Ncert Solutions For Class 6 Maths Ex 10.1

Ncert Solutions For Class 6 Maths Chapter 10 Ex 10.1

Q.1 Find the perimeter of each of the following figures

(a)

1

Ans.

Perimeter is the distance covered along the boundary of a closed figure.

Perimeter = Sum of the distances or lengths of all the sides

= PQ + QR + RS + SP.

= 3 cm + 2 cm + 1 cm + 4 cm.

= (3 + 2 + 1 + 4) cm

= 10 cm

The perimeter of the given figure is 10 cm.

 

(b)

2

Ans.

Perimeter = Sum of the distances or lengths of all the sides

= EF + FG + GH + HE.

= 20 cm + 30 cm + 20 cm + 45 cm.

= (20 + 30 + 20 + 45) cm

= 105 cm

The perimeter of the given figure is 105 cm.

 

(c)

3

Ans.

Perimeter = Sum of the distances or lengths of all the sides.

= LM + MN + NO + OL

= 20 cm + 20 cm + 20 cm + 20 cm

= (20 + 20 + 20 + 20) cm

= 80 cm

The perimeter of the given figure is 80 cm.

 

(d)

4

Ans.

Perimeter = Sum of the distances or lengths of all the sides.

= AB + BC + CD + DE + EA

= 5 cm + 5 cm + 5 cm + 5 cm + 5 cm

= (5 + 5 + 5 + 5 + 5) cm

= 25 cm

The perimeter of the given figure is 25 cm.

 

(e)

5

Perimeter = Sum of the distances or lengths of all the sides.

We are simplifying the problem by considering a single ‘L’ shaped section and finding its perimeter. There are four such ‘L’ shaped sections and the total perimeter is four times the perimeter of a single section.

6

Perimeter of the ‘L’ shaped section = AB + BC + CD + DE + EF

= 3 + 4 + 1 + 3 + 2

= 13 cm

The perimeter of a single ‘L’ section is 13 cm

The perimeter of the entire figure = 4 x perimeter of each ‘L’ section.

= 4 x 13 cm

= 52 cm

The perimeter of the entire figure is 52 cm.

 

 

Q2. The lid of a rectangular box of sides 50 cm by 10 cm is sealed all around with a tape. What is the length of the tape required?

Ans.

7

Dimensions of the given box:

Length of the rectangular box (l) = 50 cm

Breadth of the rectangular box (b) = 10 cm

Perimeter of a rectangle = 2 x (l + b)

= 2 x (50 cm + 10 cm)

= 2 x (60 cm)

= 120 cm

The perimeter of the rectangular box = Length of the tape required to measure the box = 120 cm

 

 

Q3. A table-top measures 3 m 35 cm by 1 m 40 cm. What is the perimeter of the table-top?

Ans.

8

The length of the rectangle should be greater than the breadth (l \(>\) b)

Length of the rectangle (l) = 3 m 35 cm

= (300 + 35) cm (∵ 1 m = 100 cm)

= 335 cm

Breadth of the rectangle (b) = 1 m 40 cm

= (100 + 40) cm (∵ 1 m = 100 cm)

= 140 cm

Perimeter of the table top = Perimeter of the rectangle

= 2 x (length + breadth)

= 2 x (l + b)

= 2 x (335 cm + 140 cm)

= 2 x (475 cm)

= 950 cm

The perimeter of the table top = 950 cm = 9.5 m

 

Q4. A photo of sides 17.8 cm by 12.7 cm is to be framed with wood. What is the length of wooden strip required?

Ans:

Length of photograph (I) = 17.8 cm

Breadth of photograph (b) = 12.7 cm

Length of required wooden strip = Perimeter of Photograph

= 2 x (/ + b)

= 2 x (17.8 + 12.7)

= 2 x 30.5 = 61 cm

 

Q5. Each side of rectangular piece of land which measures 0.8km by 0.7km is to be fenced with 5 rows of wires. How much length of wire is needed?

Ans:

Length of land (I) = 0.8 km

Breadth of land (b) = 0.7 km

Perimeter = 2 x (I + b) = 2 x (0.8 + 0.7) = 2 x 1.5 = 3.0 km

Length of wire required = 5 x 3 = 15 km

 

Q6. Find the perimeter of the following shapes:

(a) An equilateral triangle of side 5 cm.

(b) An Isosceles triangle with equal sides 6 cm each and the third side 5 cm.

(c) A triangle of sides 6 cm, 3 cm and 2 cm

Ans:

(a) Perimeter = of an equilateral triangle = 3 x Side of triangle

= (3 x 5) cm = 15 cm

(b) Perimeter = (6 cm + 3 cm + 2 cm) = 11 cm.

