## 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**

Question 1.

Find the perimeter of the following figures:

(a) Perimeter = AB + BC + CD + DA

= ___ + ___ + ___ + ____

= ____

(b) Perimeter = AB + BC + CD + DA

= ___ + ___ + ___ + ____

= _____

(c) Perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= ____ + ____ + ___ + _____

____ + ____ + ___ + _____

____ + ____ + ___ + _____

= _____

(d) Perimeter = AB + BC + CD + DE + EF + FA

= ___ +____ + ____ + ____ + ___ + ____

= _______

Note: For finding the perimeter of any closed figure made up entirely of line segment, simply find the sum of the lengths of all the sides.

Solution:

(a) ∵ AB = 40 cm, BC = 10 cm,

AD = 10 cm and DC = 40 cm

∴ Perimeter = AB + BC + CD + DA

= 40 cm + 10 cm + 40 cm + 10 cm

= 100 cm

(b) ∵ AB = 5 cm, BC = 5 cm,

CD = 5 cm and AD = 5 cm

∴ Perimeter = AB + BC + CD + AD

= 5 cm + 5 cm + 5 cm + 5 cm = 20 cm

(c) Here, perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm = 28 cm

(d) Here, perimeter = AB + BC + CD + DE + EF + FA

= 100 m + 120 m + 90 m + 45 m + 60 m + 80 m

= 495 m

Question 2.

Find the perimeter of each of the following figures:

Solution:

(a) Perimeter of the given figure = sum of all the sides

= 4 cm + 5 cm + 1 cm + 2 cm = 12 cm

(b) Perimeter of the given figure = sum of all the sides

= 23 cm + 35 cm + 40 cm + 35 cm = 133 cm

(c) Perimeter of the given figure = sum of all the sides

= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d) Perimeter of the given figure = sum of all the sides

= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e) Perimeter of the given figure = sum of all the sides

= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm

= 15 cm

(f) Perimeter of the given figure = sum of all the sides = 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm = 52 cm

Question 3.

Find the perimeter of the following rectangles:

Solution:

Question 4.

Find the perimeter of each of the following shapes:

(a) A triangle of sides 3 cm, 4 cm and 5 cm.

(b) An equilateral triangle of side 9 cm.

(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution:

(a) Sides of the triangle are: 3 cm, 4 cm, 5 cm

∴ Perimeter = sum of all the sides

= 3 cm + 4 cm + 5 cm

= 12 cm

(b) Length of each side = 9 cm

∵ Perimeter of an equilateral triangle = 3 × length of one side

∴ Perimeter of the given equilateral triangle = 3 × 9 cm = 27 cm

(c) Sides of the given triangle are: 8 cm, 6 cm and 8 cm

∴ Perimeter of the triangle = sum of all the sides

= 8 cm + 6 cm + 8 cm = 22 cm

Question 5.

Find the perimeter of a piece of land measuring 2 km × 60 m.

Solution:

Here, Length = 2 km = 2 × 1000 m

[∵ 1 km = 1000 m]

= 2000 m

Breadth = 60 m

∵ The piece of land is in the form of a rectangle, and perimeter of rectangle

= 2[Length + Breadth]

∴ Perimeter of the given piece of land

= 2[2000 m + 60 m] = 2[2060 m] = 4120 m

Question 6.

Find the distance walked by Mohan if he takes three rounds of a square park of side 85 m.

Solution:

∵ The side of the square park = 85 m

∴ its perimeter = 4 × Side = 4 × 85 m = 340 m

∵ Distance walked in 1 round = 340 m

∴ Distance walked by Mohan in 3 rounds

= 3 × 340 m = 1020 m

Question 7.

A rectangular park is 55 m wide and 89 m long. Find the cost of fencing it at the rate of ₹ 22.50 per 1oo metres.

Solution:

Length 89 m and Breadth = 55 m

∵ Perimeter of a rectangle

= 2 × [Length + Breadth]

Perimeter of the rectangular park

= 2[89 m + 55 m] = 2[144 m] = 288 m

∵ Length of the wire = 288 m

∴ Cost of fencing

= [Perimeter] × [Cost of fencing]

= [288 m] × [₹22.50 per 100 metre]

= ₹ 288 × \(\frac{225}{10}\) × \(\frac{1}{100}\) = ₹ 64.80

Question 8.

Which has a greater perimeter: a square plot of side 75 m or a rectangular plot with sides 100 m and 65 m and by how much?

Solution:

∵ Perimeter of a square = 4 × Side

∴ Perimeter of the square plot = 4 × 75 m = 300 m

On the other hand, perimeter of rectangle = 2 × [Length + Breadth]

∴ Perimeter of the given rectangular plot = 2[100 m + 65 m] = 2[165 m] = 330 m

∵ 330 m > 300 m and 330 m – 300 m = 30 m

∴ The perimeter of the rectangular plot is more than square plot by 30 m.

Question 9.

