs (182××50)
= Rs 9100

Page No 230:

Question 12:

The area of a rectangle is 540 cm² and its length is 36 cm. Find its width and perimeter.

ANSWER:

Let the width of the rectangle be x cm.
Length of the rectangle = 36 cm
Area of the rectangle = (Length × WidthLength × Width) = (36 × x36 × x) cm²
It is given that the area of the rectangle is 540 cm².

⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm

Perimeter of the rectangle = 2(Length + Width) cm
= 2(36 + 15) cm
= 102 cm

Page No 230:

Question 13:

A marble tile measures 12 cm × 10 cm. How many tiles will be required to cover a wall of size 4 m by 3 m? Also, find the total cost of the tiles at Rs 22.50 per tile.

ANSWER:

Length of the wall = 4 m = 400 cm
Breadth of the wall = 3 m = 300 cm
Area of the wall = (400×300) cm² = 120000 cm²

Length of the tile = 12 cm
Breadth of the tile = 10 cm
Area of one tile = (12×10) cm² = (120) cm²

Number of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tilesNumber of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tiles
Cost of 1 tile = Rs 22.50
Cost of 1000 tiles = (1000 × 22.50) = Rs 22500

Thus, the total cost of the tiles is Rs 22500.

Page No 230:

Question 14:

Find the perimeter of a rectangle whose area is 600 cm² and breadth is 25 cm.

ANSWER:

Let the length of the rectangle be x cm.
Breadth of the rectangle is 25 cm.
Area of the rectangle = (Length × Breadth) cm²
                                   = (x×25) cm²
                                   =25x cm²

It is given that the area of the rectangle is 600 cm².
⇒25x=600⇒x=60025=24⇒25x=600⇒x=60025=24
So, the length of the rectangle is 24 cm.
Perimeter of the rectangle = 2(Length + Breadth) units
                                           = 2(25 + 24) cm
                                           = 98 cm

Page No 230:

Question 15:

Find the area of a square whose diagonal is 52–√52 cm.

ANSWER:

Area of the square ={12× (Diagonal)2} sq units 12× (Diagonal)2 sq units 
 = {12×(52–√)2} cm2={12×(5)2×(2–√)2} cm2={12×25×2} cm2=(12×50) cm2= 25 cm2 = 12×(52)2 cm2=12×(5)2×(2)2 cm2=12×25×2 cm2=(12×50) cm2= 25 cm2

Page No 230:

Question 16:

Calculate the area of each one of the shaded regions given below:
Figure

ANSWER:

(i) Area of rectangle ABDC = Length ×× Breadth
                                         = AB××AC                    (AC = AE − CE)
                                         = (1×8)m21×8m2
                                         = 8 m²
   Area of rectangle CEFG = Length ×× Breadth
                                         = CG××GF               (CG = GD + CD)           
                                         = (9×2)m29×2m2
                                         = 18 m²
   Area of the complete figure = Area of rectangle ABDC + Area of rectangle CEFG
                                                 = (8 + 18) m²
                                                 =  26 m²


(ii) Area of rectangle AEDC = Length ×× Breadth
                                         = ED ×× CD              
                                         = (12×2)m212×2m2
                                         = 24 cm²
   Area of rectangle FJIH = Length ×× Breadth
                                         = HI ×× IJ                   
                                         = (1×9)m21×9m2
                                         = 9 m²
Area of rectangle ABGF = Length ×× Breadth
                                         = AB ×× AF                                  {(AB = FJ − GJ) and AF = EH − (EA + FH)}                 
                                         = (7×1.5)m27×1.5m2
                                         = 10.5 m²

   Area of the complete figure = Area of rectangle AEDC + Area of rectangle FJIH + Area of rectangle ABGF
                                         = (24 + 9 + 10.5) m²
                                         = 43.5 m²



(iii) Area of the shaded portion = Area of the complete figure − Area of the unshaded figure
                                           = Area of rectangle ABCD − Area of rectangle GBFE
                                           =(CD××AD) − (GB××BF)
                                           ={(12×9)−(7.5×10)}m2(12×9)-(7.5×10)m2                                   (BF = BC − FC)
                                           =(108 − 75) m²

                                          =33 m²

Page No 231:

Question 17:

Calculate the area of each one of the shaded regions given below (all measures are given in cm):
Figure

ANSWER:

(i) Area of square BCDE= (Side)²
                                        = (CD)²
                                        = (3)² cm²        
                                        = 9 cm²
      Area of rectangle ABFK = Length × BreadthLength × Breadth
                                             = AK××AB             [(AB = AC − BC) and (AK = AL + LK)
                                             = (5××1) cm²  
                                             = 5 cm²

     Area of rectangle MLKG = Length × BreadthLength × Breadth
                                             = ML ×× MG
                                             = (2 ×× 3) cm²
                                             = 6 cm²
     Area of rectangle JHGF= Length × BreadthLength × Breadth
                                             = JH××HG
                                             = (2××4) cm²
                                             = 8 cm²
       Area of the figure = Area of rectangle ABFK + Area of rectangle MLKG + Area of rectangle JHGF + Area of square BCDE
                               = (9 + 5 + 6 + 8) cm²
                               = 28 cm²
                   
(ii) Area of rectangle CEFG= Length × BreadthLength × Breadth
                                             = EF××CE
                                             = (1××5) cm²          (CE = EA − AC)
                                             = 5 cm²
      Area of rectangle ABDC = Length × BreadthLength × Breadth
                                             = AB××BD
                                             = (1××2) cm²  
                                             = 2 cm²

     Area of rectangle HIJG = Length × BreadthLength × Breadth
                                             = HI ×× IJ
                                             = (1××2) cm²
                                             = 2 cm²
       Area of the figure = Area of rectangle CEFG + Area of rectangle HIJG + Area of rectangle ABDC
                               = (5+2+2) cm²
                               = 9 cm²       
                      
(iii) In the figure, there are 5 squares, each of whose sides are 6 cm in length.
     Area of the figure = 5 ×× Area of square
                                   = 5××(side)²
                                   = 5××(6)² cm²
                                   = 180 cm²

Page No 231:

Exercise 21E

Question 1:

The sides of a rectangle are in the ratio 7 : 5 and its perimeter is 96 cm. The length of the rectangle is
(a) 21 cm
(b) 28 cm
(c) 35 cm
(d) 14 cm

ANSWER:

(b) 28 cm

Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively.
It is given that the perimeter of the rectangle is 96 cm.
Perimeter of the rectangle = 2(7x+5x) cm

⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm

Page No 231:

Question 2:

The area of a rectangle is 650 cm² and its breadth is 13 cm. The perimeter of the rectangle is
(a) 63 cm
(b) 130 cm
(c) 100 cm
(d) 126 cm

ANSWER:

(d) 126 cm
Let length of the rectangle be L cm.
Area of the rectangle = 650 cm²
Area of the rectangle = (L×13L×13) cm²
⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm

Perimeter of the rectangle = 2(Length + Breadth) cm = 2(50+13) cm = 126 cm

Page No 231:

Question 3:

The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 22.50 per metre is
(a) Rs 2430
(b) Rs 2340
(c) Rs 2400
(d) Rs 3340

ANSWER:

(b) Rs 2340
Perimeter of the rectangular field = 2(Length + Breadth)
                                                  = 2(34 + 18) m = 104 m
Cost of fencing 1 metre = Rs 22.50
Cost of fencing 104 m = Rs (22.50××104) = Rs 2340

Page No 231:

Question 4:

The cost of fencing a rectangular field at Rs 30 per metre is Rs 2400. If the length of the field is 24 m, then its breadth is
(a) 8 m
(b) 16 m
(c) 18 m
(d) 24 m

ANSWER:

(b) 16 m
Total cost of fencing = Rs 2400
Rate of fencing = Rs 30/m
Perimeter of the rectangular field = Total costRate=Rs 2400Rs 30/m=80 mTotal costRate=Rs 2400Rs 30/m=80 m
Let the breadth of the rectangular field be x m.
Perimeter of the rectangular field = 2(24 + x) m
⇒2(24+x)=80⇒48+2x=80⇒2x=(80−48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.⇒2(24+x)=80⇒48+2x=80⇒2x=(80-48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.

Page No 231:

Question 5:

The area of a rectangular carpet is 120 m² and its perimeter is 46 m. The length of its diagonal is
(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m

ANSWER:

(c) 17 m
Let the length and the breadth of the rectangle be L m and B m, respectively.

Area of the rectangular carpet = (L×BL×B) m²
⇒LB=120           … (i)⇒LB=120           … (i)
Perimeter of the rectangular carpet = 2(L+B)2(L+B)
⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23      …(ii)⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23      …(ii)

Diagonal of the rectangle = L2+B2−−−−−−−√ m                                   =(L+B)2−2LB−−−−−−−−−−−−−√ m                                                =(23)2−240−−−−−−−−−√ m                            (from equations (i) and (ii))                                  =529−240−−−−−−−−√ m                                  =289−−−√ m=17 mDiagonal of the rectangle = L2+B2 m                                   =(L+B)2-2LB m                                                =(23)2-240 m                            (from equations (i) and (ii))                                  =529-240 m                                  =289 m=17 m

Page No 231:

Question 6:

The length of a rectangle is three times its width and the length of its diagonal is 610−−√610 cm. The perimeter of the rectangle is
(a) 48 cm
(b) 36 cm
(c) 24 cm
(d) 2410 −−−√2410  cm

ANSWER:

(a) 48 cm
Let the width  and the length of the rectangle be x cm and 3x cm, respectively.

Applying Pythagoras theorem:

(Diagonal)2=(Length)2+(Width)2⇒(610−−√)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect −6.(Diagonal)2=(Length)2+(Width)2⇒(610)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect -6.

So, width of the rectangle is 6 cm.
Length of the rectangle = (3×6)=18 cm3×6=18 cm
Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm

Page No 231:

Question 7:

If the ratio between the length and perimeter of a rectangular plot is 1 : 3, then the ratio between the length and breadth of the plot is
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3

ANSWER:

(b) 2 : 1
Let the breadth of the plot be b cm.

Let the length of the plot be x cm.
Perimeter of the plot = 3x cm

Perimeter of the plot =2(Length + Breadth)= 2(x + b) cm
⇒2(x+b)=3x2x+2b=3x⇒2b=3x−2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1⇒2(x+b)=3x2x+2b=3x⇒2b=3x-2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1

Page No 232:

Question 8:

The length of the diagonal of a square is 20 cm. It area is
(a) 400 cm²
(b) 200 cm²
(c) 300 cm²
(d) 1002–√1002 cm²

ANSWER:

(b) 200 cm²
Area of the square = {12×(Diagonal)2 } sq units12×(Diagonal)2  sq units
={12×(20)2 } cm2={12×(20)×(20)} cm2=(20×10) cm2=200 cm2=12×(20)2  cm2=12×(20)×(20) cm2=(20×10) cm2=200 cm2

Page No 232:

Question 9:

The cost of putting a fence around a square field at Rs 25 per metre is Rs 2000. The length of each side of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) none of these

ANSWER:

(c) 20 m
Let one side of the square field be x m.
Total cost of fencing a square field = Rs 2000
Rate of fencing the field = Rs 25/m

Perimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 mPerimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 m

Perimeter of the square field = (4×side4×side) = 4x m
⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.

Page No 232:

Question 10:

The diameter of a circle is 7 cm. Its circumference is
(a) 44 cm
(b) 22 cm
(c) 28 cm
(d) 14 cm

ANSWER:

(b) 22 cm
Radius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cmRadius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cm

Page No 232:

Question 11:

The circumference of a circle is 88 cm. Its diameter is
(a) 28 cm
(b) 42 cm
(c) 56 cm
(d) none of these

ANSWER:

(a) 28 cm
Circumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cmCircumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cm

Page No 232:

Question 12:

The diameter of a wheel of a car is 70 cm. How much distance will it cover in making 50 revolutions?
(a) 350 m
(b) 110 m
(c) 165 m
(d) 220 m

ANSWER:

(b) 110 m
Radius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 mRadius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 m

Page No 232:

Question 13:

A lane 150 m long and 9 m wide is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. How many bricks are required?
(a) 65000
(b) 70000
(c) 75000
(d) 80000

ANSWER:

(d) 80000
Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length × Breadth)
= (15000 × 900) cm²
= 13500000 cm²

Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length × Breadth)
= ( 22.5 × 7.5 ) cm²
= 168.75 cm²

Number of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricksNumber of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricks

Page No 232:

Question 14:

A room is 5 m 40 cm long and 4 m 50 cm broad. Its area is
(a) 23.4 m²
(b) 24.3 m²
(c) 25 m²
(d) 98.01 m²

ANSWER:

(b) 24.3 m²

Length of the room = 5 m 40 cm = 5.40 m
Breadth of the room = 4 m 50 cm = 4.50 m

Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=(540100×450100)m2=(275×92)m2=24310m2=24.3 m2Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=540100×450100m2=275×92m2=24310m2=24.3 m2

Page No 232:

Question 15:

How many envelopes can be made out of a sheet of paper 72 cm by 48 cm, if each envelope requires a paper of size 18 cm by 12 cm?
(a) 4
(b) 8
(c) 12
(d) 16

ANSWER:

(d) 16

Length of the sheet of paper = 72 cm
Breadth of the sheet of paper = 48 cm
Area of the sheet = (Length × Breadth)
⇒ ( 72 × 48 ) cm²  = 3456 cm²

Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
Area of the piece of paper required to make 1 envelope = (18 × 12) cm²
= 216 cm²

No. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopesNo. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopes

Page No 233:

Exercise 21F

Question 1:

Find the perimeter of the following shapes:
(i) a triangle whose sides are 5.4 cm, 4.6 cm and 6.8 cm
(ii) a regular hexagon of side 8 cm
(iii) an isosceles triangle with equal sides 6 cm each and third side 4.5 cm.

