Mensuration
Class 6th Mathematics NCERT Exemplar Solution
- Following figures are formed by joining six unit squares. Which figure has the…
- A square shaped park ABCD of side 100m has two equal rectangular flower beds each…
- The side of a square is 10cm. How many times will the new perimeter become if the…
- Length and breadth of a rectangular sheet of paper are 20cm and10cm, respectively.…
- Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The…
- In Fig. 6.8 which of the following is a regular polygon? All have equal side…
- Match the shapes (each sides measures 2cm) in column I with the corresponding…
- Match the following
- Perimeter of the shaded portion in Fig. 6.9 is AB + _ + _ + _ + _ + _ + _ + HA…
- The amount of region enclosed by a plane closed figure is called its _________.…
- Area of a rectangle with length 5cm and breadth 3cm is _________.…
- A rectangle and a square have the same perimeter (Fig. 6.10). A. The area of the…
- (a) 1m = _________ cm. (b) 1sqcm = _________ cm × 1cm. (c) 1sqm = 1m × _________…
- If length of a rectangle is halved and breadth is doubled then the area of the…
- Area of a square is doubled if the side of the square is doubled.…
- Perimeter of a regular octagon of side 6cm is 36cm.
- A farmer who wants to fence his field, must find the perimeter of the field.…
- An engineer who plans to build a compound wall on all sides of a house find the…
- To find the cost of painting a wall we need to find the perimeter of the wall.…
- To find the cost of a frame of a picture, we need to find the perimeter of the…
- Four regular hexagons are drawn so as to form the design as shown in Fig.11.1. If…
- Perimeter of an isosceles triangle is 50cm. If one of the two equal sides 8cm,…
- Length of a rectangle is three times its breadth. Perimeter of the rectangle is…
- There is a rectangular lawn 10m long and 4m wide in front of Meena’s house (Fig.…
- The region given in Fig. 6.13 is measured by taking 1 as a unit. What is the area…
- Tahir measured the distance around a square field as 200 rods (lathi). Later he…
- The length of a rectangular field is twice its breadth. Jamal jogged around it…
- Three squares are joined together asshown in Fig. 6.14. Their sides are 4cm, 10cm…
- In Fig. 6.15 all triangles are equilateraland AB = 8 units. Other triangles have…
- Length of a rectangular field is 250m and width is 150m. Anuradha runs around…
- Bajinder runs ten times around a squaretrack and covers 4km. Find the length of…
- The lawn in front of Molly’s house is 12m × 8m, whereas the lawn in front of…
- The perimeter of a regular pentagon is 1540cm. How long is its each side?…
- The perimeter of a triangle is 28cm. One of it’s sides is 8cm. Write all the…
- The length of an aluminium strip is 40cm. If the lengths in cm are measured in…
- Base of a tent is a regular hexagon of perimeter 60cm. What is the length of each…
- In an exhibition hall, there are 24 display boards each of length 1m 50cm and…
- In the above question, how many square metres of cloth is required to cover all…
- What is the length of outer boundary of the park shown in Fig. 6.16? What will be…
- Total cost of fencing the park shown in Fig. 6.17 is Rs 55000. Find the cost of…
- In Fig. 6.18 each square is of unit length (a) What is the perimeter of the…
- Rectangular wall MNOP of a kitchen is covered with square tiles of 15cm length…
- Length of a rectangular field is 6 times its breadth. If the length of the field…
- Anmol has a chart paper of measure 90cm × 40cm, whereas Abhishek has one which…
- A rectangular path of 60m length and 3m width is covered by square tiles of side…
- How many square slabs each with side 90cm are needed to cover a floor of area…
- The length of a rectangular field is 8m and breadth is 2m. If a square field has…
- Parmindar walks around a square park once and covers 800m. What will be the area…
- The side of a square is 5cm. How many times does the area increase, if the side…
- Amita wants to make rectangular cards measuring 8cm × 5cm. She has a square chart…
- A magazine charges Rs 300 per 10sqcm area for advertising.A company decided to…
- The perimeter of a square garden is 48m. A small flower bed covers 18sqm area…
- Perimeter of a square and a rectangle is same. If a side of the square is 15cm…
- A wire is cut into several small pieces. Each of the small pieces is bent into a…
- Divide the park shown in Fig. 6.17 of question 40 into two rectangles. Find the…
- The area of a rectangular field is 1600sqm. If the length of the field is 80m,…
- The area of each square on a chess board is 4sqcm. Find the area of the board.…
- Find all the possible dimensions (in natural numbers) of a rectangle with a…
- Find all the possible dimensions (in natural numbers) of a rectangle with an…
- Find the area and Perimeter of each of the following figures, if area of each…
- What is the area of each small square in the Fig. 6.21 if the area of entire…
Exercise
Question 1.Following figures are formed by joining six unit squares. Which figure has the smallest perimeter in Fig. 6.4?
