Class 6 Maths Chapter 6 Extra Questions Perimeter and Area
Class 6 Maths Perimeter and Area Extra Questions
NCERT Class 6 Maths Chapter 6 Perimeter and Area Extra Questions and Answers
Very Short Answer Type Questions
Question 1.
Find the area of square of side 9 cm.
Solution :
Area = (side)2
= (9)2 = 81 cm2
Question 2.
Find the perimeter of pentagon of side 3 cm.
Solution :
Perimeter of pentagon = 5 × side = 5 × 3 = 15 cm
Short Answer Type Questions
Question 1.
Find the area of a square with perimeter 32 cm.
Solution :
Perimeter = 4 × side
32 = 4 × side
side = 32 + 4 = 8 cm
Area = (side)2
= (8)2 = 64 cm2
Question 2.
Find the length of a rectangle whose perimeter is 60 cm and breadth is 14 cm.
Solution :
Perimeter = 2 × (l + b)
60 = 2 × (l + 14)
60 = 2 l + 28
2 l = 60-28
2 l = 32
l = \(\frac{32}{2}\) = 16 cm
Question 3.
Find the perimeter of given figure.
Solution :
Perimeter of figure = sum of all the sides
= AB + BC + CD + DE + EF + EFG + GH + HA
= 3 m + 3 m + 7 m + 3 m + 3 m + 5 m + 13 m + 5 m
= 42 m
Question 4.
A room is 3 m long and 2 m 25 cm wide. How many squake meters of carpet is needed to cover the floor of the room?
Solution :
Length of the floor = 3 m
Wide of the floor = 2 m 25 cm = 2.25 m
Area of the floor = length × breadth
= 3 × 2.25 sq. m = 6.75 sq. m
6.75 sq. m carpet is needed to cover the floor of the room.
Long Answer Type Questions
Question 1.
The area of a square of side 14 cm is same as that of rectangle of length 56 cm. What is the breadth of rectangle?
Solution :
Area of square = Area of rectangle
side × side = length × breadth 14 × 14 = 56 × breadth
196 = 56 × breadth
breadth = \(\frac{196}{56}\) = 3.5 cm
Question 2.
Find the area of a triangle whose base is 18 cm and height is 7 cm.
Solution :
Area of Triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 18 × 7
= 63 cm2
Question 3.
The length and breadth of a rectangular park are 152 m and 98 m , respectively. A policeman walked 4 rounds around the park. Find the distance covered by him.
Solution :
Length of the rectangular park = 152 m
Breadth of the rectangular park = 98 m
Perimeter of the park = 2 × (length + breadth)
= 2 × (152 + 98) m
= 2 × 250 m = 500 m
Total distance covered by the policeman for 4 rounds = 4 × 500 m = 2000 m
The distance covered by him is 2000 m.
Passage Based Questions
Passage 1: Class 6 students while studying mensuration were given a test of measuring a field using metre tape. The students found out the length of field as 112 m and breadth as 85 m.
Question 1.
The area of the field is
(a) 197 m2
(b) 9520 m2
(c) 394 m2
(d) None of these
Answer:
(b) 9520 m2
Explanation: Area of field = l × b
= 112 × 85
= 9520 m2
Question 2.
The cost of levelling at ₹ 2.00 per square metre is
(a) ₹ 394
(b) ₹788
(c) ₹19040
(d) None of these
Answer:
(c) ₹19040
Explanation: Cost of levelling at the rate of ₹ 2.00 / m2
= Area × Cost
= 9520 × 2
= ₹ 19040
Question 3.
Distance travelled to go round the field 3 times is:
(a) 197 m
(b) 394 m
(c) 1182 m
(d) 591 m
Answer:
(c) 1182 m
Explanation: Distance travelled in taking 1 round
= 2(l + b)
= 2(112 + 85)
= 2(197) = 394 m
Distance covered in 3 rounds = 394 × 3 = 1182 m
Question 4.
What time will a child take if he walks at the rate of 3 m / s (3 rounds)?
(a) 295 s
(b) 354 s
(c) 394 s
(d) 284 s
Solution :
Option (c) is correct.
Explanation: Time = \(\frac{\text { Distance }}{\text { Speed }}\) = \(\frac{1182}{3}\)
= 394 s
Passage 2: The figure below shows the football field for a school tournament.
This should represent TS only
Length = 88 m, breadth = 76 m
Question 1.
What is the perimeter (in m ) of WURQ?
(a) 60
(b) 240
(c) 960
(d) 3344
Answer:
Option (b) is correct.