ANSWER:

(i) Sides of the triangle are 5.4 cm, 4.6 cm and 6.8 cm. 
    Perimeter of the triangle = (First side + Second side + Third Side)
                                   = (5.4 + 4.6 + 6.8) cm = 16.8 cm

(ii) Length of each side of the given hexagon = 8 cm
   ∴ Perimeter of the hexagon = (6 × 8) cm = 48 cm

(iii) Length of the two equal sides = 6 cm
     Length of the third side = 4.5 cm
    ∴ Perimeter of the triangle = {(2 × equal sides) + third side} cm = (2 × 6) + 4.5 = 16.5 cm

Page No 233:

Question 2:

The perimeter of a rectangular field is 360 m and its breadth is 75 m. Find its length.

ANSWER:

Let the length of the rectangle be x m.
Breadth of the rectangle = 75 m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(x + 75) m = (2x + 150) m
It is given that the perimeter of the field is 360 m.
⇒2x+150=360⇒2x=360−150⇒2x=210⇒x=2102=105 ⇒2x+150=360⇒2x=360-150⇒2x=210⇒x=2102=105 
So, the length of the rectangle is 105 m.

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Question 3:

The length and breadth of a rectangular field are in the ratio 5 : 4. If its perimeter is 108 m, find the dimensions of the field.

ANSWER:

Let the length of the rectangle be 5x m.
Breadth of the rectangle = 4x m
Perimeter of the rectangle = 2(Length + Breadth)
                                  = 2(5x + 4x) m = 18x m
It is given that the perimeter of the field is 108 m.
∴ 18 x = 108
⇒ x = 10818=610818=6
∴ Length of the field = ( 5 × 6 )m = 30 m
Breadth of the field = ( 4 × 6 )m = 24 m

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Question 4:

Find the area of a square whose perimeter is 84 cm.

ANSWER:

Let one side of the square be x cm.
Perimeter of the square = (4×side)=(4×x) cm =4x cm(4×side)=(4×x) cm =4x cm
It is given that the perimeter of the square is 84 cm.
⇒4x=84⇒x=844=21Thus, one side of the square is  21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2⇒4x=84⇒x=844=21Thus, one side of the square is  21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2

Page No 233:

Question 5:

The area of a room is 216 m² and its breadth is 12 m. Find the length of the room.

ANSWER:

Let the length of the room be x m.
Breadth of the room = 12 m
Area of the room = (Length × Breadth) = (x × 12) m²
It is given that the area of the room is 216 m².
⇒ x × 12 = 216
⇒ x = 21612=1821612=18
∴ Length of the rectangle = 18 m

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Question 6:

Find the circumference of a circle of radius 7 cm. [Take π = 227π = 227]

ANSWER:

Radius(r) of the given circle = 7 cm
Circumference of the circle, C = 2 πr
                                                 = (2×227×7) cm= 44 cm= 2×227×7 cm= 44 cm
Hence, the circumference of the given circle is 44 cm.

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Question 7:

The diameter of a wheel of a car is 77 cm. Find the distance covered by the wheel in 500 revolutions.

ANSWER:

Radius of the wheel =Diameter of the wheel2Diameter of the wheel2
⇒ r = 772772cm
Circumference of the wheel = 2 πr
= (2×227×772)2×227×772 cm
= 242 cm

In 1 revolution, the wheel covers a distance equal to its circumference.
∴ Distance covered by the wheel in 1 revolution = 242 cm
∴ Distance covered by the wheel in 500 revolutions = ( 500 × 242 ) cm
                                                                               = 121000 cm       (100 cm =1 m)
                                                                               = 1210 m

Page No 233:

Question 8:

Find the diameter of a wheel whose circumference is 176 cm.

ANSWER:

Let the radius be r cm.
Diameter = 2 × Radius(r) = 2r cm
Circumference of the wheel = 2πr
∴ 2πr = 176
⇒ 2r=176π⇒ 2r=176×722=56⇒ 2r=176π⇒ 2r=176×722=56
⇒ 2r = 56
Thus, the diameter of the given wheel is 56 cm.

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Question 9:

Find the area of a rectangle whose length is 36 cm and breadth 15 cm.

ANSWER:

Length of the rectangle = 36 cm 
Breadth of the rectangle = 15 cm
Area of the rectangle = (Length × Breadth) sq units
                             = (36 × 15) cm² = 540 cm²

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Question 10:

Perimeter of a square of side 16 cm is
(a) 256 cm
(b) 64 cm
(c) 32 cm
(d) 48 cm

ANSWER:

(b) 64 cm
Side of the square = 16 cm
 Perimeter of the square = (4 × side)
                               = (4 × 16) cm 
                               = 64 cm

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Question 11:

The area of a rectangle is 240 m² and its length is 16 m. Then, its breadth is
(a) 15 m
(b) 16 m
(c) 30 m
(d) 40 m

ANSWER:

(a) 15 m

Let the breadth of the rectangle be x m.
Length of the rectangle = 16 m
Area of rectangle = (Length × Breadth) = (16 × x) m²
It is given that the area of the rectangle is 240 m².
⇒ 16 × x = 240
⇒ x = 24016=1524016=15
So, the breadth of the rectangle is 15 m.

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Question 12:

The area of a square lawn of side 15 m is
(a) 60 m²
(b) 225 m²
(c) 45 m²
(d) 120 m²

ANSWER:

(b) 225 m²

Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2

Page No 233:

Question 13:

The area of a square is 256 cm². The perimeter of the square is
(a) 16 cm
(b) 32 cm
(c) 48 cm
(d) 64 cm

ANSWER:

(a) 16 cm
Let one side of the square be x cm.
Area of the square = (Side )² cm² = x² cm²
It is given that the area of the square is 256 cm².
⇒ x² = 256
⇒ x = 256−−−√=±16256=±16
We know that the side of a square cannot be negative.
So, we will neglect −16.
Therefore, the side of the square is 16 cm.

Perimeter of the square = (4×side)=(4×16)cm=64 cm4×side=4×16cm=64 cm

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Question 14:

The area of a rectangle is 126 m² and its length is 12 m. The breadth of the rectangle is
(a) 10 m
(b) 10.5 m
(c) 11 m
(d) 11.5 m

ANSWER:

(b) 10.5 m
Let the breadth of the rectangle be x m.
Length of the rectangle = 12 m
Area of the rectangle = 126 m²
Area of the rectangle = (length×breadth)sq units=(12×x)m2=12x m2length×breadthsq units=(12×x)m2=12x m2
It is given that the area of the rectangle is 126 m².
⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.

Page No 233:

Question 15:

Fill in the blanks.
(i) A polygon having all sides equal and all angles equal is called a …… polygon.
(ii) Perimeter of a square = …… × side.
(iii) Area of a rectangle = (……) × (……).
(iv) Area of a square = …… .
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is …… .

ANSWER:

(i) A polygon having all sides equal and all angles equal is called a regular polygon
(ii) Perimeter of a square = 4 × side
(iii) Area of a rectangle = (length) × (breadth)
(iv) Area of a square = (side)²
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is 20 m²
Area of a rectangle = (length) × (breadth) = (5×4) m² = 20 m²

Page No 233:

Question 16:

Match the following:

(a) Area of a rectangle	(i) πr2
(b) Area of a square	(ii) 4 × side
(c) Perimeter of a rectangle	(iii) l × b
(d) Perimeter of a square	(iv) (side)2
(e) Area of a circle	(v) 2(l + b)

ANSWER:

(a) Area of a rectangle	(iii) l × b
(b) Area of a square	(iv) (side)2
(c) Perimeter of a rectangle	(v) 2(l + b)
(d) Perimeter of a square	(ii) 4 × side
(e) Area of a circle	(i) πr2

Exercise 22A

Page No 235:

Question 1:

Define the terms:
(i) Data
(ii) Raw data
(iii) Array
(iv) Tabulation of data
(v) Observations
(vi) Frequency of an observation
(vii) Statistics

ANSWER:

(i) Data: It refers to the information in the form of numerical figures.

The marks obtained by 5 students of a class in a unit test are 34, 45, 65, 67, 87.

We call it the data related to the marks obtained by 5 students of a class in a unit test.
             
(ii) Raw Data: Data obtained in the original form is called raw data.

(iii) Array: Arranging the numerical figures in an ascending or a descending order is called an array.

(iv) Tabulation of data: Arranging the data in a systematic form in the form of a table is called tabulation or presentation of the data.

(v) Observations: Each numerical figure in a data is called an observation.

(vi) Frequency of an observation: The number of times a particular observation occurs is called its frequency.

(viii) Statistics: It is the science that deals with the collection, presentation, analysis and interpretation of numerical data.

Page No 235:

Question 2:

The number of children in 25 families of a colony are give below:
2, 0, 2, 4, 2, 1, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 1, 2, 2, 3, 2, 4, 1, 2, 2.
Represent the above data in the form of a frequency distribution table.

ANSWER:

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Question 3:

The sale of shoes of various sizes at a shop on a particular day is given below:
6, 9, 8, 5, 5, 4, 9, 8, 5, 6, 9, 9, 7, 8, 9, 7, 6, 9, 8, 6, 7, 5, 8, 9, 4, 5, 8, 7.
Represent the above data in the form of a frequency distribution table.

ANSWER:

Page No 235:

Question 4:

Construct a frequency table for the following:
3, 2, 5, 4, 1, 3, 2, 2, 5, 3, 1, 2, 1, 1, 2, 2, 3, 4, 5, 3, 1, 2, 3.

ANSWER:

Page No 235:

Question 5:

Construct a frequency table for the following:
7, 8, 6, 5, 6, 7, 7, 9, 8, 10, 7, 6, 7, 8, 8, 9, 10, 5, 7, 8, 7, 6.

ANSWER:

Page No 235:

Question 6:

Fill in the blanks:
(i) Data means information in the form of …… .
(ii) Data obtained in the …… form is called raw data.
(iii) Arranging the numerical figures in an ascending or a descending order is called an ……. .
(iv) The number of times a particular observation occurs is called its …… .
(v) Arranging the data in the form of a table is called …… .

ANSWER:

(i) Data means information in the form of numerical figures.
(ii) Data obtained in the original form is called raw data.
(iii) Arranging the numerical figures in an ascending or a descending order is called an  array.
(iv) The number of times a particular observation occurs is called its frequency.
(v) Arranging the data in the form of a table is called tabulation of data.

Page No 192:

Exercise 15A

Question 1:

Which of the following are simple closed figures?
Figure

ANSWER:

(a) It is a simple closed figure because it does not intersect itself
(b) It is a simple closed figure because it does not intersect itself.
(c) It is not a simple closed figure because it intersects itself.
(d) It is a simple closed figure because it does not intersect itself.
(e) It is not a simple closed figure because it intersects itself.
(f) It is a simple closed figure because it does not intersect itself.

Page No 192:

Question 2:

Which of the following are polygons?
Figure

ANSWER:

(a) It is formed by four line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is a polygon named quadrilateral.

(b) It is formed by three line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon named triangle.

(c) It is formed by twelve line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon.

(d) It is not a polygon because it contains curves.

Page No 192:

Question 3:

Fill in the blanks:
(i) A polygon is a simple closed figure formed by more than …… line segments.
(ii) A polygon formed by three line segments is called a ……. .
(iii) A polygon formed by four line segments is called a …… .
(iv) A triangle has …… sides and …… angles.
(v) A quadrilateral has …… sides and …… angles.
(vi) A figure which ends at the starting point is called a ………. .

