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
Page No 230:
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.
Page No 233:
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
Page No 233:
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
Page No 233:
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.
Page No 233:
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.
Page No 233:
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
Page No 233:
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
Page No 233:
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.
Page No 233:
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
Page No 233:
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 |
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