Find the area of the following figures by counting squares:

Solution:

Here, area of one square = 1 sq. unit and area of half square = \(\frac{1}{2}\) sq. unit

(a) ∵ Number of full squares = 9

∴ Area of the portion covered by the figure = 9 × 1 sq. unit = 9 sq. units

(b) ∵ Number of full squares = 5

∴ Area of the figure = 5 × 1 sq. unit = 5 sq. units

(c) Number of full squares = 2

Half squares = 4

∴ Area covered by the figure

= [(2 × 1) + (\(\frac{1}{2}\) + 4)] sq. unit

= [2 + 2] sq. units = 4 sq. units

(d) Number of full squares = 8

∴ Area covered by the figure

= 8 × 1 sq. unit = 8 sq. units

(e) ∵ Number of full squares = 10

∴ Area covered by the figure

= 10 × 1 sq. unit = 10 sq. units

(f) ∵ Number of full squares = 2

Half squares = 4

∴ Area of the figure

= (2 × 1) sq. units + (\(\frac{1}{2}\) × 4) sq. units

= 2 sq. units + 2 sq. units = 4 sq. units

(g) Number of full squares = 4

Half squares = 4

∴ Area of the figure

= (4 × 1) sq. units + (\(\frac{1}{2}\) × 4) sq. units

= 4 sq. units + 2 sq. units = 6 sq. units

(h) ∵ Number of full squares = 5

∴ Area of the figure = 5 × sq. unit = 5 sq. units

(i) ∵ Number of full squares = 9

∴ Area of the figure = 9 × 1 sq. unit = 9 sq. units

(j) ∵ Number of full squares = 2

Half squares = 4

∴ Area of the figure

= (2 × 1) sq.units + (\(\frac{1}{2}\) × 4)sq. units

= 2 sq. units + 2 sq. units = 4 sq. units

(k) ∵ Number of full squares = 4

Half squares = 2

∴ Area of the figure

= 4 × 1 sq.unit + \(\frac{1}{2}\) × 2 sq.unit

= 4 sq.units + 1 sq.unit = 5 sq.units

Question 10.

Find the area of the figure by splitting it into rectangles.

Solution:

By splitting the given figure into 2 rectangles by a dotted line, we get

Area of rectangle I:

Length = 12 cm and Breadth = 10 cm

Area = Length × Breadth

12 cm × 10 cm = 12o sq cm

Area of rectangle II:

Length = 28 cm and Breadth = 8 cm

Area = Length × Breadth = 28 cm × 8 cm = 224 sq cm

∴ Area of the given figure

= [Area of rectangle I] + [Area of rectangle II]

= 120 sq cm + 224 sq cm = 344 sq cm

Question 11.

By splitting the following figures into rectangles, find their areas. (The measures are given in centimetres.)

Solution:

(a) Let us split the given figure into various squares and rectangles as shown in the following figure.

Now, area of rectangle I = length × breadth

= 8 cm × 4 cm = 32 sq. cm

Area of square II = side × side

= 6 cm × 6 cm = 36 sq. cm

Area of rectangle III = length × breadth

= 4 cm × 2 cm = 8 sq. cm

Area of square IV = side × side

= 6 cm × 6 cm = 36 sq. cm

∴ Total area of the figure = 32 sq. cm + 36 sq. cm + 8 sq. cm + 36 sq. cm

= 112 sq. cm

(b) Splitting the given figure into various rectangles, we have:

Area of rectangle I = length × breadth

= 6 cm × 2 cm = 12 sq. cm

Area of rectangle’ II = length × breadth

= 6 × 2 sq. cm = 12 sq. cm

Area of rectangle III = length × breadth

= 6 × 2 sq. cm = 12 sq. cm

∴ Total area of the figure = Area of rectangle I + Area of rectangle II + Area of rectangle III

= 12 sq. cm + 12 sq. cm + 12 sq. cm = 36 sq. cm

Question 12.

Split the following shapes into rectangles and find their areas. (The measures are given in centimetres.)

Solution:

(a) Splitting the given figure into rectangles, we have

Area of rectangle I = length × breadth

= 10 cm × 2 cm = 20 sq.cm

Area of rectangle II = length × breadth

= 10 cm × 2 cm

= 20 sq. cm

∴ Total area of the figure = 20 sq. cm +20 sq. cm

= 40 sq. cm

(b) Splitting the given figure into rectangle and squares, we get

Area of square I = 7 cm × 7 cm = 49 sq. cm

Area of rectangle II = (7 + 7 + 7) cm × 7 cm

= 21 cm × 7 cm

= 147 sq. cm

Area of square III = 7 cm × 7 cm = 49 sq. cm

∴ Total area of the given figure

= Area of square I + Area of rectangle II + Area of square III

= 49 sq. cm + 147 sq. cm + 49 sq. cm = 245 sq. cm

Question 13.

Assertion: Side of the square whose area is 225 sq m is 15 m.

Reason: The standard unit of area is sq units.

In the given question, a statment of Assertion is followed by a statement of Reason. Choose the correct option as:

(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) Assertion is true and the reason is false.

(d) Assertion is false and the reason is true.

Solution:

(b) Since, side of the square = 15 m

So, area of the square is 15m × 15m = 225 sq m

Hence the assertion is true, further the standard unit of area is sq units.

So, reason is also true.

But here, the reason is not the correct explanation of the assertion, so the correct answer is (b).

Question 14.

Case Based Question

An excursion trip of students of class 6 visited the Jungle Safari. For their safety, the guide provides the map of the park to each student. The map is as given below:

Based on the above information answer the following questions.

(i) What is the perimeter of the Jungle area?

(ii) If there is fencing around the resort area for safety, find the length of the fencing.

(iii) What is the area of the parking, if the area of parking and the area of the ticket and officers room are same?

Solution:

(i) Perimeter of any rectangular area = 2 (Length + breadth)

So, perimeter of Jungle area = 2 (10 + 8) = 36 km.

(ii) Fencing around the resort area = perimeter of resort area = 6 km + 5 km + 10 km = 21 km.

(iii) Since, parking area and the ticket and officers room makes a rectangle.

And length of this rectangle = 8 km

And breadth of this rectangle = 2 km

So, area of this rectangle length × breadth

= 8 km × 2 km = 16 sq km

So, area of parking = \(\frac{16}{2}\) = 8 sq km.