ANSWER:

(i) A polygon is a simple closed figure formed by more than two line segments.
(ii) A polygon formed by three line segments is called a triangle.
(iii) A polygon formed by four line segments is called a quadrilateral .
(iv) A triangle has three sides and three angles.
(v) A quadrilateral has four sides and four angles.
(vi) A figure which ends at the starting point is called a closed figure.

Page No 196:

Exercise 16A

Question 1:

Take three noncollinear points A, B and C on a page of your notebook. Join AB, BC and CA. What figure do you get?
Name: (i) the side opposite to ∠C
           (ii) the angle opposite to the side BC
           (iii) the vertex opposite to the side CA
           (iv) the side opposite to the vertex B
Figure

ANSWER:

We get a triangle by joining the three non-collinear points A, B, and C.
(i) The side opposite to ∠C is AB.
(ii) The angle opposite to the side BC is ∠A.
(iii) The vertex opposite to the side CA is B.
(iv) The side opposite to the vertex B is AC.

Page No 196:

Question 2:

The measures of two angles of a triangle are 72° and 58°. Find the measure of the third angle.

ANSWER:

The measures of two angles of a triangle are 72° and 58°. 
Let the third angle be x. 
Now, the sum of the measures of all the angles of a triangle is 180^o​.
 ∴∴    x + 72^o + 58^o = 180^o
    ⇒ x + 130^o = 180^o                   
    ⇒ x = 180^o​ −- 130^o
    ⇒ x = 50^o
​The measure of the third angle of the triangle is 50^o​.

Page No 196:

Question 3:

The angles of a triangle are in the ratio 1 : 3 : 5. Find the measure of each of the angles.

ANSWER:

The angles of a triangle are in the ratio 1:3:5. 
Let the measures of the angles of the triangle be (1x), (3x) and (5x)
Sum of the measures of the angles of the triangle = 180^o
      ∴ 1x + 3x + 5x = 180^o
        ⇒ 9x = 180^o
        ⇒ x = 20^o
 1x = 20^o
3x = 60^o
​5x = 100^o
The measures of the angles are 20^o, 60^o and 100^o. 

Page No 196:

Question 4:

One of the acute angles of a right triangle is 50°. Find the other acute angle.

ANSWER:

In a right angle triangle, one of the angles is 90^o.
It is given that one of the acute angled of the right angled triangle is 50^o.
We know that the sum of the measures of all the angles of a triangle is 180^o.
Now, let the third angle be x.
​Therefore, we have:
            90^o​ + 50^o + x = 180^o
   ⇒        140^o + x = 180^o
​   ⇒                    x = 180^o −- 140^o
  ⇒                     x =  40^o
​ The third acute angle is 40^o​.

Page No 196:

Question 5:

One of the angles of a triangle is 110° and the other two angles are equal. What is the measure of each of these equal angles?

ANSWER:

Given:
∠A = 110^o and ​∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180^o .
              ∠A + ∠B + ∠C = 180^o
      ⇒    110^o + ​∠B + ∠B = 180^o
​      ⇒    110^o  + 2​∠B = 180^o
​       ⇒                2​∠B = 180^o −- 110^o
       ⇒                2∠B =  70^o
      ⇒                  ∠B = 70^o / 2
      ⇒                  ∠B = 35^o

      ∴ ​∠C = 35^o
The measures of the three angles:
∠A = 110^o, ∠B = 35^o, ​∠C = 35^o

Page No 196:

Question 6:

If one angle of a triangle is equal to the sum of other two, show that the triangle is a right triangle.

ANSWER:

Given:
 ∠A = ∠B + ∠C
​We know:
       ∠A + ∠B + ∠C = 180^o
    ⇒ ∠B +∠C + ∠B + ∠C = 180^o
​    ⇒ 2∠B + 2∠C = 180^o
​    ⇒ 2(∠B +∠C) = 180^o
​    ⇒ ∠B + ∠C = 180/2
    ⇒ ​∠B + ∠C = 90^o
∴∴ ∠A = 90^o
This shows that the triangle is a right angled triangle.

Page No 196:

Question 7:

In a ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate the angles.

ANSWER:

Let 3∠A = 4 ∠B = 6 ∠C = x
Then, we have:
 ∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°

Page No 196:

Question 8:

Look at the figures given below. State for each triangle whether it is acute, right or obtuse.
Figure

ANSWER:

(i) It is an obtuse angle triangle as one angle is 130^o, which is greater than 90^o.

(ii) It is an acute angle triangle as all the angles in it are less than 90^o.

(iii) It is a right angle triangle as one angle is 90^o.

(iv) It is an obtuse angle triangle as one angle is 92^o, which is greater than 90^o.

Page No 197:

Question 9:

In the given figure some triangles have been given. State for each triangle whether it is scalene, isosceles or equilateral.
Figure

ANSWER:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60^o.
Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.
Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.

(i) Isosceles
   AC = CB = 2 cm
(ii) Isosceles
   DE = EF = 2.4 cm
(iii) Scalene
   All the sides are unequal.
(iv) Equilateral
    XY = YZ = ZX = 3 cm
(v) Equilateral
     All three angles are 60^o.
(vi) Isosceles
     Two angles are equal in measure.
(vii) Scalene
    All the angles are unequal.

Page No 197:

Question 10:

Draw a ∆ABC. Take a point D on BC. Join AD. How many triangles do you get? Name them.
Figure

ANSWER:

In ∆ABC, if we take a point D on BC, then we get three triangles, namely ∆ADB, ∆ADC and ∆ABC.

Page No 197:

Question 11:

Can a triangle have
(i) two right angles?
(ii) two obtuse angles?
(iii) two acute angles?
(iv) each angle more than 60°?
(v) each angle less than 60°?
(vi) each angle equal to 60°?

ANSWER:

(i) No
     If the two angles are 90^o each, then the sum of two angles of a triangle will be 180^o​, which is not possible.
(ii) No
      For example, let the two angles be 120^o and 150^o. Then, their sum will be 270^o​, which cannot form a triangle.
(iii) Yes
       For example, let the two angles be 50^o and 60^o​, which on adding, gives 110^o. They can easily form a triangle whose third angle is 180^o −- 110^o = 70^o​.
(iv) No
      For example, let the two angles be 70^o​ and 80^o, which on adding, gives 150^o. They cannot form a triangle whose third angle is 180^o​ −- 150^o = 30^o, which is less than 60^o.
(v) No
      For example, let the two angles be 50^o and 40^o, which on adding, gives 90^o . Thus, they cannot form a triangle whose third angle is 180^o −- 90^o = 90^o​, which is greater than 60^o.
(vi) Yes
      Sum of all angles = 60^o + 60^o + 60^o​ = 180^o

Page No 197:

Question 12:

Fill in the blanks.
(i) A triangle has …… sides, …… angles and …… vertices.
(ii) The sum of the angles of a triangle is …… .
(iii) The sides of a scalene triangle are of ……. lengths.
(iv) Each angle of an equilateral triangle measures …… .
(v) The angles opposite to equal sides of an isosceles triangle are ……. .
(vi) The sum of the lengths of the sides of a triangle is called its ………. .

ANSWER:

(i) A triangle has 3 sides 3 angles and 3 vertices.
(ii) The sum of the angles of a triangle is 180^o. 
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60^o.
(v) The angles opposite to equal sides of an isosceles triangle are equal.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.

Page No 197:

Exercise 16B

Question 1:

How many parts does a triangle have?
(a) 2
(b) 3
(c) 6
(d) 9

ANSWER:

Correct option: (c)
A triangle has 6 parts: three sides and three angles.

Page No 197:

Question 2:

With the angles given below, in which case the construction of triangle is possible?
(a) 30°, 60°, 70°
(b) 50°, 70°, 60°
(c) 40°, 80°, 65°
(d) 72°, 28°, 90°

ANSWER:

Correct option: (b) 
(a) Sum = 30° + 60° + 70° = 160^o
     This is not equal to the sum of all the angles of a triangle. 
(b) Sum = 50° + 70° + 60° = 180^o
     Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185^o
      This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190^o​
     This is not equal to the sum of all the angles of a triangle.

Page No 197:

Question 3:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is
(a) 60°
(b) 80°
(c) 76°
(d) 84°

ANSWER:

(b) 80^o
Let the measures of the given angles be (2x)^o, (3x)^o​ and (4x)^o.
∴∴ (2x)^o + (3x)^o + (4x)^o = 180^o
  ⇒ (9x)^o = 180^o
​  ⇒ x = 180 / 9
  ⇒ x = 20^o
∴∴ ​2x =  40^o, 3x = 60^o, 4x = 80^o

Hence, the measures of the angles of the triangle are 40^o​, 60^o, 80^o.
Thus, the largest angle is 80^o.

Page No 198:

Question 4:

The two angles of a triangle are complementary. The third angle is
(a) 60°
(b) 45°
(c) 36°
(d) 90°

ANSWER:

Correct option: (d)
The measure of two angles are complimentary if their sum is 90^o degrees. 
Let the two angles be x and y, such that x + y = 90^o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180^o​.
   x + y + z​ = 180^o
 ⇒ 90^o + z = 180^o
 ⇒ z  = 180^o −- 90^o 
        = 90^o
​The third angle is 90^o.

Page No 198:

Question 5:

One of the base angles of an isosceles triangle is 70°. The vertical angle is
(a) 60°
(b) 80°
(c) 40°
(d) 35°

ANSWER:

Correct option: (c)
Let ∠A = 70^o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
∴∴ ​∠B = 70^o
​We need to find the vertical angle ​∠C.
Now, sum of all the angles of a triangle is 180^o.
    ∠A + ∠B + ∠C = 180^o
⇒​ 70^o + 70^o​ + ∠C = 180^o
⇒ 140^o + ​∠C = 180^o
⇒ ∠C = 180^o −- 140^o
⇒ ∠C = 40^o

Page No 198:

Question 6:

A triangle having sides of different lengths is called
(a) an isosceles triangle
(b) an equilateral triangle
(c) a scalene triangle
(d) a right triangle

ANSWER:

Correct option: (c)
A triangle having sides of different lengths is called a scalene triangle.

Page No 198:

Question 7:

In an isosceles ∆ABC, the bisectors of ∠B and ∠C meet at a point O. If ∠A = 40°, then ∠BOC =
?
(a) 110°
(b) 70°
(c) 130°
(d) 150°

ANSWER:

Correct option: (a)

In the isosceles ABC, ​the bisectors of ∠B and ∠C meet at point O.
Since the triangle is isosceles, the angles opposite to the equal sides are equal.
∠B = ∠C
∴∴ ∠A + ∠B + ∠C = 180^o
  ⇒  40^o + 2∠B = 180^o
  ⇒ 2∠B = 140^o
  ⇒ ∠B = 70^o
Bisectors of an angle divide the angle into two equal angles.
So, in  ∆BOC:
∠OBC = 35^o and ∠OCB = 35^o
∠BOC + ∠OBC + ∠OCB = 180​^o
  ⇒ ∠BOC + 35^o + 35^o = 180^o
  ⇒ ∠BOC = 180^o​ – 70^o
  ⇒ ∠BOC = 110^o

Page No 198:

Question 8:

The sides of a triangle are in the ratio 3 : 2 : 5 and its perimeter is 30 cm. The length of the longest side is
(a) 20 cm
(b) 15 cm
(c) 10 cm
(d) 12 cm

ANSWER:

Correct option: (b)
The sides of a triangle are in the ratio 3:2:5.
Let the lengths of the sides of the triangle be (3x), (2x), (5x).
We know:
 Sum of the lengths of the sides of a triangle = Perimeter
   (3x) + (2x) + (5x) = 30
  ⇒ 10x = 30
  ⇒ x =  30 
            10
  ⇒ x = 3
First side = 3x = 9 cm
Second side = 2x = 6 cm
Third side = 5x = 15 cm
The length of the longest side is 15 cm.

Page No 198:

Question 9:

Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is
(a) 35°
(b) 45°
(c) 65°
(d) 125°

ANSWER:

Correct option: (d)
Two angles of a triangle measure 30° and 25°, respectively.
  Let the third angle be x.
   x + 30^o + 25^o = 180^o
​                    x = 180^o −- 55^o
                     x = 125^o

Page No 198:

Question 10:

Each angle of an equilateral triangle measures
(a) 30°
(b) 45°
(c) 60°
(d) 80°

ANSWER:

Correct option: (c)
Each angle of an equilateral triangle measures 60^o.

Page No 198:

Question 11:

In the adjoining figure, the point P lies
(a) in the interior of ∆ABC
(b) in the exterior of ∆ABC
(c) on ∆ABC
(d) outside ∆ABC
Figure

ANSWER:

Correct option: (c)
Point P lies on ∆ABC.