Explanation:
l = \(\frac{88}{2}\) = 44 m, b = 76 m
Perimeter = 2(44 + 76) = 240 m
Question 2.
“The area enclosed by QRUW is equal to the area enclosed by WUST” Is the statement true? Give reason.
Solution :
Yes, because areas of QRUW and WUST are the same which is 3344 sQuestion m
= 44 m, b = 76 m
Area = 44 × 76 = 3344 sq m
Question 3.
What is the area (in m2) of the penalty area?
(a) 30
(b) 272
(c) 480
(d) 720
Answer:
Option (d) is correct.
Explanation: Given, 1 unit = 4 m
Thus, Length = 9 × 4 = 36 m
and, Breadth = 5 × 4 = 20 m
Area of rectangle = length × breadth
= 36 × 20 = 720 m2
Question 4.
Is the perimetsr of the penalty area of the football field double the perimeter of the goal area? Give reason.
Solution :
No
Perimeter of penalty area = 2(36 + 20)
= 2 × 56 = 112 m
Perimeter of goal area = 2(12 + 20)
= 2 × 32 = 64 m
64 ≠ 112
Question 5.
Does the school football field meet the FIFA standards? Give reason.
[FIFA standards: Length = 90 m to 120 m , Breadth = 45 m to 90 m]
Solution :
Yes
The standard dimensions for football field by FIFA are l = 120 m to 90 m , breadth = 90 m to 45 m And, School football dimensions lie within these limits. l = 88 m(120 < 88 < 90) and b = 76 m(90 < 76 < 45)
Question 6.
Find the perimeter of the shape WURMNOPQW.
Solution :
280 m
WU + UR + Rm + MN + NO + OP + PQ + QW
= 76 + 44 + 20 + 20 + 36 + 20 + 20 + 44 = 280 m
Passage 3: The figure below shows a square wooden frame enclosing a square picture.
Question 1.
What is the area (in cm2) of the frame?
(a) 81
(b) 144
(c) 181
(d) 225
Answer:
(a) 81
Explanation: Area of frame
= area of outer square – area of inner square
= 152-122
= 225-144
= 81 cm2
Question 2.
What is the perimeter (in cm) of the picture?
(a) 12
(b) 48
(c) 60
(d) 144
Answer:
(b) 48
Explanation: Perimeter of square = 4 a = 4 × 12 = 48 cm
In the following questions, out of the four options, only one is correct. Write the correct answer
Question 1.
Following figures are formed by joining six unit squares. Which figure has the smallest perimeter in the given figures ?
(a) (ii)
(b) (iii)
(c) (iv)
(d) (i)
Answer :
Option (d) is correct.
Explanation: In (i) figure, only 10 edges of squares are at the boundary.
So, it has the smallest perimeter.
Question 2.
A square shaped park ABCD of side 100 m has two equal rectangular flower beds each of size 10 m × 5 m. Length of the boundary of the remaining park is :
(a) 360 m
(b) 400 m
(c) 340 m
(d) 460 m
Answer :
Option (b) is correct.
Explanation: Length of the boundary of the remaining park = Perimeter of remaining park
= (90 + 5 + 10 + 95 + 90 + 5 + 10 + 95) m
= 400 m
Question 3.
The side of a square is 10 cm . How many times will the new perimeter become if the side of square is doubled?
(a) 2 times
(b) 4 times
(c) 6 times
(d) 8 times
Answer :
Option (a) is correct.
Explanation: We know that, When side get doubled then the perimeter also gets doubled (i.e. two times).
Question 4.
Length and breadth of a rectangular sheet of paper are 20 cm and 10 cm , respectively. A rectangular piece is cut from the sheet as shown in given Fig. 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 :
Option (a) is correct.
Explanation: We know that, On cutting the rectangular piece the perimeter remains the same but the area of the remaining sheet changes.
Question 5.
Two regular hexagons of perimeter 30 cm each are joined as shown in Fig. The perimeter of the new figure is :
(a) 65 cm
(b) 60 cm
(c) 55 cm
(d) 50 cm
Answer :
Option (d) is correct.
Explanation: Perimeter of new figure
= (25 + 25) = 50 cm
Match the Column.
I. Match the shapes (each sides measures 2 cm) in column I with the corresponding perimeters in columns II :
Answer :
(a) – (iv)
(b) – (i)
(c) – (ii)
(d) – (iii)
II. Match the following :
Answer :
(a) – (iii)
(b) – (iv)
(c) – (ii)
(d) – (i)
Fill in the Blanks
Question 1.