A. (ii)
B. (iii)
C. (iv)
D. (i)
Answer:
For First figure:
Perimeter = Total number of sides × length of each side
= 10 × 1
= 10 cm
For second figure:
Perimeter = Total number of sides × length of each side
= 12 × 1
= 12 cm
For third figure:
Perimeter = Total number of sides × length of each side
= 14 × 1
= 14 cm
For fourth figure:
Perimeter = Total number of sides × length of each side
= 14 × 1
= 14 cm
So correct answer is D [i].
Question 2.
A square shaped park ABCD of side 100m has two equal rectangular flower beds each of size10m × 5m (Fig. 6.5). Length of the boundary of the remaining park is
A. 360m
B. 400m
C. 340m
D. 460m
Answer:
Given:
Side = 100 m
Flower Bed length = 10 m
Flower Bed breadth = 5 m
∴ Length of boundary of remaining park = Sum of all sides
= [90 + 5 + 10 + 95 + 90 + 5 + 10 + 95]
= 400 m
So correct answer is B [400 m].
Question 3.
The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
A. 2 times
B. 4 times
C. 6 times
D. 8 times
Answer:
Given:
Length of side = 10 cm.
Perimeter of Square, P1 = 10 × 4
= 40 cm
If the side is doubled, so new side = 20 cm
So new perimeter, P2 = 20 × 4
= 80 cm
So it is clearly seen that the new perimeter is twice that of the old perimeter.
So correct answer is A [two times].
Question 4.
Length and breadth of a rectangular sheet of paper are 20cm and10cm, respectively. A rectangular piece is cut from the sheet as shown in Fig. 6.6. Which of the following statements is correct for the remaining sheet?
A. Perimeter remains same but area changes.
B. Area remains the same but perimeter changes.
C. Both area and perimeter are changing.
D. Both area and perimeter remain the same.
Answer:
Perimeter of the rectangle = 2× (l + b)
Area of the rectangle = l × b
After looking at the formula it can be interpreted as that the area depend upon the length and breadth of the rectangle.
Perimeter of old rectangle = 2 × (l + b)
= 2 × (20 + 10)
= 60 cm
Perimeter of new figure = 20 + 10 + 15 + 2 + 5 + 8
= 60 cm.
So perimeter remains same.
Area will change because the new figure’s area = Area of old rectangle – Area of smaller rectangle
So correct answer is A.
Question 5.
Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is
A. 65cm
B. 60cm
C. 55cm
D. 50cm
Answer:
Perimeter for any figure is the sum of all sides.
So the perimeter of hexagon = 30 cm.
So each side of hexagon = 30 / 6
= 5 cm
When two hexagons are joined, each side of each hexagon overlaps.
So total number of sides of the new figure is 10.
So the new perimeter = 10 × 5
= 50 cm
So the correct answer is D [50 cm].
Question 6.
In Fig. 6.8 which of the following is a regular polygon? All have equal side except (i)
A. (i)
B. (ii)
C. (iii)
D. (iv)
Answer:
In regular polygon all sides and all angles are equal
In Figure I all sides are not equal.
So it is not regular polygon.
Figure II has all sides and angles are equal.