Exercise 17 A

Page No 202:

Question 1:

In the adjacent figure, a quadrilateral has been shown.
Name: (i) its diagonals,
           (ii) two pairs of opposite sides,
           (iii) two pairs of opposite angles,
           (iv) two pairs of adjacent sides,
           (v) two pairs of adjacent angles.
Figure

ANSWER:

(i) The diagonals are AC, and BD.
(ii) AB  and CD, and AD and BC are the two pairs of opposite sides.
(iii) ∠A and ∠C, and ∠B and ∠D are the two pairs of opposite angles.
(iv) AB and BC, and AD and DC are the two pairs of adjacent sides.
(v) ​∠A and ∠B, and ∠C and ∠D are the two pairs of adjacent angles.

Page No 202:

Question 2:

Draw a parallelogram ABCD in which AB = 6.5 cm, AD = 4.8 cm and ∠BAD = 70°. Measure its diagonals.

ANSWER:

Since ABCD is a parallelogram, AB = DC = 6.5 cm and AD = BC = 4.8 cm.
Given:
∠A = 70°∠A = 70°

Steps of construction :
1) Draw AD equal to 4.8 cm.
2) Make an angle of 70^° at A and cut an arc of 6.5 cm. Name it B.
3) Cut an arc of 4.8 cm from B and 6.5 cm from D. Name it C.
4) Join AB, BC and CD.
5) Measuring the diagonals AC and BD, we get AC equal to 9.2 cm and BD equal to 6.6 cm.

Page No 202:

Question 3:

Two sides of a parallelogram are in the ratio 4 : 3. If its perimeter is 56 cm, find the lengths of its sides.

ANSWER:

Two sides of a parallelogram are in the ratio 4:3.
Let the two sides be 4x and 3x.
In a parallelogram, opposite sides are equal and parallel. So, they are also in the ratio of 4:3, i.e. 4x and 3x.
Perimeter = 4x + 3x + 4x +3x 
           56 = 14x
             x = 56145614
                        
             x = 4

∴ 4x = 16
3x = 12
Length of its sides are 16cm, 12 cm, 16cm and 12cm.

          

Page No 202:

Question 4:

Name each of the following parallelograms:
(i) The diagonals are equal and the adjacent sides are unequal.
(ii) The diagonals are equal and the adjacent sides are equal.
(iii) The diagonals are unequal and the adjacent sides are equal.

ANSWER:

(i) Rectangle
(ii) Square
(iii) Rhombus

Page No 203:

Question 5:

What is a trapezium? When do you call a trapezium an isosceles trapezium?
Draw an isosceles trapezium. Measure its sides and angles.

ANSWER:

A trapezium has only one pair of parallel sides.
A trapezium is said to be an isosceles trapezium if its non-parallel sides are equal.
Following are the measures of the isosceles trapezium:
AB = 5.4 cm
BC = 3 cm
DC = 7.4 cm
AD = 3 cm

∠A =∠B =110°∠D = ∠C = 70°∠A =∠B =110°∠D = ∠C = 70°

Page No 203:

Question 6:

Which of the following statements are true and which are false?
(a) The diagonals of a parallelogram are equal.
(b) The diagonals of a rectangle are perpendicular to each other.
(c) The diagonals of a rhombus are equal.

ANSWER:

(a) False
(b) False
(c) False

Page No 203:

Question 7:

Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A square can be thought of as a special rhombus.
(c) A rectangle can be thought of as a special parallelogram.
(d) a square is also a parallelogram.

ANSWER:

(a) This is because a rectangle with equal sides becomes a square.
(b) This is because a rhombus with each angle a right angle becomes a square.
(c) This is because a parallelogram with each angle a right angle becomes a rectangle.
(d) This is because in a square opposite sides are parallel.

Page No 203:

Question 8:

A figure is said to be regular if its sides are equal in length and angles are equal in measure. What do you mean by a regular quadrilateral?

ANSWER:

A square is a regular quadrilateral all of whose sides are equal in length and all of whose angles are equal in measure.

Page No 203:

Exercise 17 B

Question 1:

The sum of all the angles of a quadrilateral is
(a) 180°
(b) 270°
(c) 360°
(d) 400°

ANSWER:

(c) 360°
The sum of all the angles of a quadrilateral is  360^°​.

Page No 203:

Question 2:

The three angles of a quadrilateral are 80°, 70° and 120°. The fourth angle is
(a) 110°
(b) 100°
(c) 90°
(d) 80°

ANSWER:

(c) 90°

The three angles of a quadrilateral are 80°, 70° and 120°.
Let the fourth angle be x.
We know that the sum of all the angles of a quadrilateral is 360°.
                  80° + 70°​ + 120° + x = 360°
​                         ⇒  270°​ + x = 360°
^​                                 ⇒              x = 360° − 270°
^​                                   ⇒             x = 90°
Thus, the fourth angle is 90°.

Page No 203:

Question 3:

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The largest of these angles is
(a) 90°
(b) 120°
(c) 150°
(d) 102°

ANSWER:

Let the angles of a quadrilateral be (3x)^°, (4x)^°​, (5x)^° and (6x)^°.
Sum of all the angles of a quadrilateral is 360^°.

∴  3x + 4x + 5x + 6x = 360°
         ⇒              18x  = 360°
         ⇒                  x =  3601836018
          ⇒                x = 20°
So,
       3x = 60°
      4x =  80°
​      5x = 100°
​      6x = 120°
The largest of these angles is 120°​.
So, the correct answer is given in option (b).

Page No 203:

Question 4:

A quadrilateral having one and only one pair of parallel sides is called
(a) a parallelogram
(b) a kite
(c) a rhombus
(d) a trapezium

ANSWER:

(d) a trapezium
A trapezium is a quadrilateral that has only one pair of parallel sides.

Page No 203:

Question 5:

A quadrilateral whose opposite sides are parallel is called
(a) rhombus
(b) a kite
(c) a trapezium
(d) a parallelogram

ANSWER:

(d) a parallelogram
A parallelogram is a quadrilateral whose opposite sides are parallel.

Page No 203:

Question 6:

An isosceles trapezium has
(a) equal parallel sides
(b) equal nonparallel sides
(c) equal opposite sides
(d) none of these

ANSWER:

(b) equal nonparallel sides
The non-parallel sides of an isosceles trapezium are equal.

Page No 203:

Question 7:

If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is
(a) a rectangle
(b) a rhombus
(c) a kite
(d) none of these

ANSWER:

(b) a rhombus
The diagonals of a rhombus bisect each other at right angle.

Page No 203:

Question 8:

A square has
(a) all sides equal and diagonals unequal
(b) all sides equal and diagonals equal
(c) all sides unequal and diagonals equal
(d) none of these

ANSWER:

(b) all sides equal and diagonals equal
In a square, all the sides are equal. All of its diagonals are also equal. 

Page No 203:

Question 9:

A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides, is called a
(a) trapezium
(b) parallelogram
(c) kite
(d) rectangle

ANSWER:

(c) kite

A kite has two pairs of equal adjacent sides, but unequal opposite sides.

Page No 203:

Question 10:

What do you mean by a regular quadrilateral?
(a) A rectangle
(b) A rhombus
(c) A square
(d) A trapezium

ANSWER:

(c) A square
The only regular quadrilateral is a square. This is because all of its sides and angles are equal.

Page No 207:

Exercise 18A

Question 1:

Take a point O on your notebook and draw circles of radii 4 cm, 5.3 cm and 6.2 cm, each having the same centre O.

ANSWER:

This is the required diagram as asked in the question.

 

Page No 207:

Question 2:

Draw a circle with centre C and radius 4.5 cm. Mark points P, Q, R such taht P lies in the interior of the circle, Q lies on the circle, and R lies in the exterior of the circle.

ANSWER:

This is the required diagram as asked in the question.

Page No 207:

Question 3:

Draw a circle with centre O and radius 4 cm. Draw a chord AB of the circle. Indicate by marking points X and Y, the minor arc AXB and the major arc AYB of the circle.

ANSWER:

This is the required diagram as asked in the question.

Page No 207:

Question 4:

Which of the following statements are true and which are false?
(i) Each radius of a circle is also a chord of the circle.
(ii) Each diameter of a circle is also a chord of the circle.
(iii) The centre of a circle bisects each chord of the circle.
(iv) A secant of a circle is a segment having its end points on the circle.
(v) A chord of a circle is a segment having its end points on teh circle.

ANSWER:

(i)  False
Diameter of a circle is a chord of the circle, not radius.
(ii) True
It is the longest chord of the circle.
(iii) False
A perpendicular drawn from the centre of the circle to the chord, bisects the chord.
(iv) False
It is a line passing through the circle that intersects the circle at two points.
(v) True.

Page No 207:

Question 5:

Draw a circle with centre O and radius 3.7 cm. Draw a sector having the angle 72°.

ANSWER:

Therefore, the required arc is arc OACB.

Page No 207:

Question 6:

Fill in the blanks by using <, >, = or ≤.
(i) OP …… OQ, where O is the centre of the circle, P lies on the circle and Q is in the interior of the circle.
(ii) OP …… OR, where O is the centre of the circle, P lies on the circle and R lies in the exterior of the circle.
(iii) Major arc …… minor arc of the circle.
(iv) Major arc …… semicircumference of the circle.

ANSWER:

(i) >
(ii) <
(iii) >
(iv) >
This is because the major arc covers more than half of the circumference of the circle.

Page No 207:

Question 7:

Fill in the blanks:
(i) A diameter of a circle is a chord that ………. the centre.
(ii) A radius of a circle is a line segment with one end point ……….. and the other end point …… .
(iii) If we join any two points of a circle by a line segment, we obtain a ……….. of the circle.
(iv) Any part of a circle is called an ……….. of the circle.
(v) The figure bounded by an arc and the two radii joining the end points of the arc with the centre is called a ……….. of the circle.

ANSWER:

(i)  passes through
(ii)  on the circle, at the centre of the circle
(iii)  chord
(iv)  arc
(v)  sector

Page No 209:

Exercise 18B

Question 1:

Define each of the following:
(a) Closed figures
(b) Open figures
(c) Polygons

ANSWER:

(i) A closed figure is a figure that can be traced with the same starting and ending points, and that too without crossing or retracing any part of the figure.
For example: Polygon, circle, etc.

(ii) A figure having no boundary and no starting or ending points is called as an open figure.

(iii)  A polygon is a plane shape with 3 or more straight sides. It is a 2 dimensional closed figure.

Page No 209:

Question 2:

Define each of the following:
(a) A scalene triangle
(b) An isosceles triangle
(c) An obtuse triangle

ANSWER:

(a) A triangle having no sides or angles equal is called a scalene triangle.

(b) A triangle having two sides and the corresponding opposite angles equal is called an isosceles triangle.

(c) A triangle having one of the angles more than 90°° is called an obtuse triangle.

Page No 209:

Question 3:

(i) What do you mean by a convex quadrilateral?
(ii) Define a regular polygon.

ANSWER:

(i)  A quadrilateral with no interior angles greater than 180° is known as a convex quadrilateral.

(ii)  A regular polygon is a polygon all of whose sides are of the same lengths and all of whose interior angles are of the same measures.

Page No 209:

Question 4:

The angles of a triangle are in the ratio 3 : 5 : 7. Find the measures of these angles.

ANSWER:

The angles are in ratio 3:5:7.
Suppose the angles are 3x, 5x and 7x.

∴ 3x + 5x + 7x = 180                (angle sum property of a triangle)
                  15x =180
                     x = 12^o

Therefore, the angles are of the measures 36°, 60° and 84°.

Page No 209:

Question 5:

The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find the measures of these angles.

ANSWER:

Suppose the angles are 2x, 3x, 4x, and 6x.
We know that the sum of the angles of a quadrilateral is 360°.
∴ 2x + 3x + 4x + 6x = 360
                         15x = 360
                            x = 24

Therefore, the measures of the angles are 48°, 72°, 96° and 144°.

Page No 209:

Question 6:

State the properties of a rhombus.

ANSWER:

1. A rhombus is a parallelogram whose opposite sides are parallel.
2. All four of its sides are equal in length. Also, the opposite angles are equal.
3. The diagonals bisect each other at right angles.

Page No 209:

Question 7:

Define (i) a trapezium (ii) a kite.

ANSWER:

(i) A trapezium is a quadrilateral with only one pair of parallel sides.   

(ii) A kite is a quadrilateral that has two pairs of equal adjacent sides, but unequal opposite sides.

Page No 209:

Question 8:

Draw a circle with centre O and radius 3 cm. Draw a sector having an angle of 54°.