Perimeter of the shaded portion in Fig is
Answer :
Bm+ MD + DE + EN + NG + GH}
Question 2.
The amount of region enclosed by a plane closed figure is called its …… –
Answer :
area.
Question 3.
4 Area of a rectangle with length 5 cm and breadth 3 cm is …… +
Answer :
15 sq cm
Question 4.
A rectangle and a square have the same perimeter.
(a) The area of the rectangle is ……
(b) The area of the square is ……
Answer :
(a) 12 sq units (b) 16 -sq units
Question 5.
(a) 1 m = …… cm.
(b) 1 sq cm = …… cm × 1 cm.
(c) 1 sq m = 1 m × …… m = 100 cm × …. cm.
(d) 1 sq m = …… sq cm.
Answer :
(a) 100
(b) 1
(c) 1,100
(d) 10000.
True/False
Question 1.
If length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.
Answer :
True
Question 2.
Area of a square is doubled if the side of the square is doubled.
Answer :
False
Question 3.
Perimeter of a regular octagon of side 6 cm is 36 cm .
Answer :
False
Question 4.
A farmer who wants to fence his field, must find the perimeter of the field.
Answer :
True
Question 5.
An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
Answer :
False
Short Answer Type Questions
Question 1.
Four regular hexagons are drawn, so as to form the design as shown in figure, if the perimeter of the design is 28 cm. Then, find the length of each side of the hexagon.
Solution :
Given, four regular hexagons as shown in the figure:
Perimeter of the design = 28 cm
Perimeter of the given design
= Sum of all outer sides of the four hexagon
Here, this figure has 14 outer equal sides.
∴ Perimeter of the design = 14 × Length of one side of hexagon
28 = 14 × Length of one side of hexagon Length of one side of hexagon = \(\frac{28}{14}\) = 2 cm
Hence, the length of each side of the hexagon is 2 cm.
Question 2.
Perimeter of an isosceles triangle is 50 cm . If one of the two equal sides is 18 cm , find the third side.
Solution :
Given, perimeter = 50 cm
Perimeter of an isosceles triangle
= sum of its all side
Perimeter = a + b + c
50 = 18 + 18 + Third side
50 – 36 = Third side
Third side = 14 cm
Question 3.
There is a rectangular lawn 10 m long and 4 m wide in front of Meena’s house. It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance. Find the length of fencing,
Solution :
Given, width of the lawn AB = EF = 4 m and length of the lawn, BE = 10 m
Total fength of fencing $=A B+ (B C+ D E)+ EF
[Here, we subtract the length of gap CD from BE]
= 4 cm + (10-1) cm + 4 cm
= (4+ 9+ 4) cm = 17 m
Hence, the length of fencing of the lawn is 17 m.
Question 4.
Tahir measured the distance around a square field as 200 rods (lathi). Later, he found that the length of this rod was 140 cm , Find the side of this field in metres.
Sol. Given, Tahir measured the distance around a square field as 200 rods (lathi).
Distance covered by Tahir in one round = Perimeter of the square field i.e., Perimeter of a square field = 200 rods Later on, Tahir found that the length of this rod was 140 cm.
∴ Perimeter of a square field (in cm)
= 200 × 140 cm
and perimeter of a square field in meters
= \(\frac{200 × 140}{100}\) m
[1 cm = \(\frac{1}{100}\) m
=2 × 140 m = 280 m
We know that, perimeter of a square field
= 4 × Length of one side
280 = 4 × Length of one side
Length of one side = \(\frac{280}{4}\) = 70 m
Hence, the side of a square field is 70 m.
Question 5.
Three squares are joined together as shown in Fig. Their sides are 4 cm, 10 cm and 3 cm. Find the perimeter of the figure.
Solution :
Given, sides of three squares are 4 cm, 10 cm and 3 cm , respectively.
Total perimeter of given squares
= sum of all outer sides of the figure
= 4 + 4 + 4 + 6 + 10 + 7 + 3 + 3 + 3 + 10
= 54 cm
Question 6.
In the given figure, all triangles are equilateral and 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, Δ ABC is an equilateral triangle,
Here,
AB = 8 units
∴ AB = BC = CA = 8 units
Thus, Δ ADE is an equilateral triangle.
Here, E is the mid-point of A B.
∴ AE = BE = \(\frac{AB}{2}\) = 4 units
Now, in Δ ADE,
AD = DE = EA = 4 units
Similarly, equilateral triangles are Δ BOT and Δ UPC, having each sides i.e., BO = OT = BT = UC = PC = PU = 4 units.