(c) Perimeter = (2 x 6) + 5 = 17 cm

 

Q7. A triangle has sides measuring 15 cm, 12 cm and 13 cm. Find its perimeter.

Ans:

Perimeter of the triangle = Sum of the lengths of all sides of the triangle

Perimeter = 15 + 12 + 13 = 40 cm

 

Q8. A regular hexagon of each side measuring 6 m is given. Find its perimeter.

Ans:

Perimeter of regular hexagon = 6 x Side of regular hexagon

Perimeter of the regular hexagon = 6 x 6 = 36 m

 

Q9. Find the side of the square whose perimeter is 16 m.

Ans:

Perimeter of square = 4 x Side

16 = 4 x Side

\(Side = \frac{16}{4} = 4\ cm\)

Side = 4

 

Q10. Find the side of a regular pentagon if its perimeter is 150 cm.

Ans:

Perimeter of regular pentagon = 5 x Length of side

150 = S x Side

\(Side = \frac{150}{5} = 30\ cm\)

Therefore, Side = 30 cm

 

Q11. A piece of string is 120 cm long. What will be the length of each side if the string is used to form

(a) A regular hexagon?

(b) A square?

(c) An equilateral triangle?

Ans:

(a) Perimeter = 4 x Side

120 = 4 x Side

\(Side = \frac{120}{4} = 30\ cm\)

(b) Perimeter = 3 x Side

120 = 3 x Side

\(Side = \frac{120}{3} = 40\ cm\)

(c) Perimeter = 6 x Side

120 = 6 x Side

\(Side = \frac{120}{6} = 20\ cm\)

 

Q12: What is the third side of a triangle if two of its sides are 15 cm and 12 cm, whose perimeter is 40 cm?

Ans:

Perimeter of triangle = Sum of all sides of the triangle

40 = 15 + 12 + Side

40 = 27 + Side

Side = 40 – 27 = 13 cm

Hence, the third side of the triangle is 13 cm.

 

Q13: A square park of side 240m is to be fenced. Find the cost of fencing if the rate of fencing per meter is Rs.10.

Ans:

Length of fence required = Perimeter of the square park

= 4 x Side = 4 x 240 = 960 m

Cost for fencing 1 m of square park = Rs 10

Cost for fencing 960 m of square park = 960 x 10 = Rs 9600

 

Q14: A rectangular park of length 150 and breadth 175 is to be fenced. Find the cost if the fencing rate is Rs.15 per meter.

Ans:

Length of rectangular park (I) = 150 m

Breadth of rectangular park (b) = 175 m

Length of wire required for fencing the park = Perimeter of the park

= 2 x (/ + b) = 2 x (150 + 175) = 2 x 325 = 650 m

Cost for fencing 1 m of the park = Rs 15

Cost for fencing 650 m of the square park = 650 x 15 = Rs 9750

 

Q15. Adam runs around a square park of side 80 m. Eve runs around a rectangular park with length 70 m and breadth 40 m. Who covers a comparatively lesser distance?

Ans:

Distance covered by Adam = 4 x Side of square park

= 4 X 80 = 240 m

Distance covered by Eve = 2 x (70 + 40)

= 2 x 110 = 220 m

Therefore, Eve covers less distance.

 

Q16: What is the perimeter of each of the following figures? What do you infer from the answers?

9

Ans:

(a) Perimeter of square = 4 x 50 = 200 cm

(b) Perimeter of rectangle = 2 x (20 + 80) = 200 cm

(c) Perimeter of rectangle = 2 x (40 + 60) = 200 cm

(d) Perimeter of triangle = 60 + 60 + 80 = 200 cm

It can be inferred that all the figures have the same perimeter.

 

Q17: Ram buys 9 square slabs, each having a side of 2 m. Initially, he lays them in the form of a square.

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(a) Looking at the figure, find out the perimeter of his arrangement.

(b) His friend Mani looked at the figure and asked him to lay them out in the shape of a cross. Looking at the second figure, find out the perimeter of the cross.

(c) Out of the two shapes, find out which one has a larger perimeter.

(d) Mani suggests that there’s a way to get an even greater perimeter. Can you guess a way of doing it? (The paving slabs must meet along complete edges. That is, they cannot be broken.)

Ans:

(a) Side of the square = (3 x 2) m = 6 m

Perimeter of the square = (4 x 3) m = 12 m

(b) Perimeter of the cross = 2+4+4+2+4+4+2+4+4+2+4+4 = 40m

(c) The cross formation has the greatest perimeter of all arrangement.

(d) Formation with perimeter which is greater than 40 m can’t be determined.

 

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