ANSWER:

Page No 209:

Question 9:

A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a
(a) parallelogram
(b) rectangle
(c) trapezium
(d) kite

ANSWER:

(d) kite

Page No 209:

Question 10:

If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is a
(a) rectangle
(b) parallelogram
(c) rhombus
(d) kite

ANSWER:

(c) rhombus

Page No 209:

Question 11:

A quadrilateral having one and only one pair of parallel sides is called a
(a) parallelogram
(b) a kite
(c) a trapezium
(d) a rhombus

ANSWER:

(c) a trapezium

Page No 209:

Question 12:

One of the base angles of an isosceles triangle is 70°. The vertical angle is
(a) 35°
(b) 40°
(c) 70°
(d) 80°

ANSWER:

(b) 40°

Since one base angle is 70°, the other base angle will also be 70° because the triangle is isosceles.
Vertical angle:
180 − 70 − 70 = 40°°                 (angle sum property of a triangle)

Page No 209:

Question 13:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) The diagonals of a rhombus are equal.
(ii) The diagonals of a parallelogram bisect each other.
(iii) The centre of a circle bisects each chord of a circle.
(iv) Each diameter of a circle is a chord of the circle.
(v) The diagonals of a rhombus bisect each other at right angles.

ANSWER:

(i) False
Diagonals are perpendicular and bisect each other.

(ii) True
The diagonals of a parallelogram bisect each other.

(iii) False
A perpendicular drawn from the centre of a circle to the chord, bisects the chord.

(iv)  True
It divides the circle in two equal parts.

(v) True
The diagonals of a rhombus bisect each other at right angles.

Exercise 14A

Page No 186:

Question 1:

Draw a line segment PQ = 6.2 cm. Draw the perpendicular bisector of PQ.

ANSWER:

Steps for construction:
1. Draw a line segment PQ, which is equal 6.2 cm.
2. With P as the centre and radius more than half of PQ, draw arcs, one on each side of PQ.
3. With Q as the centre and the same radius as before, draw arcs cutting the previously drawn arcs at A and B, respectively.
4. Draw AB, meeting PQ at R.

Page No 186:

Question 2:

Draw a line segment AB = 5.6 cm. Draw the perpendicular bisector of AB.

ANSWER:

Steps for construction:
1. Draw a line segment AB = 5.6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at P and Q, respectively.
4. Draw PQ, meeting AB at R.

Page No 186:

Question 3:

Draw an angle equal to ∠AOB, given in the adjoining figure.

ANSWER:

Here ∠∠AOB is given.
Steps for construction:
1. Draw a ray QP.
2. With O as the centre and any suitable radius, draw an arc cutting OA and OB at C and E, respectively..
3. With Q as the centre and the same radius as in step (2), draw an arc cutting QP at D.
4. With D as the centre and radius equal to CE, cut the arc through D at F.
5. Draw QF and produce it to point R.
∴∴ ∠∠PQR = ∠∠AOB

Page No 186:

Question 4:

Draw an angle of 50° with the help of a protractor. Draw a ray bisecting this angle.

ANSWER:

Steps for construction:

1. Draw ∠∠BAC = 50°° with the help of protractor.
2. With A as the centre and any convenient radius, draw an arc cutting AB and AC at Q and P, respectively.
3. With P as the centre and radius more than half of PQ, draw an arc.
4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point S.
5. Draw SA and produce it to point R.
Then, ray AR bisects ∠∠BAC.

Page No 186:

Question 5:

Construct ∠AOB = 85° with the help of a protractor. Draw a ray OX bisecting ∠AOB.

ANSWER:

Steps for construction:

1. Draw ∠∠AOB = 85°° with the help of a protractor.
2. With O as the centre and any convenient radius, draw an arc cutting OA and OB  at P and Q, respectively.
3. With P as the centre and radius more than half of PQ, draw an arc.
4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point R.
5. Draw RO and produce it to point X.
Then, ray OX bisects ∠∠AOB.

Page No 186:

Question 6:

Draw a line AB. Take a point P on it. Draw a line passing through P and perpendicular to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P on line AB.
3. With P as the centre, draw an arc of any radius, which intersects line AB at M and N, respectively.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius as in step (4), draw an arc that cuts the previously drawn arc at R.
6. Draw PR.
PR is the required line, which is perpendicular to AB.

Page No 186:

Question 7:

Draw a line AB. Take a point P outside it. Draw a line passing through P and perpendicular to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P outside AB.
3. With P as the centre and a convenient radius, draw an arc intersecting AB at M and N, respectively.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius, draw an arc cutting the previously drawn arc at Q.
6. Draw PQ meeting AB at S.
PQ is the required line that passes through P and is perpendicular to AB.

Page No 186:

Question 8:

Draw a line AB. Take a point P outside it. Draw a line passing through P and parallel to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P outside AB and another point O on AB.
3. Draw PO.
4. Draw ∠∠FPO such that ∠∠FPO is equal to AOP.
5. Extend FP to E.
Then, the line EF passes through the point P and EF||AB.

Page No 186:

Question 9:

Draw ∠ABC of measure 60° such that AB = 4.5 cm and AC = 5 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD.

ANSWER:

Steps for construction:

1. Draw a line BX and take a point A, such that AB is equal to 4.5 cm.
2. Draw ∠∠ABP = 60°° with the help of protractor.
3. With A as the centre and a radius of 5 cm, draw an arc cutting PB at C.
4. Draw AC.
5. Now, draw ∠∠BCY = 60°°.
6. Then, draw ∠∠ABW, such that ∠∠ABW is equal to∠∠CAX, which cut the ray CY at D.
7. Draw BD.

When we measure BD and CD, we have:
BD = 5 cm and CD = 4.5 cm

Page No 186:

Question 10:

Draw a line segment AB = 6 cm. Take a point C on AB such taht AC = 2.5 cm. Draw CD perpendicular to AB.

ANSWER:

Steps of constructions

1. Draw a line segment AB, which is equal to 6 cm.
2. Take a point C on AB such that AC is equal to 2.5 cm.
3. With C as the centre, draw an arc cutting AB at M and N.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius as before, draw another arc cutting the perviously drawn arc at S.
6. Draw SC and produce it to D.

Page No 186:

Question 11:

Draw a line segment AB = 5.6 cm. Draw the right bisector of AB.

ANSWER:

Steps for construction:

1. Draw a line segment AB, which is equal to 5.6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at M and N, respectively.
4. Draw MN, meeting AB at R.

Page No 188:

Exercise 14B

Question 1:

Using a pair of compasses construct the following angles:
(i) 60°
(ii) 120°
(iii) 90°

ANSWER:

(i) 
Steps for construction:
1. Draw a ray QP.
2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3. With N as the centre and the same radius as before, draw another arc to cut the previous arc at M.
4. Draw QM and produce it to R.
^∠∠ PQR is the required angle of 60^o.
(ii)
Steps for construction:
1. Draw a ray QP.
2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3. With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.
4. Draw QM and produce it to R.
^{∠PQR is the required angle of 120°.∠PQR is the required angle of 120°.}

(iii)
Steps for construction:
1. Draw a line PX.
2. Take a point Q on AC. With Q as the centre and any convenient radius, draw an arc cutting AX at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.
5. Draw QW produce it to R.
∠PQR is required angle of 90°.∠PQR is required angle of 90°.

Page No 188:

Question 2:

Draw an angle of 60°, using a pair of compasses. Bisect it to make an angle of 30°.

ANSWER:

Constructions steps:

1. Draw a ray QP.
2. Wth Q as the centre and any convenient radius,draw an arc cutting QP at N.
3. With N as the centre and radius same as before, draw another arc to cut the previous arc at M.
4. Draw QM and produce it to R.
∠PQR is an angle of 60°.∠PQR is an angle of 60°.
5. With M as the centre and radius more than half of MN, draw an arc.
6. With N as the centre and radius same as in step (5), draw another arc, cutting the previously drawn arc at point X.
7. Draw QX and produce it to point S.
Ray QS is the bisector of ∠PQR∠PQR.

Page No 188:

Question 3:

Draw an angle of 45°, using a pair of compasses.

ANSWER:

Construction steps:

1. Draw a line PR.
2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.
3.With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.
5. Draw QX, meeting the arc at Z. Produce it to W.
6. With Z as the centre and radius more than half of ZN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at a point Y.
8. Draw QY and produce it to point S.
∠PQS is the required angle of 45°.∠PQS is the required angle of 45°.

Page No 188:

Question 4:

Use a pair of compasses and construct the following angles:
(i) 150°
(ii) 15°
(iii) 135°
(iv) 2212°2212°
(v) 105°
(vi) 75°
(vii) 6712°6712°
(viii) 45°

ANSWER:

(i)

Steps for construction:
1. Draw a line XY and take a point O.
2. With O as the centre and any suitable radius, draw an arc cutting XY at M and N.
3.With N as the centre and the same radius,draw an arc cutting MN at R.
4.With R as the centre and the same radius as before, draw another arc cutting MN at Q .
5. With Q as the centre and radius less than MQ draw an arc.
6. With M as the centre and the same radius draw another arc cutting the previously drawn arc at P
5. Join PO.
 ∴∴ ∠∠XOP = 150°°

(ii)

Steps for construction:
1. Draw a ray XY.
2. With X as the centre and any convenient radius, draw an arc cutting XY at M.
3. With M as the centre and the same radius, draw an arc cutting the previously drawn arc at N.
4. Draw YN and produce it to B.
4. Draw the bisector AY of ∠∠XYB.
5. Again, draw the bisector YZ of ∠∠XYA.
 ∴∴ ∠∠XYZ = 15°°

(iii)

Steps for construction:
1. Draw a line XY and take a point A.
2. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.
3. With N as the centre and the same radius, draw an arc.
4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at R.
5. Draw RA.
6. Draw draw the bisector ZA of ∠∠YAR.
∴∴ ∠∠XAZ = 135°°

(iv)

Steps for construction:
1. Draw a line XY.
2. Take a point A on XY. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at P.
5. Draw PA meeting the arc at C. Produce it to E.
6. With C as the centre and radius more than half of CN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Q.
8. Draw AQ and produce it to point F.
9. Draw the bisector ZA of ∠∠XAF.
 ∴∴ ∠∠XAZ = 22.5°°

(v)


Steps for construction:
1. Draw a line XY.
2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P and Q.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at B.
5. Draw BO meeting the arc at E.
6. With Q as the centre and radius more than half of PE, draw an arc.
7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point A.
8. Draw AO and produce it to point Z.

 ∴∴ ∠∠XOZ = 105°°
(vi)

Steps for construction:
1. Draw a line XY.
2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at R.
5. Draw RO meeting the arc at E. Produce it to A.
6. With P as the centre and radius more than half of PE, draw an arc.
7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point S.
8. Draw OS and produce it to point Z.

 ∴∴ ∠∠XOZ = 75°°

(vii)

Steps for construction:
1. Draw a line XY and take a point O.
2. With O as the centre and any convinient radius, draw an arc cutting XY at M and N.
3. With N as the centre and the same radius, draw an arc.
4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at Q.
5. Draw QO.
6. Draw PO bisector of ∠∠YOA.
7. Draw ZO bisector of ∠∠POX.
 ∴∴ ∠∠XAZ = 67.5°°

(viii)

Steps for construction:
1. Draw a line PR.
2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.
5. Draw QX, meeting the arc at Z. Produce it to W.
6. With Z as the centre and radius more than half of ZN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Y.
8. Draw QY and produce it to point S.

Page No 188:

Question 5:

Draw a rectangle whose two adjacent sides are 5 cm and 3.5 cm. Make use of a pair of compasses and a ruler only.

ANSWER:

Construction steps:

1. Draw a ray AX.
2. With A as the centre, cut the ray AX at B such that AB is equal to 5 cm.
3. With B as the centre and any convenient radius, draw an arc cutting AX at M and N.
4. With N as the centre and radius more than half of MN, draw an arc.
5. With M as the centre and the same radius as before, draw another arc to cut the previous arc at Y.
6. Draw BY and produce it to W.
7. With B as the centre and a radius of 3.5 cm, cut ray BW at point C.
8. With C as the centre and a radius of 5 cm, draw an arc on the right side of BC.
9. With A as the centre and a radius of 3.5 cm, draw an arc cutting the previous arc at D.
10. Join CD and AD.
ABCD is the required rectangle.

Page No 188:

Question 6:

Draw a square, each of whose sides is 5 cm. Use a pair of compasses and a ruler in your construction.

ANSWER:

Construction steps:

1. Draw a ray AX.
2. With A as centre cut the ray AX at B such that AB=5 cm
3. With B as centre and any convenient radius,draw an arc cutting AX at M and N.
4. With N as centre and radius more than half of MN draw an arc.
5. With M as centre and the same radius as before,draw another arc to cut the previous arc at Y.
6. Join BY and produced it to W.
7. With B as centre and radius 5 cm cut ray BW at point C.
8.With C as centre and  radius  5 cm draw an arc on right side of BC.
9. With A as centre and  radius 5 cm draw an arc cutting the previous arc at D.
10.Join CD and AD.
ABCD is required square.