So figure II is regular polygon
In figure III all sides are not equal.
So it is also not regular polygon.
The figure numbered (iv) is not a regular polygon.
Because in this figure the sides are not equal.
Question 7.
Match the shapes (each sides measures 2cm) in column I with the corresponding perimeters in column II:
Answer:
Perimeter for any figure is the sum of all sides.
So figure labelled A has 14 sides and so the perimeter = 14× 2
= 28 cm.
So figure labelled B has 8 sides and so the perimeter = 8× 2
= 16 cm.
So figure labelled C has 10 sides and so the perimeter = 10× 2
= 20 cm.
So figure labelled D has 12 sides and so the perimeter = 12× 2
= 24 cm.
Question 8.
Match the following
Answer:
Perimeter of rectangle = 2 × (l + b)
= 2 × (6 + 4)
= 2 × 10
= 20 cm
Perimeter of square = 4 × side
= 4 × 5
= 20 cm
Perimeter of equilateral triangle = 3 × 6
= 18 cm
Perimeter of isosceles triangle = a + b + c
= 2 + 4 + 4
= 10 cm.
Question 9.
Perimeter of the shaded portion in Fig. 6.9 is
AB + _ + _ + _ + _ + _ + _ + HA
Answer:
The perimeter of shaded region is sum of all sides of shaded region.
AB + BM + MD + DE + EN + NG + GH + HA
Question 10.
The amount of region enclosed by a plane closed figure is called its _________.
Answer:
The amount region enclosed by any plane closed figure constitutes its area.
The amount of region enclosed by a plane closed figure is called its area.
Question 11.
Area of a rectangle with length 5cm and breadth 3cm is _________.
Answer:
Given:
Length of rectangle, l = 5 cm.
Breadth of rectangle, b = 3 cm.
Area of rectangle = l × b
= 5 × 3
= 15 cm2
Area of a rectangle with length 5cm and breadth 3cm is 15 cm2.
Question 12.
A rectangle and a square have the same perimeter (Fig. 6.10).
A. The area of the rectangle is _________.
B. The area of the square is _________. Fig. 6.10
Answer:
Given:
Length of rectangle, l = 5 cm.
Breadth of rectangle, b = 3 cm.
Perimeter of rectangle = 2 × (l + b)
= 2 × (6 + 2)
= 2 × 8
= 16 cm
Perimeter of square = 4 × side
16 = 4 × side
Side = 16 / 4
∴ Side of Square = 4 cm.
Area of rectangle = l × b
= 2 × 6
= 12 cm2
Area of square = side2
= 42
= 16 cm2
A. The area of the rectangle is 12 cm2.
B. The area of the square is 16 cm2.
Question 13.
(a) 1m = _________ cm.
(b) 1sqcm = _________ cm × 1cm.
(c) 1sqm = 1m × _________ m = 100cm × _________ cm.
(d) 1sqm = _________ sqcm.
Answer:
(a) For metric conversions,
1 metre = 100 cm.
(b) 1 square centimetre or cm2 is always equal to 1 cm × 1 cm.
1sqcm = 1 cm × 1cm.
(c) 1 square metre or m2 is always equal to 1 m × 1 m.
1 m2 = 100 cm × 100 cm
1sqm = 1m × 1 m = 100cm × 100 cm.
(d) 1 square metre or m2 is always equal to 1 m × 1 m.
We know that 1 m =100 cm
So 1 m2 = 100 cm × 100 cm
1sqm = 10000 sqcm
Question 14.
If length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.
Answer:
Let us try this using formula.
Area of rectangle = l × b
New condition length = l / 2
Breadth = 2b
So new area = l× b
So area remain unchanged.
So statement is true.
Question 15.
Area of a square is doubled if the side of the square is doubled.
Answer:
Let us try this using formula.
Area of square = side2
So new condition, side = 2× side
So new area = (2 × side) 2
= 4 × side2
So the new area becomes 4 times the old area.
So statement is false.
Question 16.
Perimeter of a regular octagon of side 6cm is 36cm.