It is also clear that OC = PA = 4 units.
Also, Δ D I F is an equilateral triangle.
Here, F is the mid-point of D E.
∴ DF = FE = \(\frac{D E}{2}\) = \(\frac{4}{2}\) = 2 units
Now, in Δ DIF, DI = IF = DF = 2 units
Similarly, in Δ TKN and Δ R Q U,
TK = KN = TN = RQ = UQ = UR
= 2 units
It is also clear that NO = RP = 2 units Also, Δ HIG is an equilateral triangle. Here, G is the mid-point of IF.
∴ IG = GF = \(\frac{IF}{2}\) = \(\frac{2}{2}\) = 1 unit
Now, in Δ HIG, HG = HI = GI = 1 unit
Similarly in Δ MLK and Δ XQS.
ML = MK = SQ = XS = QX
= 1 unit
It is also clear that G F=L N=X R=1 unit
Now, perimeter of the given figure
= Sum of all outer sides of the given figure
= AD + DI + IH + HG + GF + FE + E B + B T
TKM + LM + LN + NO + OC + CU + UQ + QS + XS + XR + PR + PA
= 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4
= 45 units
Long Answer Type Questions
Question 1.
Length of a rectangular field is 250 m and width is 150 m . 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 4 km ?
Solution :
Given, length of rectangular field (l) = 250 m and width of rectangular field (b) = 150 m
Perimeter of the fields = 2(l+b)
= 2(250+150) m
= (2 × 400) m
= 800 m
Since, Distance covered in one round
= perimeter of the fields
= 800 m
So, Distance covered in three rounds =3 × 800
= 2400 m
Now, no. of rounds to cover 4 km , i.e., 4000 m
= \(\frac{4000}{800}\)
= 5
1 km = 1000 m
Hence, she should run 5 times around the field to cover the distance of 4 km.
Question 2.
In an exhibition hall, there are 24 display boards each of length, 1 m 50 cm and breadth 1 m . There is a 100 m 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.
Solution :
Given, total display boards = 24
Length of one display boards = 1 m + 50 cm
= 1 m + \(\frac{50}{100}\) m
= 1.5 m
Breadth of one display board = 1 m
Perimeter of display board
= 2 × (length + Breadth)
= 2 × (1.5+1) m
= 2 × 2.5 m
= 5 m
Length of strip = 100 m
Now, no. of boards will be framed
= \(\frac{\text { Length of strip }}{\text { Perimeter of one board }}\)
= \(\frac{100}{5}\) = 20
This means that out of 24 only 20 boards will be framed.
No. of boards left unframed = 24-20 = 4
∴ length of the strip required for remaining boards
= 4 × perimeter of one board
= 4 × 2(1.5+1)
= 4 × 2 × 2.5
= 20 m
Question 3.
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 120 cm ?
Answer:
Area of one display board =1.5 × 1 = 1.5 sq. m
Area of 24 display boards = 24 × 1.5 = 36 sq. m
Hence 36 sq. m of cloth is required to cover all the display boards.
Given that the breadth of the cloth is 120 cm = 1.2 m
Length × Breadth = area
Length × 1.2 =36
Length = \(\frac{36}{1.2}\)
= 30 m
Question 4.
What is the length of outer boundary of the park shown in figure? What will be the total cost of fencing it at the rate of ₹ 20 per meter? There is a rectangular flower bed in the centre of the park. Find the cost of manuring the flower bed at the rate of ₹ 50 per square meter.
Solution :
Let the park be ABCDEF.
Here, AB = 300 m, BC = 80 m, CD = 300 m,
DE = 200 m, E F=260 m and F A = 200 m.
The total length of outer boundary of the park or perimeter of the park = Sum of outer length of all sides
= AB + BC + CD + DE + EF + FA
= 300 m + 80 m + 300 m + 200 m + 260 m + 200 m
= 1340 m
Hence, the total length of outer boundary of the park is 1340 m.
Given, cost of fencing per meter = ₹ 20
Cost of fencing the park of 1340 m = ₹ 20 × 1340
= ₹ 26800
Given, a rectangular flower bed in the centye of the park.
Length of rectangular flower bed = 100 m and breadth of rectangular flower bed = 80 m
Area of the rectangle flower bed
= Length × Breadth
= 100 m × 80 m = 8000 sq m
∴ Cost of manuring the flower bed per sq meter = ₹ 50
∴ Cost of manuring the flower bed of 8000 sq cm.
= ₹ 8000 × ₹ 50
= ₹ 400000