Page No 189:

Exercise 14C

Question 1:

How many lines can be drawn to pass through
(i) a given point
(ii) two given points
(iii) three given points

ANSWER:

(i) We can draw infinite number of lines passing through a given point.


(ii) Only one line can be drawn with two given points.


(iii) We can draw one line with three given points if all the three point are collinear.
 But, if the points are not collinear, then we cannot draw any line passing through the points.
 

Page No 189:

Question 2:

Classify the angles whose magnitudes are given below.
(i) 50°
(ii) 92°
(iii) 185°
(iv) 90°
(v) 180°

ANSWER:

(i) It is an acute angle because it is more than 0°° and less than 90°°.
(ii) It is an obtuse angle because it is more than 90°° and less than 180°°.
(iii) It is a reflex angle because it is more than 180°° and less than 360°°.
(iv) It is a right angle because it is 90°°.
(v) It is an straight angle because it is 180°°.

Page No 189:

Question 3:

Draw the perpendicular bisector of a given line segment AB of length 6 cm.

ANSWER:

Steps for construction:

1. Draw a line segment AB, which is equal to 6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and radius same as before, draw arcs, cutting the perviously drawn arcs at M and N, respectively.
4. Draw MN meeting AB at D.
MN is the required perpendicular bisector of AB.

Page No 189:

Question 4:

Construct an angle of 120° and bisect it.

ANSWER:

Steps of construction:

1. Draw a ray QP.
2.With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3.With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.
4. Draw QM and produce it to R.
∠PQR is 120°.∠PQR is 120°.
5. With M as the centre and radius more than half of MN, draw an arc.
6. With N as the centre and the same radius mentioned in step(5), draw another arc, cutting the previously drawn arc at point X.
7. Draw QX and produce it to point S.
Ray QS is a bisector of ∠PQR.Ray QS is a bisector of ∠PQR.

Page No 189:

Question 5:

Construct an angle of 90° and bisect it.

ANSWER:

Construction steps:

1. Draw a line OA.
2. Take a point B on OA. With B as the centre and any convenient radius, draw an arc cutting OA at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.
5. Draw WB, meeting the arc at S. Produce it to C.
∠ABC is the required angle of 90°∠ABC is the required angle of 90°.
6. With S as the centre and radius more than half of SN, draw an arc.
7. With N as centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at point X.
8. Draw BX and produce it to point D.
Ray BD is the angle bisctor of ∠ABC.Ray BD is the angle bisctor of ∠ABC. Ray BD is the angle bisctor of ∠ABC.Ray BD is the angle bisctor of ∠ABC.

Page No 189:

Question 6:

Draw a rectangle whose two adjacent sides are 5.4 cm and 3.5 cm.

ANSWER:

Steps of construction:

1. Draw a ray AX.
2. With A as the centre, cut the ray XA at B, such that AB is equal to 3.5 cm.
3. With B as the centre and with any convenient radius, draw an arc cutting AX at M and N.
4. With N as the centre and with radius more than half of MN, draw an arc.
5. With M as the centre and with the radius same as before, draw another arc to cut the previous arc at Y.
6. Draw BY and produced it to W.
7. With B as the centre and a radius of 5.4 cm, cut ray BW at point C.
8. With C as the centre and a radius 3.5 cm, draw an arc on the right side of BC.
9. With A as the centre and a radius 5.4 cm, draw an arc cutting the previous arc at D.
10. Join CD and AD.
ABCD is the required rectangle.

Page No 189:

Question 7:

Which of the following has no end points?
(a) A line segment
(b) A ray
(c) A line
(d) none of these

ANSWER:

(c) A line
 A line has no end points. We can produce it infinitely in both directions.

Page No 189:

Question 8:

Which of the following has one end point?
(a) A line
(b) A ray
(c) A line segment
(d) none of these

ANSWER:

(b) A ray
 A ray has one end point. We can produce a ray infinitely in one direction.

Page No 189:

Question 9:

Which of the following has two end points?
(a) A line segment
(b) A ray
(c) A line
(d) none of these

ANSWER:

(a) A line segment
 A line segment has two end points and both of them are fixed. Thus, a line segment has a fixed length.

Page No 189:

Question 10:

Two planes intersect
(a) at a point
(b) in a line
(c) in a plane
(d) none of these

ANSWER:

(b) in a line
When the common points of two planes intersect, they form a line.

Page No 189:

Question 11:

3232 right angles = …….
(a) 115°
(b) 135°
(c) 230°
(d) 270°

ANSWER:

(b) 135°°

32 right angles = 32 × 90° = 135°32 right angles = 32 × 90° = 135°

Page No 189:

Question 12:

Where does the vertex of an angle lie?
(a) in its interior
(b) in its exterior
(c) on the angle
(d) none of these

ANSWER:

(c) on the angle

Page No 189:

Question 13:

An angle measuring 270° is
(a) an obtuse angle
(b) an acute angle
(c) a straight line
(d) a reflex angle

ANSWER:

(d) a reflex angle
This is because it is more than 180°° and less than 360°°.

Page No 189:

Question 14:

Fill in the blanks.
(i) A line has …… end point.
(ii) A ray has …… end point
(iii) A line …… be drawn on a paper.
(iv) 0° …… acute angle …… 90° < obtuse angle < 180°.

ANSWER:

(i) A line has no end point.
(ii) A ray has one end point
(iii) A line cannot be drawn on a paper.
(iv) 0° < acute angle < 90° < obtuse angle < 180°.
(v) The standard unit of measuring an angle is degree(°).

Page No 189:

Question 15:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) If two line segments do not intersect, they are parallel.
(ii) If two rays do not intersect, they are parallel.
(iii) If two lines do not meet even when produced, they are called parallel lines.
(iv) Two parallel lines are everywhere the same distance apart.
(v) A ray has a finite length.
(vi) Ray AB−→−AB→ is the same as ray BA−→−BA→.

ANSWER:

(i) False
     If two line segments do not intersect, they may or may not be parallel.

(ii)False
If two rays do not intersect, they may or may not be parallel.

(iii) True

(iv)True

(v)False
    We can produce a ray in one direction.

(vi)False
     AB−→−AB→ means A is fixed and B is not fixed. In other words, we can produce AB towards B.
     BA−→−BA→ means B is fixed and A is not fixed. In other words, we can produce B towards A.

Page No 176:

Exercise 13A

Question 1:

Name three examples of angles from your daily life.

ANSWER:

Three examples from our daily life are as follows:

1)  Angle formed at the vertex of our elbow with the upper arm and the lower arm as the two rays. This angle will vary as per the position of our arm.

2)  Angle formed between the two hands of the clock that are hinged at a point.

3)  Angle formed between the two hands of a windmill. They are also hinged at a point, which is called the vertex of that angle.

Page No 176:

Question 2:

Name the vertex and the arms of ∠ABC, given in the figure below.
Figure

ANSWER:

The vertex is B.

Arms of ∠ABC are rays BA−→− and BC−→−∠ABC are rays BA→ and BC→.

Page No 176:

Question 3:

How many angles are formed in each of the figures given below? Name them.
Figure

ANSWER:

(i) Here, three angles are formed. They are ∠ABC, ∠ACB and ∠BAC.∠ABC, ∠ACB and ∠BAC.
(ii) Here, four angles are formed. They are ∠ABC, ∠BCD, ∠CDA and ∠DAB∠ABC, ∠BCD, ∠CDA and ∠DAB.
(iii) Here, eight angles are formed. They are ∠ABC, ∠BCD,∠CDA,∠DAB, ∠ABD, ∠ADB, ∠CDB, ∠CBD∠ABC, ∠BCD,∠CDA,∠DAB, ∠ABD, ∠ADB, ∠CDB, ∠CBD.

Page No 176:

Question 4:

In the given figure, list the points which
(i) are in the interior of ∠AOB
(ii) are in the interior of ∠AOB
(iii) lie on ∠AOB
Figure

ANSWER:

(i) Q and S are in the interior of ∠∠AOB.

(ii) P and R are in the exterior of ∠∠AOB.

(iii) A, O, B, N and T lie on the angle ∠∠AOB.

Page No 176:

Question 5:

See the adjacent figure and state which of the following statements are true and which are false.
(i) Point C is in the interior of ∠AOC.
(ii) Point C is in the interior of ∠AOD.
(iii) Point D is in the interior of ∠AOC.
(iv) Point B is in the interior of ∠AOD.
(v) Point C lies on ∠AOB.
Figure

ANSWER:

(i)False
Point C is on the angle ∠∠AOC.

(ii)True
Point C lies in the interior of ∠∠AOD.

(iii) False
Point D lies in the exterior of ∠∠AOC.

(iv) True
Point B lies in the exterior of ∠∠AOD.

(v) False
Point C lies in the interior of ∠∠AOB.

Page No 177:

Question 6:

In the adjoining figure, write another name for:
(i) ∠1
(ii) ∠2
(iii) ∠3
Figure

ANSWER:

(i) ∠∠EPB
(ii) ∠∠PQC
(iii) ∠∠FQD

Page No 179:

Exercise 13B

Question 1:

State the type of each of the following angles:
Figure

ANSWER:

(i) ∠∠AOB is an obtuse angle since its measure is more than 90°°.

(ii) ∠∠COD is a right angle since its measure is 90°°.

(iii) ∠∠FOE is a straight angle since its measure is 180°°.

(iv) ∠∠POQ is a reflex angle since its measure is more than 180°° but less than 360°°.

(v)   ∠∠HOG is an acute angle since its measure is more than 0 but less than 90°°.

(vi)  ∠∠POP is a complete angle since its measure is 360°°.

Page No 179:

Question 2:

Classify the angles whose magnitudes are given below:
(i) 30°
(ii) 91°
(iii) 179°
(iv) 90°
(v) 181°
(vi) 360°
(vii) 128°
(viii) (90.5)°
(ix) (38.3)°
(x) 80°
(xi) 0°
(xii) 15°

ANSWER:

(i) Acute angle
This is because its measure is less than 90°° but more than 0°°.
(ii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°
(iii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(iv)Right angle
This is because its measure is 90°°.
(v) Reflex angle
This is because its measure is more than 180°° but less than 360°°.
(vi) Complete angle
This is because its measure is 360°°.
(vii)  Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(viii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(ix) Acute angle
This is because its measure is more than 0°° but less than 90°°.
(x)  Acute angle
This is because its measure is more than 0°° but less than 90°°.
(xi)  Zero angle
This is because its measure is zero.
(xii)  Acute angle
This is because its measure is more than 0°° but less than 90°°.

Page No 179:

Question 3:

How many degrees are there in
(i) one right angle?
(ii) two right angles?
(iii) three right angles?
(iv) four right angles?
(v) 2323 right angle?
(vi) 112112 right angles?

ANSWER:

(i) One right angle has 90°°.
(ii)  Two right angles have 90°° + 90°° = 180°°.
(iii) Three right angles have 90°° + 90°° + 90°° = 270°°.
(iv)  Four right angles have 90°° + 90°° + 90°° + 90°° = 360°°.
(v) 23×90=60°23×90=60°
(vi) (1+12)right angles =32×90=135°1+12right angles =32×90=135°

Page No 179:

Question 4:

How many degrees are there in the angle between the hour hand and the minute hand of a clock, when it is
(i) 3 o’clock?
(ii) 6 o’clock?
(iii) 12 o’clock?
(iv) 9 o’clock?

ANSWER:

(i) At 3 o’clock the angle formed between the hour hand and the minute hand is right angle, i.e. 90°°.
(ii) At 6 o’clock the angle formed between the hour hand and the minute hand is a straight angle, i.e. 180°°.
(iii) At 12 o’clock the angle formed between the hour hand and the minute hand is a complete angle, i.e. 0°°.
       This is because the hour hand and minute hand coincides to each other at 12 o’clock.
(iv) At 9 o’clock the angle formed between the hour hand and the minute hand is a right angle, i.e. 90°°.

Page No 179:

Question 5:

Using only a ruler, draw an acute angle, an obtuse angle and a straight angle.

ANSWER:

(i) Acute angle

(ii) Obtuse angle
 
(iii) Straight angle

Page No 182:

Exercise 13C

Question 1:

Measure each of the following angles with the help of a protractor and write the measure in degrees:
Figure

ANSWER:

(i) ∠AOB = 45°∠AOB = 45°
(ii) ∠PQR = 75°∠PQR = 75°
(iii)  ∠DEF = 135°∠DEF = 135°
(iv)  ∠LMN = 55°∠LMN = 55°
(v) ∠TSR = 135°∠TSR = 135°
(vi) ∠GHI = 75°∠GHI = 75°

We have measured all the above angles by placing the protractor on one of the arms of the angle and measuring the angle through the other arm that coincides with the angle on the protractor.