Answer:
The perimeter of shaded region is sum of all sides.
So an octagon has 8 sides. [“oct” means 8 as “pent” means 5 and “hex” means 6]
So perimeter = 8 × 6
Perimeter = 48 cm
So the statement is false.
Question 17.
A farmer who wants to fence his field, must find the perimeter of the field.
Answer:
The statement is true.
Perimeter is the sum of all sides of the field and a fence is also required only in the sides.
Question 18.
An engineer who plans to build a compound wall on all sides of a house find the area of the compound.
Answer:
The statement is false.
To build a compound wall, one has to know the perimeter of the compound.
Question 19.
To find the cost of painting a wall we need to find the perimeter of the wall.
Answer:
The paint will be applied on the whole area of the wall and not on the sides of the wall.
So to find the cost of paint one has to know the area of the wall.
So the statement is false.
Question 20.
To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Answer:
Frame will only cover the sides of picture and hence it is necessary to find the perimeter of the picture.
So the statement is true.
Question 21.
Four regular hexagons are drawn so as to form the design as shown in Fig.11.1. If the perimeter of the design is 28cm, find the length of each side of the hexagon.
Answer:
Given:
Perimeter of figure = 28 cm.
The total number of sides the figure contains is 14.
So length of each side = 28 / 14
= 2 cm.
Question 22.
Perimeter of an isosceles triangle is 50cm. If one of the two equal sides 8cm, find the third side.
Answer:
Perimeter of isosceles triangle = a + b + c
50 = a + 8 + 8
a = 50 – 16
a = 34 cm.
Question 23.
Length of a rectangle is three times its breadth. Perimeter of the rectangle is 40cm. Find its length and width.
Answer:
Given:
Length, l = 3 × Breadth, b
Perimeter of rectangle = 2 × (l + b)
2 × (3b + b) = 40
2 × 4b= 40
8b = 40
b = 5 cm
Length, l = 3 × 5
Length = 15 cm
Breadth = 5 cm
Question 24.
There is a rectangular lawn 10m long and 4m wide in front of Meena’s house (Fig. 6.12). It is fenced along the two smaller sides and onelonger side leaving a gap of 1m for the entrance. Find the length of fencing
Answer:
The perimeter of the fencing will give us the length of fencing.
The lawn is fenced only on three sides.
Two smaller sides and one longer side.
Perimeter = 4 + 4 + 10 – 1
Perimeter = 17 m
Therefore the length of fencing = 17 m.
Question 25.
The region given in Fig. 6.13 is measured by taking 
as a unit. What is the area of the region?
Answer:
Given:
That the region is measured using the rectangle as one unit.
There are thirteen such units.
So area of the region = 13 × 1
= 13 sq.units.
Question 26.
Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140cm. Find the side of this field in metres.
Answer:
Given:
Distance around square field = 200 rods
Perimeter of field = 200 rods
200 × 140 = 4 × side
28,000 / 4 = side
side = 7000
1m = 100 cm
∴ Side of field = 70 m
Question 27.
The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6km. What is the length of the field?
Answer:
Given:
length, l = 2 × breadth, b
Distance covered after 4 times of jogging = 6 km
Perimeter of field = 6 / 4
= 1.5 km or 1500 m
Perimeter = 2 × (l + b)
2 × (l + b) = 1500
2 × (2b + b) = 1500
2 × 3b = 1500
6b = 1500
b = 250 m
Length, l = 2 × 250
Length = 500 m
Question 28.
Three squares are joined together asshown in Fig. 6.14. Their sides are 4cm, 10cm and 3cm. Find the perimeter of the figure.
Answer:
For First Square of side 4 cm, only three sides will be considered and same is for square with 3 cm.
So perimeter of figure = 4 + 4 + 4 + 6 + 10 + 7 + 3 + 3 + 3 + 10
= 54
Question 29.
In Fig. 6.15 all triangles are equilateraland AB = 8 units. Other triangles have been formed by taking the mid points of the sides. What is the perimeter of the figure?
Answer:
Given other triangles are formed by taking midpoints
DF = FE = DE / 2
= 4 / 2
= 2 units
In Δ DIF
DI = DF = IF = 2 units
In Δ TKN and Δ RQU
TK = KN = TN = RQ = UQ = UR = 2 units
NC = RP = 2 units
IG = GF = 1 unit
In Δ HIG
HG = HI = GI = 1 unit
In Δ MLK and Δ XQS
ML = MK = LK = SQ = XS = QX = 1 unit
LN = XR = 1 unit
Perimeter = 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 +1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4
= 45 cm
Question 30.
Length of a rectangular field is 250m and width is 150m. Anuradha runs around this field 3 times. How far did she run?
How many times she should run around the field to cover a distance of 4km?
Answer:
Given:
Length, l = 250 m
Breadth, b = 150 m
Perimeter of field = 2 × (l + b)
= 2 × (250 + 150)
= 2 × 400
= 800 m
So distance covered after running three times = 3 × 800
= 2400 m or 2.4 km
To cover a distance of 4 km, Anuradha should run = 4000 / 800
= 5 times.
Question 31.
Bajinder runs ten times around a squaretrack and covers 4km. Find the length of the track.
Answer:
Given:
Number of times Bajinder run = ten times
Distance covered = 4 km
Solution:
Distance covered in one round = 4000 / 10
= 400 m
Perimeter of track = Distance covered in one round
4 × side = 400
Length of side = 100 m.
Question 32.
The lawn in front of Molly’s house is 12m × 8m, whereas the lawn in front of Dolly’s house is 15m × 5m. A bamboo fencing is built around both the lawns. How much fencing is required for both?
Answer:
For Molly’s House.
Length, l = 12 m
Breadth, b = 8 m
Perimeter of the lawn = Fencing required = 2 × (l + b)
= 2 × (12 + 8)
= 2 × 20
= 40 m.
For Dolly’s House.
Length, l = 15 m
Breadth, b = 5 m
Perimeter of the lawn = Fencing required = 2 × (l + b)
= 2 × (15 + 5)
= 2 × 20
= 40 m.
So total fencing = 40 + 40
= 80 m.
Question 33.
The perimeter of a regular pentagon is 1540cm. How long is its each side?
Answer:
Given:
Perimeter = 1540 cm
Pentagon has 5 sides
So length of each side = 1540 / 5
= 308 cm
Question 34.
The perimeter of a triangle is 28cm. One of it’s sides is 8cm. Write all the sides of the possible isosceles triangles with these measurements.
Answer:
Given:
Perimeter of triangle = 28 cm
Length of one side =8 cm
Triangle is isosceles
So let the equal side be b
Perimeter = a + 2b
8 + 2b = 28
2b = 20
b = 10 cm
So length of equal sides = 10 cm.
The two equal sides can also be 8 cm
16 + a = 28
a = 12 cm.
So new side length = 8, 8, 12 cm.
Question 35.
The length of an aluminium strip is 40cm. If the lengths in cm are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with 15cm length and 5cm breadth can be made from this strip.)
Answer:
Given:
Length of Aluminium strip = 40 cm.
Perimeter of rectangular frame = 40 cm
2 × (l + b) = 40
l + b = 20 cm
So l × b can be any set of values like 19 × 1; 18 × 2; 16 × 4 etc.
Question 36.
Base of a tent is a regular hexagon of perimeter 60cm. What is the length of each side of the base?
Answer:
Perimeter = 60 cm
A hexagon has six sides.
So length of side = 60 / 6
= 10 cm
Question 37.
In an exhibition hall, there are 24 display boards each of length 1m 50cm and breadth 1m. There is a 100m long aluminium strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.
Answer:
Given:
Total display boards = 24
Length of strip = 100 m
Length of one display board = 1 + 0.5
= 1.5 m
Breadth of one display board = 1 m
Perimeter of one display board = 2 × (l + b)
= 2 × (1.5 + 1)
= 5 m
Number of boards which will be framed = 100 / 5
= 20
Number of boards unframed = 24 – 20
= 4
Length of strip for required for remaining boards = 4 × Perimeter of one board
= 4 × 5
= 20 m.
Question 38.
In the above question, how many square metres of cloth is required to cover all the display boards? What will be the length in m of the cloth used, if its breadth is 120cm?
Answer:
Given:
Total display boards = 24
Length of display board = 1 + 0.5 [50cm =50/100 m= 0.5m]
= 1.5 m
Breadth of display board = 1 m
Area of one display board = 1 × 1.5
= 1.5 m2
Area of 24 display boards = 24 × 1.5
= 36 m2
36 m cloth is required to cover the display board.
Breadth = 120 cm
= 1.2 m
Area = l × b
l = 36 / 1.2
l = 30 m.
Question 39.
What is the length of outer boundary of the park shown in Fig. 6.16? What will be the total cost of fencing it at the rate of Rs 20 per metre? There is a rectangular flower bed in the center of the park. Find the cost of manuring the flower bed at the rate of Rs 50 per square metre.
Answer:
Perimeter = Sum of all sides
Perimeter = 200 + 300 + 80 + 300 + 200 + 260
= 1340 m
Rate of fencing = Rs 20
Total Rate = 20 × 1340
= Rs 26,800
Area of rectangular flower bed = 100 × 80
= 8000 m2
Rate of manuring = Rs 50
So total rate = 50 × 8000
= Rs 4, 00,000
Question 40.
Total cost of fencing the park shown in Fig. 6.17 is Rs 55000. Find the cost of fencing per metre.
Answer:
Perimeter of figure = Sum of all sides
= 150 + 100 + 120 + 180 + 270 + 280
= 1100 m
Total cost = Rs 55,000
Cost of fencing = 55,000 / 1100
= Rs 50
Question 41.
In Fig. 6.18 each square is of unit length
(a) What is the perimeter of the rectangle ABCD?
(b) What is the area of the rectangle ABCD?
(c) Divide this rectangle into ten parts of equal area by shading squares.
(Two parts of equal area are shown here)
(d) Find the perimeter of each part which you have divided. Are they all equal?
Answer:
Given each square is of unit length. Figure contains length of 10 squares and width of 6 squares.
Length of rectangle = AD = BC
Sum of length of side of 10 squares = 10 units
Breadth = 6 units.
(a) Perimeter of rectangle = 2 × (l + b)
= 2 × (10 + 6)
= 32 units.
(b) Area of rectangle = 10 × 6
= 60 sq.units
(c)
The rectangle is divided into 10 equal parts of 6 sq.units.
(d) Perimeter of the figure can be found as follows:
Perimeter = 1 + 1+1 +1 +1 +1 +1 +1 +1 +1 +1 +1
= 12 units.
Yes all parts have same perimeter.
Question 42.
Rectangular wall MNOP of a kitchen is covered with square tiles of 15cm length (Fig. 6.19). Find the area of the wall.
Answer:
Number of square tiles = 28
Length of square tile = 15 cm.
Area of square tile = 15 × 15
= 225 sq.cm
Area of 28 tiles = 225 × 28
= 6300 sq.cm
Area of wall = 6300 sq.cm
Question 43.
Length of a rectangular field is 6 times its breadth. If the length of the field is 120cm, find the breadth and perimeter of the field.
Answer:
Given:
Length of rectangle = 120 cm
Breadth of rectangle = b
Length = 6b
Breadth = 120 / 6
= 20 cm
Perimeter = 2 × (l + b)
= 2 × (20 + 120)
= 280 cm.
Question 44.
Anmol has a chart paper of measure 90cm × 40cm, whereas Abhishek has one which measures 50cm × 70cm. Which will cover more area on the table and by how much?
Answer:
Given:
For first chart paper:
Area of first chart paper = 90 × 40
= 3600 sq.cm
For second chart paper:
Area of second chart paper = 50 × 70
= 3500 sq.cm
Hence first chart paper will cover more area by 100 sq.cm.
Question 45.
A rectangular path of 60m length and 3m width is covered by square tiles of side 25cm. How many tiles will there be in one row along its width? How many such rows will be there? Find the number of tiles used to make this path?
Answer:
Given:
Length of path = 60 m
Width of path = 30 m
Side of square tile = 25 cm
Number of tiles in one row = 3 / 0.25
= 12
Number of rows = 60 / 0.25
= 240
Number of tiles = 12 × 240
= 2880
Question 46.
How many square slabs each with side 90cm are needed to cover a floor of area 81sqm.
Answer:
Given:
Side = 90 cm
Area of square lab = 90 × 90
= 8100 sq.cm
Area of slab = 81sq.m = 81 × 10000 = 810000 sq.cm
Number of slabs = 810000 / 8100
= 100
Question 47.
The length of a rectangular field is 8m and breadth is 2m. If a square field has the same perimeter as this rectangular field, find which field has the greater area.
Answer:
Given:
Length of field = 8 m
Breadth of field = 2 m
Perimeter = 2 × (l + b)
= 2 × (8 + 2)
= 20 m
Area of rectangle = 8 × 2
= 16 sq.m
According to question
Perimeter of Square = Perimeter of rectangle
4 × side = 20
Side = 5 m
Area of Square = side2
= 5 × 5
= 25 sq.m
Hence square has greater area.
Question 48.
Parmindar walks around a square park once and covers 800m. What will be the area of this park?
Answer:
Given:
Distance covered in once = Perimeter of square park = 800 m
Perimeter of square = 4 × side
4 × side = 800
side = 200 m
Area of square park = side2
= 200 × 200
= 40,000sq.m
Question 49.
The side of a square is 5cm. How many times does the area increase, if the side of the square is doubled?
Answer:
Side of square = 5 cm
Area of square = 5 × 5
= 25 sq.m
If side doubles = 10 cm
New area = 10 × 10
= 100 sq.m
So the area becomes four times the old area.
Question 50.
Amita wants to make rectangular cards measuring 8cm × 5cm. She has a square chart paper of side 60cm. How many complete cards can she make from this chart? What area of the chart paper will be left?
Answer:
Let ABCD be square chart of side 60 cm.
Cut out rectangular cards measuring 8 cm × 5 cm
Now if we cut along AB then we can cut 12 rectangular cards.
So we can cut 7 rows with 12 rectangular cards each
Area of cut outs of rectangular cards = 7 × 12 × 8 × 5
= 3360 sq.cm
Area of left part = 4 × 60
= 240 sq.cm
Question 51.
A magazine charges Rs 300 per 10sqcm area for advertising. A company decided to order a half page advertisement. If each page of the magazine is 15cm × 24cm, what amount will the company has to pay for it?
Answer:
Charges of 10 sq.cm of advertisement = Rs 300
Charges for 1 sq.cm = 300 / 10
= Rs 30
Area of 1 page of magazine = 15 × 24
= 360 sq.cm
Area of half page = 360 / 2
= 180 sq.cm
So Charges for half page advertisement = 180 × 30
= Rs 5400
Question 52.
The perimeter of a square garden is 48m. A small flower bed covers 18sqm area inside this garden. What is the area of the garden that is not covered by the flower bed? What fractional part of the garden is covered by flower bed? Find the ratio of the area covered by the flower bed and the remaining area.
Answer:
Perimeter of garden = 48 m
4 × side = 48
side = 12 m
Area of garden = side2
= 12 × 12
= 144 sq.m
Area of flower bed = 18 sq.m
Area not covered by flower bed = 144 – 18
= 126 sq.m
Fractional part of garden covered by flower bed = 18 / 144
= 1 / 8
Ratio of area covered by flower and remaining area = 18 / 126
= 1 / 7
= 1:7
Question 53.
Perimeter of a square and a rectangle is same. If a side of the square is 15cm and one side of the rectangle is 18cm, find the area of the rectangle.
Answer:
Length of rectangle = 18 cm
Side of square = 15 cm
Perimeter of square = 4 × 15
= 60 cm
Perimeter of rectangle = Perimeter of square
2 × (l + b) = 60
18 + b = 30
b = 12 cm
Area of rectangle = 18 × 12
= 216 sq.cm
Question 54.
A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Answer:
Side of square by bending small piece of wire = 2 cm.
Total area of small squares = 28 sq.cm
Number of small squares = 28 / 4
= 7
Perimeter of small square = 4 × 2
= 8 cm
Perimeter of 7 small squares = 8 × 7
= 56 cm
Hence original length of wire = 56 cm.
Question 55.
Divide the park shown in Fig. 6.17 of question 40 into two rectangles. Find the total area of this park. If one packet of fertilizer is used for 300sqm, how many packets of fertilizer are required for the whole park?
Answer:
Area of rectangle ABCG = l × b
= 150 × 100
= 15000 sq.m
Area of rectangle DEFG = 270 × 180
= 48600 sq.m
Total area of park =15000 + 48600
= 63600 sq.m
Area used by one packet of fertilizer = 300 sq.m
So total packets required = 63600 / 300
= 121
Question 56.
The area of a rectangular field is 1600sqm. If the length of the field is 80m, find the perimeter of the field.
Answer:
Given:
Area of field = 1600 sq.m
Length of field = 80 m
Area = l × b
b = 1600 / 80
= 20 m
Perimeter = 2 × (l + b)
= 2 × (80 + 20)
= 200 m.
Question 57.
The area of each square on a chess board is 4sqcm. Find the area of the board.
(a) At the beginning of game when all the chess men are put on the board, write area of the squares left unoccupied.
(b) Find the area of the squares occupied by chess men.
Answer:
Area of 1 square = 4 sq.cm
Area of 64 squares = 4 × 64
= 256 sq.cm
(a) Number of chessmen at beginning = 32
Number of squares occupied = 32
Number of unoccupied squares = 32
Area of unoccupied squares = 32 × 4
= 128 sq.cm
(b) Area occupied by chessmen = 32 × 4
= 128 sq.cm
Question 58.
Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter 36cm and find their areas.
Answer:
Perimeter of rectangle = 36 cm.
2 × (l + b) = 36
l + b = 18
So l × b = 16 × 2; 14 × 4; 12 × 6 etc.
Area for l × b = 16 × 2 = 32 sq.cm
Area for second case = 14 × 4
= 56 sq.cm
Area for third case = 12 × 6
= 72 sq.cm
Question 59.
Find all the possible dimensions (in natural numbers) of a rectangle with an area of 36sqcm, and find their perimeters.
Answer:
Area = 36 sq.cm
l × b = 36
Case I = 18 × 2
Case II = 9 × 4
Perimeter for Case I = 2 × (18 + 2)
= 2 × 20
= 40 sq.cm
Perimeter for Case II = 2 × (9 + 4)
= 2 × 13
= 26 sq.cm
Question 60.
Find the area and Perimeter of each of the following figures, if area of each small square is 1sqcm.
Answer:
(i) There are 11 small squares
Area of 11 squares = 11 × 1
= 11 sq.cm
Perimeter = 1 + 1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1
= 16 cm
(ii) In the given figure there are 13 small squares
Area of 13 squares = 13 × 1
= 13 sq.cm
Perimeter = 1 + 1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1+1 + 1 +1 +1 +1 +1 +1 +1 +1 +1 +1
= 27 cm
(iii) In the given figure there are 13 small squares
Area of 13 squares = 13 × 1
= 13 sq.cm
Perimeter = 1 + 1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1+1 + 1 +1 +1 +1 +1 +1 +1 +1 +1 +1 + 1
= 28 cm
Question 61.
What is the area of each small square in the Fig. 6.21 if the area of entire figure is 96sqcm. Find the perimeter of the figure.
Answer:
Area of entire figure = 96 sq.cm
Number of square = 24
Area of small square = 96 / 24
= 4 sq.cm
Area = side2
Side = 2 cm
Perimeter = Total number of sides × length of each side
= 34 × 2
= 68 cm.