Page No 182:

Question 2:

Construct each of the following angles with the help of a protractor:
(i) 25°
(ii) 72°
(iii) 90°
(iv) 117°
(v) 165°
(vi) 23°
(vii) 180°
(viii) 48°

ANSWER:

Steps to follow:

1. Draw a ray QP with Q as the initial point.
2. Place the protractor on QP. With its centre on Q, mark a point R against the given angle mark of the protractor.
3. Join RQ. Now, PQR is the required angle.

(i)  

(ii) 

(iii) 

(iv) 

(v) 

(vi)

(vii) 

(viii) 

Page No 182:

Question 3:

Measure ∠ABC given in the adjoining figure and construct an angle DEF equal to ∠ABC.

ANSWER:

We can see that ∠ABC = 47°∠ABC = 47°.
Steps to follow to construct angle ∠∠DEF equal to ∠∠ABC:

• Draw a ray EF with E as the initial point.
• Place the protractor on EF. With its centre at E, mark a point D against the angle 47°° of the protractor.
• Join DE.  ∠∠DEF = 47°° = ∠∠ABC is the required angle.

Page No 182:

Question 4:

Draw a line segment AB = 6 cm. Take a point C on AB such that AC = 4 cm. From C, draw CD ⊥ AB.

ANSWER:

1. Draw a line segment AB of length 6 cm.
2. Mark point C on AB such that AC is equal to 4 cm.
3. Place the protractor on AB such that the centre of the protractor is on C and its base lies along AB.
4. Holding the protractor, mark a point D on the paper against the 90°° mark of the protractor.
5. Remove the protractor and draw a ray CD with C as the initial point.

Now, CD ⊥⊥ AB

Page No 182:

Exercise 13C

Question 1:

Where does the vertex of an angle lie?
(a) In its interior
(b) In its exterior
(c) On the angle
(d) None of these

ANSWER:

(c) On the angle
Vertex is the initial point of two rays between which the angle is formed. Therefore, it lies on the angle.

Page No 182:

Question 2:

The figure formed by two rays with the same initial point is called
(a) a ray
(b) a line
(c) an angle
(d) none of these

ANSWER:

(c) an angle
The initial point is called the vertex.

Page No 182:

Question 3:

An angle measuring 180° is called
(a) a complete angle
(b) a reflex angle
(c) a straight angle
(d) none of these

ANSWER:

(c) straight angle

Page No 182:

Question 4:

An angle measuring 90° is called
(a) a straight angle
(b) a right angle
(c) a complete angle
(d) a reflex angle

ANSWER:

(b) right angle

Page No 182:

Question 5:

An angle measuring 91° is
(a) an acute angle
(b) an obtuse angle
(c) a reflex angle
(d) none of these

ANSWER:

(b) an obtuse angle
This is because it is more than 90°° but less than 180°°.

Page No 182:

Question 6:

An angle measuring 270° is
(a) an obtuse angle
(b) an  acute angle
(c) a straight angle
(d) a reflex angle

ANSWER:

(d) a reflex angle
This is because it is more than 180°° but less than 360°°.

Page No 182:

Question 7:

The measure of a straight angle is
(a) 90°
(b) 150°
(c) 180°
(d) 360°

ANSWER:

(c) 180°°

Page No 183:

Question 8:

An angle measuring 200° is
(a) an obtuse angle
(b) an acute angle
(c) a reflex angle
(d) none of these

ANSWER:

(c) a reflex angle
This is because it is more than 180°° but less than 360°°.

Page No 183:

Question 9:

An angle measuring 360° is
(a) a reflex angle
(b) an obtuse angle
(c) a straight angle
(d) a complete angle

ANSWER:

(d) a complete angle
This is because it completes the rotation of 360°°.

Page No 183:

Question 10:

A reflex angle measures
(a) more than 180° but less than 270°
(b) more than 180° but less than 360°
(c) more than 90° but less than 180°
(d) none of these

ANSWER:

(b) more than 180° but less than 360°° but less than 360°

Page No 183:

Question 11:

2 right angles = ?
(a) 90°
(b) 180°
(c) 270°
(d) 360°

ANSWER:

(b)
2 right angles =  2×90°= 180°2×90°= 180° (straight angle)

Page No 183:

Question 12:

3232 right angles = ?
(a) 115°
(b) 135°
(c) 270°
(d) 230°

ANSWER:

(b) 135°°

32 right angle =32× 90°=135°32 right angle =32× 90°=135°

Page No 183:

Question 13:

If there are 36 spokes in a bicycle wheel, then the angle between a pair of adjacent spokes is
(a) 15°
(b) 12°
(c) 10°
(d) 18°

ANSWER:

( c)  10°°

Number of spokes = 36
Measure of the angle of the wheel = Complete angle = 360°°
Angle between a pair of adjacent spokes=Measure of angleNumber of spokes=360°36=10°Measure of angleNumber of spokes=360°36=10°

Page No 172:

Exercise 12A

Question 1:

In the figure of a table given below, name the pairs of parallel edges of the top.
Figure

ANSWER:

Following are the parallel edges of the top:

AD∥BCThis is because AD and BC will not intersect even if both these line segments are produced indefinitely in both the directions.AB∥DCThis is because AB and DC will not intersect even if both these line segments are produced indefinitely in both the directions.AD∥BCThis is because AD and BC will not intersect even if both these line segments are produced indefinitely in both the directions.AB∥DCThis is because AB and DC will not intersect even if both these line segments are produced indefinitely in both the directions.

Page No 172:

Question 2:

Name the groups of all possible parallel edges of the box whose figure is shown below.
Figure

ANSWER:

The groups of parallel edges are (AD∥GH∥BC∥FE), (AB∥DC∥GF∥HE) and (AH∥BE∥CF∥DG).The above mentioned groups of edges are parallel because they will not meet each other if produced infinitely to both sides.The groups of parallel edges are (AD∥GH∥BC∥FE), (AB∥DC∥GF∥HE) and (AH∥BE∥CF∥DG).The above mentioned groups of edges are parallel because they will not meet each other if produced infinitely to both sides.

Page No 173:

Question 3:

Identify parallel line segments in each of the figures given below:
(i) Figure
(ii) Figure
(iii) Figure
(iv) Figure
(v) Figure

ANSWER:

(i)
DE∥BCThis is because they do not intersect each other.DE∥BCThis is because they do not intersect each other.
(ii)
AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.
(iii)
  AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB does not intersect DC and AD does not intersect BC.AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB does not intersect DC and AD does not intersect BC.

(iv)
LM∥RQ, RS∥PM and LS∥PQLM∥RQ, RS∥PM and LS∥PQ
These pairs of line segments are non-intersecting.
So, these pairs of lines are parallel.

(v)
AB∥DC, AB∥EF. DC∥EFAC∥BD, CE∥DFAB∥DC, AB∥EF. DC∥EFAC∥BD, CE∥DF
These pairs of line segments are non-intersecting.
So, these pairs of lines are parallel.

Page No 173:

Question 4:

Find the distance between the parallel lines l and m, using a set square.
(i) Figure
(ii) Figure

ANSWER:

(i) Distance between l and m is 1.3 cm.

Place the ruler so that one of its measuring edges lies along the line l. Hold it with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw the line segment PM along the edge of the set square, as shown in the figure. Then, measure the distance (PM) between l and m, which will be equal to 1.3 cm.


(ii) Distance between l and m is 1 cm.

Place the ruler so that one of its measuring edges lies along the line l. Hold it with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw the line segment PM along the edge of the set square, as shown in figure. Then, measure the distance (PQ) between l and m as 1 cm.

Page No 173:

Question 5:

In the figure, l||m. If AB ⊥ l, CD ⊥ l and AB = 2.3 cm, find CD.
Figure

ANSWER:

At point A, AB is the perpendicular distance between l and m.
At point C, CD is the perpendicular distance between l and m.
The perpendicular distance between two parallel lines is same at all points.
∴ CD = AB = 2.3 cm

Page No 173:

Question 6:

In the figure, do the segments AB and CD intersect? Are they parallel? Give reasons for your answer.
Figure

ANSWER:

Line segments AB and CD will intersect if they are produced endlessly towards the ends A and C, respectively.

Therefore, they are not parallel to each other.

Page No 173:

Question 7:

Using a set square and a ruler, test whether l || m in each of the following cases:
(i) Figure
(ii) Figure

ANSWER:

(i) Place the ruler so that one of its measuring edges lies along the line l. Hold it firmly with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw line segments between l and m (say PM, RS, AB) with the set square.

Now, we see that PM = AB = RS.
Thus, we can say that l ∥∥m.



(ii) In this case, we see that when we draw line segments between l and m, they are unequal, i.e. PM≠RSPM≠RS.
Therefore, l is not parallel to m.

Page No 173:

Question 8:

Which of the following statements are true and which are false?
(i) Two lines are parallel if they do not meet, even when produced.
(ii) Two parallel lines are everywhere the same distance apart.
(iii) If two line segments do not intersect, they are parallel.
(iv) If two rays do not intersect, they are parallel.

ANSWER:

(i) True
The statement is true because such lines do not intersect even when produced.

(ii) True
Perpendicular distance between two parallel lines is same at all points on the lines.

(iii) True
If the corresponding lines are produced infinitely, they will not intersect. Hence, they are parallel.

(iv) True
The corresponding lines determined by them will not intersect. Hence, they are parallel to each other.

Exercise 11A

Page No 166:

Question 1:

Name all the line segments in each of the following figures:
(i) Figure
(ii) Figure
(iii) Figure

ANSWER:

(i)  The line segments are
YX ¯¯¯¯¯¯This is because it has two end points Y and X.YZ ¯¯¯¯¯¯ This is because it has two end points Y an Z.YX ¯This is because it has two end points Y and X.YZ ¯ This is because it has two end points Y an Z.

(ii)  
AD¯¯¯¯¯This is because it has two end points A and D.AB¯¯¯¯¯ This is because it has two end points A and B.AC¯¯¯¯¯ This is because it has two end points A and C.AE¯¯¯¯¯ This is because it has two end points A and E.DB¯¯¯¯¯ This is because it has two end points B and D.BC¯¯¯¯¯ This is because it has two end points B and C.CE¯¯¯¯¯ This is because it has two end points C and E.AD¯This is because it has two end points A and D.AB¯ This is because it has two end points A and B.AC¯ This is because it has two end points A and C.AE¯ This is because it has two end points A and E.DB¯ This is because it has two end points B and D.BC¯ This is because it has two end points B and C.CE¯ This is because it has two end points C and E.

(iii)
PS¯¯¯¯¯ This is because it has two end points P and S.PQ¯¯¯¯¯ This is because it has two end points P and Q.QR¯¯¯¯¯ This is because it has two end points Q and R.RS¯¯¯¯¯ This is because it has two end points R and S.PR¯¯¯¯¯ This is because it has two end points P and R.QS¯¯¯¯¯ This is because it has two end points Q and S.PS¯ This is because it has two end points P and S.PQ¯ This is because it has two end points P and Q.QR¯ This is because it has two end points Q and R.RS¯ This is because it has two end points R and S.PR¯ This is because it has two end points P and R.QS¯ This is because it has two end points Q and S.

Page No 166:

Question 2:

Identify and name the line segments and rays in each of the following figures:
(i) Figure
(ii) Figure
(iii) Figure

ANSWER:

(i)  Line segment is AB¯¯¯¯¯. This is because it has two end points A and B.AB¯. This is because it has two end points A and B.
Rays are:
−→AC This is because it has only one end point A.−→BD This is because it has only one end point B.→AC This is because it has only one end point A.→BD This is because it has only one end point B.


(ii) Line segments are:
EP¯¯¯¯¯  This is because it has two end points Eand P.EG¯¯¯¯¯ This is because it has two end points E and G.GP¯¯¯¯¯ This is because it has two end points G and P.EP¯  This is because it has two end points Eand P.EG¯ This is because it has two end points E and G.GP¯ This is because it has two end points G and P.

Rays are:
EF−→− This is because it has only one end point, i.e. E.GH−→− This is because it has only one end point, i.e. G.PQ−→− This is because it has only one end point, i.e. P.EF→ This is because it has only one end point, i.e. E.GH→ This is because it has only one end point, i.e. G.PQ→ This is because it has only one end point, i.e. P.

(iii) Line segments are:
OL¯¯¯¯¯ This is because it has two end points O and L.OP¯¯¯¯¯ This is because it has two end points O and P.OL¯ This is because it has two end points O and L.OP¯ This is because it has two end points O and P.

Rays are:
LM−→− This is because it has only one end point, i.e. L.PQ−→−  This is because it has only one end point, i.e. P.LM→ This is because it has only one end point, i.e. L.PQ→  This is because it has only one end point, i.e. P.

Page No 167:

Question 3:

In the adjoining figure, name
(i) four line segments;
(ii) four rays;
(iii) two non-intersecting line segments.
Figure

ANSWER:

(i)
PR¯¯¯¯¯ This is because it has two end points P and R.QS¯¯¯¯¯ This is because it has two end points Q and S.PQ¯¯¯¯¯ This is because it has two end points P and Q.RS¯¯¯¯¯ This is because it has two end points R and S.PR¯ This is because it has two end points P and R.QS¯ This is because it has two end points Q and S.PQ¯ This is because it has two end points P and Q.RS¯ This is because it has two end points R and S.


(ii)
PA−→− This is because it has only one end point, i.e. P.RB−→− This is because it has only one end point, i.e. R.QC−→− This is because it has only one end point, i.e. Q.SD−→− This is because it has only one end point, i.e. S.PA→ This is because it has only one end point, i.e. P.RB→ This is because it has only one end point, i.e. R.QC→ This is because it has only one end point, i.e. Q.SD→ This is because it has only one end point, i.e. S.

(iii)
PR ¯¯¯¯¯¯and QS¯¯¯¯¯ are the two non−intersecting line segments as they do not have any point in common.PR ¯and QS¯ are the two non-intersecting line segments as they do not have any point in common.

Page No 167:

Question 4:

What do you mean by collinear points?
(i) How many lines can you draw passing through three collinear points?
(ii) Given three collinear points A, B, C. How many line segments do they determine? Name them.
Figure

ANSWER:

 COLLINEAR POINTS :
 Three or more points in a plane are said to be collinear if they all lie in the same line. This line is called the line of collinearity for the given points.

(i) We can draw only one line passing through three collinear points.

(ii) 3 Line segments are:
AB¯¯¯¯¯ This is because it has two end points A and B.BC¯¯¯¯¯ This is because it has two end points B and C.AC¯¯¯¯¯ This is because it has two end points A and C.AB¯ This is because it has two end points A and B.BC¯ This is because it has two end points B and C.AC¯ This is because it has two end points A and C.

Page No 167:

Question 5:

In the adjoining figure, name:
(i) four pairs of intersecting lines
(ii) four collinear points
(iii) three noncollinear points
(iv) three concurrent lines
(v) three lines whose point of intersection is P
Figure

ANSWER:

(i)
PS←→ and AB ←→− intersecting at S.CD←→and RS←→ intersecting at R.PS←→ and CD←→ intersecting at P.AB←→ and RS←→ intersecting at S.PS↔ and AB ↔ intersecting at S.CD↔and RS↔ intersecting at R.PS↔ and CD↔ intersecting at P.AB↔ and RS↔ intersecting at S.

(ii) A, Q, S and B are four collinear points as they all lie on the same line AB ←→−AB ↔.

(iii) A, C and B are non-collinear points as they do not lie on the same line. 

(iv)

PS←→ , RS←→ and AB←→ are three concurrent lines passing through the same point SPS↔ , RS↔ and AB↔ are three concurrent lines passing through the same point S.

(v)

PS←→ , PQ←→ and CD←→ have common point of intersection PPS↔ , PQ↔ and CD↔ have common point of intersection P.

Page No 167:

Question 6:

Mark three noncollinear points A, B, C, as shown. Draw lines through these points taking two at a time. Name the lines. How many such different lines can be drawn?
Figure

ANSWER:

Taking points A and B, we can draw only one line AB ←→−AB ↔.
Taking points B and C, we can draw only one line BC←→ BC↔ .
Taking points A and C, we can draw only one line AC ←→−AC ↔.
We can draw only three lines through these non-collinear points A ,B and C.

Page No 167:

Question 7:

Count the number of line segments drawn in each of the following figures and name them.
(i) Figure
(ii) Figure
(iii) Figure
(iv) Figure

ANSWER:

(i) There are 6 line segments. These are:
AB¯¯¯¯¯ (with end points A and B)AC¯¯¯¯¯ (with end points A and C)AD¯¯¯¯¯ (with end points A and D)BC¯¯¯¯¯ (with end points B and C)BD¯¯¯¯¯ (with end points B and D)CD¯¯¯¯¯ (with end points C and D)AB¯ (with end points A and B)AC¯ (with end points A and C)AD¯ (with end points A and D)BC¯ (with end points B and C)BD¯ (with end points B and D)CD¯ (with end points C and D)

(ii) There are 10 line segments. These are:
AB¯¯¯¯¯  (with end points A and B)BC ¯¯¯¯¯¯ (with end points B and C)CD¯¯¯¯¯  (with end points C and D)AD¯¯¯¯¯  (with end points A and D)AC¯¯¯¯¯  (with end points A anc C)BD ¯¯¯¯¯¯ (with end points B and D)AO ¯¯¯¯¯¯ (with end points A and O)CO¯¯¯¯¯  (with end points C and O)BO¯¯¯¯¯  (with end points B and O)DO¯¯¯¯¯  (with end points D and O)AB¯  (with end points A and B)BC ¯ (with end points B and C)CD¯  (with end points C and D)AD¯  (with end points A and D)AC¯  (with end points A anc C)BD ¯ (with end points B and D)AO ¯ (with end points A and O)CO¯  (with end points C and O)BO¯  (with end points B and O)DO¯  (with end points D and O)

(iii) There are 6 line segments. They are:
AB¯¯¯¯¯, AF¯¯¯¯¯, FB¯¯¯¯¯, EC¯¯¯¯¯, ED¯¯¯¯¯, DC¯¯¯¯¯AB¯, AF¯, FB¯, EC¯, ED¯, DC¯

(iv) There are 12 line segments. They are:
AB¯¯¯¯¯, AD¯¯¯¯¯, AE¯¯¯¯¯BC¯¯¯¯¯, BF¯¯¯¯¯ CG¯¯¯¯¯, CD¯¯¯¯¯HG¯¯¯¯¯¯, HE¯¯¯¯¯¯ , DH¯¯¯¯¯¯EF¯¯¯¯¯, GF¯¯¯¯¯AB¯, AD¯, AE¯BC¯, BF¯ CG¯, CD¯HG¯, HE¯ , DH¯EF¯, GF¯

Page No 167:

Question 8:

Consider the line PQ←→PQ↔ given below and find whether the given statements are true or false:
(i) M Is a point on ray NQ−→−NQ→.
(ii) L is a point on ray MP−→−MP→.
(iii) Ray MQ−→−MQ→ is different from ray NQ−→−NQ→.
(iv) L, M, N are points on line segment LN¯¯¯¯¯LN¯.
(v) Ray LP−→LP→ is different from ray LQ−→−LQ→.
Figure

ANSWER:

(i) False
M is outside ray NQ.

(ii) True
L is placed between M and P.

(iii) True
Ray MQ is extended endlessly from M to Q and ray NQ is extended endlessly from N to Q.

(iv) True

(v) True
  LP −→−is extended endlessly from L to P.LQ−→− is extended endlessly from L to Q.LP →is extended endlessly from L to P.LQ→ is extended endlessly from L to Q.

Page No 168:

Question 9:

Write ‘T’ for true and ‘F’ for false in case of each of the following statements:
(i) Every point has a size.
(ii) A line segment has no length.
(iii) Every ray has a finite length.
(iv) The ray AB−→−AB→ is the same as the ray BA−→−BA→.
(v) The line segment AB¯¯¯¯¯AB¯ is the same as the line segment BA¯¯¯¯¯BA¯.
(vi) The line AB←→AB↔ is the same as the line BA←→BA↔.
(vii) Two points A and B in a plane determine a unique line segment.
(viii) Two intersecting lines intersect at a point.
(ix) Two intersecting planes intersect at a point.
(x) If points A, B, C are collinear and points C, D, E are collineaer then the pints A,B, C, D, E are collinear.
(xi) One and only one ray can be drawn with a given end point.
(xii) One and only one line can be drawn to pass through two given points.
(xiii) An unlimited number of lines can be drawn to pass through a given point.

ANSWER:

(i)  False
A point does not have any length, breadth or thickness.

(ii)   False
A line segment has a definite length.

(iii) False
A ray has no definite length.

(iv) False
Ray AB has initial point A and is extended endlessly towards B, while ray BA has initial point B and is extended endlessly towards A.

(v) True
This is because both the line segments have definite length with end points A and B.

(vi)  True
This is because it neither has a definite length nor any end point.

(vii) True
Only one line segment can pass through the two given points.

(viii) True

(ix) False
Two intersecting planes intersect at a line.

(x) False
Different set of collinear points need not be collinear.


(xi) False
    With point P, endless rays (like PA, PB, PC, PD, PE, PF) can be drawn.


(xii) True
Two points define one unique line.
(xiii) True

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Question 10:

Fill in the blanks:
(i) A line segment has a ………….. length.
(ii) A ray has ………….. end point.
(iii) A line has ………….. end point.
(iv) A ray has no ………….. length.
(v) A line ………….. be drawn on a paper.

ANSWER:

 (i) definite
(ii) one
(iii) no
(iv) definite
(v) cannot

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Exercise 11B

Question 1:

Which of the following has no end points?
(a) A line segment
(b) A ray
(c) A line
(d) None of these

ANSWER:

(c) A line does not have any end point. It is a line segment that is extended endlessly on both sides.

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Question 2:

Which of the following has one end point?
(a) A line
(b) A ray
(c) A line segment
(d) None of these

ANSWER:

(b) A ray has one end point, which is called the initial point. It is extended endlessly towards the other direction.

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Question 3:

Which of the following has two end points?
(a) A line segment
(b) A ray
(c) A line
(d) None of these

ANSWER:

(a) A line segment has two end points and a definite length that can be measured.

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Question 4:

Which of the following has definite length?
(a) A line
(b) A line segment
(c) A ray
(d) None of these

ANSWER:

(b) A line segment has a definite length that can be measured by a ruler and, therefore, it can be drawn on a paper.

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Question 5:

Which of the following can be drawn on a piece of paper?
(a) A line
(b) A line segment
(c) A ray
(d) A plane

ANSWER:

(b) A line segment has a definite length that can be measured by a ruler. So, it can be drawn on a paper.

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Question 6:

How many lines can be drawn passing through a given point?
(a) One only
(b) Two
(c) Three
(d) Unlimited number

ANSWER:

(d) Unlimited number of lines can be drawn.

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Question 7:

How many lines can be drawn passing through two given point?
(a) One only
(b) Two
(c) Three
(d) Unlimited number

ANSWER:

(a) Only one line can be drawn that passes through two given points.

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Question 8:

Two planes intersect
(a) at a point
(b) in a plane
(c) in a line
(d) none of these

ANSWER:

(c)  Two intersecting planes intersect in a line.

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Question 9:

Two lines intersect
(a) at a point
(b) at two points
(c) at an infinite number of points
(d) in a line

ANSWER:

(a) Two lines intersect at a point.

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Question 10:

Two points in a plane determine
(a) exactly one line segment
(b) exactly two line segments
(c) an infinite number of line segments
(d) none of these

ANSWER:

(a) exactly one line segment

Two points in a plane determine exactly one line segment with those two points as its end points.

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Question 11:

The minimum number of points of intersection of three lines in a plane is
(a) 1
(b) 2
(c) 3
(d) 0

ANSWER:

(d) 0
Three lines will not necessarily intersect in a plane. Thus, the minimum point of intersection will be 0.

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Question 12:

The maximum number of points of intersection of three lines in a plane is
(a) 0
(b) 1
(c) 2
(d) 3

ANSWER:

(d) 3

The maximum number of points of intersection of three lines that intersect in a plane are three.

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Question 13:

Choose the correct statement:
(a) every line has a definite length
(b) every ray has a definite length
(c) every line segment has a definite length
(d) none of these

ANSWER:

(c) Every line segment has a definite length.

Every line segment has a definite length, which can be measured using a ruler.

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Question 14:

Choose the false statement:
(a) Line AB←→AB↔ is the same as line BA←→BA↔
(b) Ray AB−→−AB→ is the same as ray BA−→−BA→
(c) Line segment AB¯¯¯¯¯AB¯ is the same as teh line segment BA¯¯¯¯¯BA¯
(d) None of these

ANSWER:

(b) Ray AB−→− is same as ray BA−→− AB→ is same as ray BA→ 
This is because the initial points in these rays are A and B, respectively, and are extended endlessly towards B and A, respectively.

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Question 15:

How many rays can be drawn with a given point as the initial point?
(a) One
(b) Two
(c) An unlimited number
(d) A limited number only

ANSWER:

(c) An unlimited number of rays can be drawn with a given point as the initial point. For example: