RD Sharma 2020 2021 Solutions for Class 6 Maths Chapter 20 - Mensuration

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RD Sharma 2020 2021 Solutions for Class 6 Maths Chapter 20 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among class 6 students for Maths Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2020 2021 Book of class 6 Maths Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2020 2021 Solutions. All RD Sharma 2020 2021 Solutions for class 6 Maths are prepared by experts and are 100% accurate.

Page No 20.10:

Question 1:

Find the perimeters of the rectangles whose lengths and breadths are given below:

(i) 7 cm, 5 cm
(ii) 5 cm, 4 cm
(iii) 7.5 cm, 4.5 cm

Answer:

(i) Perimeter of a rectangle = 2 $\times$ (Length + Breadth)
$∵$ Length = 7 cm, Breadth = 5 cm
$∴$ Perimeter = 2 $\times$ (7 + 5) = 2 $\times$ (12) = 24 cm

(ii) Perimeter of a rectangle = 2 $\times$ (Length + Breadth)
$∵$ Length = 5 cm, Breadth = 4 cm
$∴$ â€‹Perimeter = 2 $\times$ (5 + 4) = 2 $\times$ (9) = 18 cm

(iii) Perimeter of a rectangle = 2 $\times$ (Length + Breadth)
$∵$ Length = 7.5 cm, Breadth = 4.5 cm
$∴$ Pâ€‹erimeter = 2 $\times$ (7.5 + 4.5) = 2 $\times$ (12) = 24 cm

Page No 20.10:

Question 2:

Find the perimeters of the squares whose sides are given below:

(i) 10 cm
(ii) 5 m
(iii) 115.5 cm

Answer:

Perimeter of a square = 4 $\times$ (Length of one side)

(i) Length of one side = 10 cm
Perimeter = 4 $\times$ 10 = 40 cm

(ii) Length of one side = 5 m
Perimeter = 4 $\times$ 5 = 20 m

(iii) Length of one side = 115.5 cm
Perimeter = 4 $\times$ 115.5 = 462 cm

Page No 20.10:

Question 3:

Find the side of the square whose perimeter is:

(i) 16 m
(ii) 40 cm
(iii) 22 cm

Answer:

Side of a square = $\frac{P e r i m e t e r}{4}$
(i) Perimeter = 16 m
$∴$ Side of this square = $\frac{16}{4}$ = 4 m
(ii) Perimeter = 40 cm
$∴$ Side of this square = $\frac{40}{4}$ = 10 cm
(iii) Perimeter = 22 cm
$∴$ Side of this square = $\frac{22}{4}$ = 5. 5cm

Page No 20.10:

Question 4:

Find the breadth of the rectangle whose perimeter is 360 cm and whose length is

(i) 116 cm
(ii) 140 cm
(iii) 102 cm

Answer:

Perimeter of a rectangle = 2 (Length + Breadth)

$∴$ Breadth of the rectangle = $\frac{Perimeter}{2} -$ Length

(i)
Perimeter = 360 cm
Length = 116 cm

$∴$ Breadth = $\frac{360}{2} - 116$ = 180 $-$ 116 = 64 cm

(ii)
Perimeter = 360 cm
Length = 140 cm

$∴$ Breadth = $\frac{360}{2} - 140$ = 180 $-$ 140 = 40 cm
(iii)
Perimeter = 360 cm
Length = 102 cm

Breadth = $\frac{360}{2} - 102$ = 180 $-$ 102 = 78 cm

Page No 20.10:

Question 5:

A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.

Answer:

Length of the lawn = 98 m
Breadth of the lawn = 55 m
Length of the fence around the lawn = Perimeter of the lawn = 2 $\times$ (Length + Breadth)
Perimeter of the lawn = 2 $\times$ (98 + 55) m = 2 $\times$ (153) = 306 m
Thus, the length of the fence around the lawn = 306 m

Page No 20.10:

Question 6:

The side of a square field is 65 m. What is the length of the fence required all around it?

Answer:

Side of the square field = 65 m
Length of the fence around the square field = Perimeter of the square field = 4 $\times$ (Side of the square)
Perimeter of the square field = 4 $\times$ 65 = 260 m
Thus, the length of the fence around the square filed = 260 m

Page No 20.10:

Question 7:

Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third side?

Answer:

Given:
Perimeter = 50 cm
Length of the first side = 15 cm
Length of the second side = 20 cm
We have to find the length of the third side.
Perimeter of a triangle = Sum of all three sides of the triangle
$∴$ Length of the third side = (Perimeter of the triangle) $-$ (Sum of the length of the other two sides)
= 50 $-$ (15 + 20)
= 50 $-$ 35 = 15 cm

Page No 20.10:

Question 8:

A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres ?

Answer:

It is given that a wire of length 20 m is to be folded in the form of a rectangle; therefore, we have:
Perimeter of the rectangle = 20 m
$\Rightarrow$ 2(Length + Breadth) = 20 m
$\Rightarrow$ (Length + Breadth ) = 20/2 = 10 m

Since, length and breadth are positive integers in metres, therefore, the possible dimensions are:
(1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) and (5m, 5m)

Thus, five rectangles can be formed with the given wire.

Page No 20.10:

Question 9:

A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, what is the length of the wire needed?

Answer:

Side of the square field = 100 m
Wire required to fence the square field = Perimeter of the square field = 4 $\times$ Side of the square field
Perimeter = 4 $\times$ 100 = 400 m
This perimeter is the length of wire required to fence one layer.
Therefore, the length of wire required to fence three layers = 3 $\times$ 400 m = 1200 m

Page No 20.10:

Question 10:

Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?

Answer:

Shikha and Priya, while running around the square and rectangular field respectively, actually cover a distance equal to the perimeters of these fields.
$∴$ Distance covered by Shikha = Perimeter of the square = 4 $\times$ 75 m = 300 m
Similarly, distance covered by Priya = Perimeter of the rectangle = 2 $\times$ (60 + 45) = 2 $\times$ 105 = 210 m
Thus, it is evident that the distance covered by Priya is less than that covered by Shikha.

Page No 20.10:

Question 11:

The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame the picture?

Answer:

Dimensions of the photograph = 30 cm $\times$ 20 cm
So, the required length of wooden frame = Perimeter of the photograph = 2 (Length + Breadth)
= 2 $\times$ (30 + 20) cm
= 2 $\times$ 50 cm
= 100 cm

Page No 20.10:

Question 12:

The length of a rectangular fields is 100 m. If its perimeter is 300 m, what is its breadth?

Answer:

Length of the rectangular field = 100 m
Perimeter of the rectangular field = 300 m
Perimeter of a rectangle = 2 (Length + Breadth)
Applying the above formula, we get:
Breadth of the rectangular field = $\frac{P e r i m e t e r}{2} - L e n g t h$ = $\frac{300}{2} - 100$ = 150 $-$ 100 = 50 m

Page No 20.10:

Question 13:

To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for bosts, he fixed a post every 5 metres apart. Each post was 2 m long. What is total length of the pipes he bought for the posts?

Answer:

Length of the garden = 70 m
Breadth of the garden = 50 m
Perimeter of the garden = 2 $\times$ (Length + Breadth)
= 2 $\times$ (70 + 50)
= 2 $\times$ 120 = 240 m
On the perimeter of the gardenâ€‹, it is given that Arvind fixes a post every 5 metres apart.

So, the number of posts required = $\frac{240}{5}$ = 48
$∵$ Length of each post = 2 m
$∴$ Total length of the pipe required = 48 $\times$ 2 = 96 m

Page No 20.11:

Question 14:

Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per meter.

Answer:

Length of the park = 175 m
Breadth of the park = 125 m
Perimeter of the park = 2 $\times$ (Length + Breadth)
= 2 $\times$ (175 + 125)
= 2 $\times$ 300 = 600 m
Rate of fencing = Rs. 12 per meter
Cost of fencing = Rs. 12 $\times$ 600 = Rs. 7,200

Page No 20.11:

Question 15:

The perimeter of a regular pentagon is 100 cm. How long is each sides?

Answer:

A regular pentagon is a closed polygon having five sides of equal length.
Perimeter of the regular pentagon = 100 cm
Perimeter of the regular pentagon = 5 $\times$ Side of the regular pentagon
Therefore, side of the regular pentagon = $\frac{P e r i m e t e r}{5}$ = $\frac{100}{5}$ = 20 cm

Page No 20.11:

Question 16:

Find the perimeter of a regular hexagon with each side measuring 8 m.

Answer:

A regular hexagon is a closed polygon having six sides of equal lengths.
Side of the hexagon = 8 m
Perimeter of the hexagon = 6 $\times$ Side of the hexagon
= 6 $\times$ 8 = 48 m

Page No 20.11:

Question 17:

A rectangular piece of land measure 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. What length of the wire is needed?

Answer:

Dimensions of the rectangular land = 0.7 km $\times$ 0.5 km
Perimeter of the rectangular land = 2 (Length + Breadth)
= 2 (0.7 + 0.5) km = 2 $\times$ 1.2 km = 2.4 km
This perimeter is equal to one row of wire required to fence the land.
Therefore, length of wire required to fence the land with four rows of wire = 4 $\times$ 2.4 km
= 9.6 km

Page No 20.11:

Question 18:

Avneet buys 9 square paving slabs, each with a side of $\frac{1}{2}$ m. He lays them in the form of a square.

(i) What is the perimeter of his arrangement?
(ii) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?
(iii) Which has greater perimeter?
(iv) Avneet wonders, If there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges they cannot be broken)

Answer:

(i) Length of the side of one slab = 1/2 m
    In the square arrangement, one side of the square is formed by three slabs.
    So, length of the side of the square = 3 $\times \frac{1}{2}$ = 3/2 m
    The perimeter of the square arrangement = 4 $\times \frac{3}{2}$ = 6 m
(ii) The cross arrangement consists of 8 sides. These sides form the periphery of the arrangement and measure 1 m each.
   Also, this arrangement consists of other 4 sides that measure 1/2 m each.
     So, the perimeter of the cross arrangement = (1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1)
= (8 + 2) = 10 m

(iii) Perimeter of the cross arrangement = 10 m
      Perimeter of the square arrangement = 6 m
      Thus, the perimeter of the the cross arrangement is more than that of the square arrangement.

(iv) No, there is no way of arranging these slabs where the perimeter is more than 10 m.

Page No 20.14:

Question 1:

The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each squares as 1 cm². (Fig.20.25).

Answer:

(i) There are 16 complete squares in the given shape.
     $∵$ Area of one square = 1 cm²
$∴$ Area of this shape = 16 $\times$ 1 = 16 cm²
(ii) There are 36 complete squares in the given shape.
     $∵$ Area of one square = 1 cm²
$∴$ Area of 36 squares = 36 $\times$ 1 = 36 cm²
(iii) There are 15 complete and 6 half squares in the given shape.
     $∵$ Area of one square = 1 cm²
$∴$ Area of this shape = (15 + 6 $\times$ $\frac{1}{2}$ ) = 18 cm²
(iv) There are 20 complete and 8 half squares in the given shape.
     $∵$ Area of one square = 1 cm²
     $∴$ Area of this shape = (20 + 8 $\times$ $\frac{1}{2}$ ) = 24 cm²
(v) There are 13 complete squares, 8 more than half squares and 7 less than half squares in the given shape.
     $∵$ Area of one square = 1 cm²
     $∴$ Area of this shape = (13 + 8 $\times$ 1) = 21 cm²

(vi) There are 8 complete squares, 6 more than half squares and 4 less than half squares in the given shape.
     $∵$ Area of one square = 1 cm²
     $∴$ Area of this shape = (8 + 6 $\times$ 1) = 14 cm²

Page No 20.14:

Question 2:

On a squared paper, draw (i) a rectangle, (ii) a triangle (iii) any irregular closed figure. Find the approximate area of each by counting the number of squares complete, more than half and exactly half.

Answer:

(i) A rectangle: This contains 18 complete squares.
If we assume that the area of one complete square is 1 cm², then the area of this rectangle will be 18 cm².

(ii) A triangle: This triangle contains 4 complete squares, 6 more than half squares and 6 less than half squares.
If we assume that the area of one complete square is 1 cm², then the area of this shape = (4 + 6 $\times$ 1) = 10 cm²

(iii) Any irregular figure: This figure consists of 10 complete squares, 1 exactly half square, 7 more than half squares and 6 less than half squares.
If we assume that the area of one complete square is 1 cm², then the area of this shape = (10 + $1 \times \frac{1}{2}$ + 7 $\times$ 1) = 17.5 cm²

Page No 20.14:

Question 3:

Draw any circle on the graph paper. Count the squares and use them to estimate the area of the circular region.
Figure

Answer:

This circle on the squared paper consists of 21 complete squares, 15 more than half squares and 8 less than half squares.
Let us assume that the area of 1 square is 1 cm².
If we neglect the less than half squares while approximating more than half square as equal to a complete square, we get:
Area of this shape = (21 + 15) = 36 cm²

Page No 20.15:

Question 4:

Use tracing paper and centimeter graph paper to compare the areas of the following pairs of figures:

Answer:

                  (i)                                                                                        (ii)
Using tracing paper, we traced both the figures on a graph paper.
(i) This figure contains 4 complete squares, 9 more than half squares and 9 less than half squares.
      Let us assume that the area of one square is 1 cm².
      If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square, we get:
      Area of this shape = (4 + 9) = 13 cm²
(ii) This figure contains 8 complete squares, 11 more than half squares and 10 less than half squares.
      Let us assume that the area of one square is 1 cm²â€‹.
      If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square, we get:
      Area of this shape = (8 + 11) = 19 cm²

      On comparing the areas of these two shapes, we get that the area of Fig. (ii) is more than that of Fig. (i).

Page No 20.21:

Question 1:

Find the area of a rectangle, whose

(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.

Answer:

(i)
Area of a rectangle = Length $\times$ Breadth
Length = 6 cm
Breadth = 3 cm
$∴$ Area of rectangle = 6 $\times$ 3 = 18 cm²
(ii)
Area of a rectangle = Length $\times$ Breadth
Length = 8 cm
Breadth = 3 cm
$∴$ Area of rectangle = 8 $\times$ 3 = 24 cm²
(iii)
Area of a rectangle = Length $\times$ Breadth
Length = 4.5 cm
Breadth =2 cm
$∴$ Area of rectangle = 4.5 $\times$ 2 = 9 cm²

Page No 20.21:

Question 2:

Find the area of a square whose side is:

(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm

Answer:

Area of a square = Side $\times$ Side
(i) Side of the square = 5 cm
    Area of the square = 5 $\times$ 5 = 25 cm²
(ii) Side of the square = 4.1 cm
    Area of the square = 4.1 $\times$ 4.1 = 16.81 cm²
⁽ⁱⁱⁱ⁾Side of the square = 5.5 cm
    Area of the square = 5.5 $\times$ 5.5 = 30.25 cm²
^(iv)Side of the square = 2.6 cm
    Area of the square = 2.6 $\times$ 2.6 = 6.76 cm²

Page No 20.21:

Question 3:

The area of a rectangle is 49 cm² and its breadth is 2.8 cm, find the length of the rectangle.

Answer:

Area = 49 cm²
Breadth = 2.8 cm
Area of the rectangle = Length $\times$ Breadth
$∴$ Length = $\frac{A r e a}{B r e a d t h}$ = $\frac{49}{2.8}$ = 17.5 cm

Page No 20.21:

Question 4:

The side of a square is 70 cm. Find its area and perimeter.

Answer:

Side of the square = 70 cm
Area of the square = Side $\times$ Side
= 70 $\times$ 70 = 4900 cm²
Perimeter of the square = 4 $\times$ Side = 4 $\times$ 70 = 280 cm

Page No 20.21:

Question 5:

The area of a rectangle is 225 cm² and its one side is 25, find its other side.

Answer:

Area = 225 cm²
One of the sides = 25 cm
Area of the rectangle = Product of the lengths of its two sides
Other side = $\frac{A r e a}{S i d e}$ = $\frac{225}{25}$ = 9 cm

Page No 20.21:

Question 6:

What will happen to the area of a rectangle if its

(i) Length and breadth are trebled.
(ii) Length is doubled and breadth is same.
(iii) Length is doubled and breadth is halved.

Answer:

(i) If the length and breadth of a rectangle are trebled.
Let the initial length and breadth be l and b, respectively.
Original area = l $\times$ b = lb
Now, the length and breadth are trebled which means they become three times of their original value.
Therefore
New length = 3l
New breadth = 3b
New area = 3l $\times$ 3b = 9 lb
Thus, the area of the rectangle will become 9 times that of its original area.

(ii) If the length is doubled and the breadth is same.
Let the initial length and breadth be l and b, respectively.â€‹
Original area = l $\times$ b = lb
Now, length is doubled and breadth remains same.
Therefore
New length = 2l
New breadth = b
New area = 2l $\times$ b = 2 lb
Thus, the area of the rectangle will become 2 times that of its original area.

(iii) If the Length is doubled and breadth is halved.
Let the initial length and breadth be l and b, respectively.â€‹
Original area = l $\times$ b = lb
Now, length is doubled and breadth is halved.
Therefore
New length = 2l
New breadth = b/2
New area = 2l $\times$ b/2 = lb
New area is also lb. This means that the areas remain the same.

Page No 20.21:

Question 7:

What will happen to the area of a square if its side is

(i) Tripled
(ii) increased by half of it.

Answer:

(i) Let the original side of the square be s.
Original area = s $\times$ s = s²
If the side of a square is tripled, new side will be equal to 3s.
New area = 3s $\times$ 3s = 9s²
This means that the area becomes 9 times that of the original area.
(ii) Let the original side of the square be s.
Original area = s $\times$ s = s²
If the side of a square is increased by half of it, new side = (s + $\frac{1}{2}$ s) = $\frac{3}{2}$ s
New area = $\frac{3}{2}$ s $\times$ $\frac{3}{2}$ s = $\frac{9}{4}$ s
This means that the area becomes $\frac{9}{4}$ times that of the original area.

Page No 20.21:

Question 8:

Find the perimeter of a rectangle whose area is 500 cm² and breadth is 20 cm.

Answer:

Area = 500 cm²
Breadth = 20 cm
Area of rectangle = Length $\times$ Breadth
Therefore
Length = $\frac{A r e a}{B r e a d t h}$ = $\frac{500}{20}$ = 25 cm
Perimeter of a rectangle = 2 (Length + Breadth)
= 2 (25 + 20) cm = 2 $\times$ 45 cm = 90 cm

Page No 20.21:

Question 9:

A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm, find its length.

Answer:

Side of the square = 80 cm
Area of square = Side $\times$ Side = 80 $\times$ 80 = 6400 cm²
Given that:
Area of the rectangle = Area of the square = 6400 cm²
Breadth of the rectangle = 20 cm
Applying the formula:
Length of the rectangle = $\frac{A r e a}{B r e a d t h}$
We get:
Length of the rectangle = $\frac{6400}{20}$ = 320 cm

Page No 20.21:

Question 10:

Area of a rectangle of breadth 17 cm is 340 cm². Find the perimeter of the rectangle.

Answer:

Area of the rectangle = 340 cm²
Breadth of the rectangle = 17 cm
Applying the formula:
Length of a rectangle = $\frac{A r e a}{B r e a d t h}$
We get:
Length of the rectangle = $\frac{340}{17}$ =20 cm
Perimeter of rectangle = 2 (Length + Breadth)
= 2 (20 + 17)
= 2 $\times$ 37
= 74 cm

Page No 20.21:

Question 11:

A marble tile measures 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

Answer:

Dimensions of the tile = 15 cm × 20 cm
Dimensions of the wall = 4 m × 6 m = 400 cm × 600 cm (Since, 1 m = 100 cm, so, 4 m = 400 cm and 6 m = 600 cm)
Area of the tile = 15 cm × 20 cm = 300 cm²
Area of the wall = 400 cm × 600 cm = 2,40,000 cm²

Number of tiles required to cover the wall = $\frac{A r e a o f w a l l}{A r e a o f o n e t i l e}$ = $\frac{240000}{300}$ = 800 tiles

Page No 20.21:

Question 12:

A marble tile measures 10 cm × 12 cm. How many tiles will be required to cover a wall of size
3 m × 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

Answer:

Dimension of the tile = 10 cm × 12 cm
Dimension of the wall =3 m × 4 m = 300 cm × 400 cm (Since, 1 m = 100 cm, so, 3 m = 300 cm and 4 m = 400 cm)
Area of the tile = 10 cm × 12 cm = 120 cm²
Area of the wall = 300 cm × 400 cm =1,20,000 cm²
Number of tiles required to cover the wall = $\frac{A r e a o f w a l l}{A r e a o f o n e t i l e}$ = $\frac{120000}{120}$ = 1,000 tiles
Cost of tiles at the rate of Rs. 2 per tile = 2 × 1,000 = Rs. 2,000

Page No 20.21:

Question 13:

One side of a square plot is 250 m, find the cost of levelling it at the rate of Rs 2 per square metre.

Answer:

Side of the square plot = 250 m
Area of the square plot = Side $\times$ Side = 250 × 250 = 62,500 m²
Rate of levelling the plot = Rs. 2 per m²
Cost of levelling the square plot = Rs. 62,500 × 2 = Rs. 1,25,000

Page No 20.22:

Question 14:

The following figures have been split into rectangles. Find their areas. (The measures are given in centimetres)

Answer:

(i) This figure consists of two rectangles II and IV and two squares I and III.
     Area of square I = Side × Side = 3 × 3 = 9 cm²
     Similarly, area of rectangle II = (2 × 1) = 2 cm²
area of square III = (3× 3) = 9 cm²
area of rectangle IV = (2 × 4) = 8 cm²
   Thus, the total area of this figure = (Area of square I + Area of rectangle II + Area of square III + Area of rectangle IV) = 9 + 2 + 9 + 8 = 28 cm²


(ii) This figure consists of three rectangles I, II and III.
Area of rectangle I = Length × Breadth = 3 × 1 = 3 cm²
     Similarly, area of rectangle II = (3 × 1) = 3 cm²
area of rectangle III = (3 × 1) = 3 cm²
   Thus, the total area of this figure = (Area of rectangle I + area of rectangle II + area of rectangle III) = 3 + 3 + 3 = 9 cm²

Page No 20.22:

Question 15:

Split the following shapes into rectangles and find the area of each. (The measures are given in centimetres)

Answer:

(i) This figure consists of two rectangles I and II.
     The area of rectangle I = Length $\times$ Breadth = 10 $\times$ 2 = 20 cm²
     Similarly, area of rectangle II = 10 $\times$ $\frac{3}{2}$ = 15 cm²
     Thus, total area of this figure = (Area of rectangle I + Area of rectangle II) = 20 + 15= 35 cm²

(ii) This figure consists of two squares I and III and one rectangle II.
     Area of square I = Area of square III = Side $\times$ Side = 7 $\times$ 7 = 49 cm²
     Similarly, area of rectangle II = (21 $\times$ 7) = 147 cm²
   Thus, total area of this figure = (Area of square I + Area of rectangle II + Area of square III) = 49 + 49 + 147 = 245 cm²

(iii) This figure consists of two rectangles I and II.
     Area of rectangle I = Length $\times$ Breadth = 5 $\times$ 1 = 5 cm²
     Similarly, area of rectangle II = 4 $\times$ 1 = 4 cm²
   Thus, total area of this figure = (Area of rectangle I + Area of rectangle II) = 5 + 4 = 9 cm²

Page No 20.22:

Question 16:

How many tiles with dimensions 5 cm and 12 cm will be needed to fit a region whose length and breadth are respectively:

(i) 100 cm and 144 cm
(ii) 70 cm and 36 cm

Answer:

(i) Dimension of the tile = 5 cm × 12 cm
     Dimension of the region = 100 cm × 144 cm
   Area of the tile = 5 cm × 12 cm = 60 cm²
     Area of the region = 100 cm × 144 cm = 14,400 cm²
     Number of tiles required to cover the region = $\frac{A r e a o f t h e r e g i o n}{A r e a o f o n e t i l e}$ = $\frac{14400}{60}$ = 240 tiles
(ii) Dimension of the tile = 5 cm × 12 cm
    Dimension of the region = 70 cm × 36 cm
      Area of the tile = 5 cm × 12 cm = 60 cm²
    Area of the region = 70 cm × 36 cm = 2,520 cm²
    Number of tiles required to cover the region = $\frac{A r e a o f t h e r e g i o n}{A r e a o f o n e t i l e}$ = $\frac{2520}{60}$ = 42 tiles

Page No 20.23:

Question 1:

The side of a rectangle are in the ratio 5 : 4. If its perimeter is 72 cm, then its length is

(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm

Answer:

(b) 20 cm
   Let the sides of the rectangle be 5x and 4x. (Since, they are in the ratio 5 : 4)
   Now, perimeter of rectangle = 2 (Length + Breadth)
                                         72 = 2 (5x + 4x)
                                         72 = 2 $\times$ 9x
                                         72 = 18x
                                           x = $\frac{72}{18}$ = 4
      Thus, the length of the rectangle = 5x = 5 $\times$ 4 = 20 cm

Page No 20.23:

Question 2:

The cost of fencing a rectanglular field 34 m long and 18 m wide at Rs 2.25 per metre is

(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334

Answer:

(b) Rs. 234
For fencing the rectangular field, we need to find the perimeter of the rectangle.
Length of the rectangle = 34m
Breadth of the rectangleâ€‹ = 18m
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (34 + 18) m
= 2 $\times$ 52 m = 104 m
Cost of fencing the field at the rate of Rs. 2.25 per meter = Rs. 104 $\times$ 2.25 = Rs. 234

Page No 20.23:

Question 3:

If the cost of fencing a rectangular field at Rs. 7.50 per metre is Rs. 600, and the length of the field is 24 m, then the breadth of the field is

(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m

Answer:

(d) 16 m
Cost of fencing the rectangular field = Rs. 600
Rate of fencing the field = Rs. 7.50 per m
Therefore, perimeter of the field = $\frac{C o s t o f f e n c i n g}{R a t e o f f e n c i n g}$ = $\frac{600}{7.50}$ = 80 m
Now, length of the field = 24 m
Therefore, breadth of the field = $\frac{P e r i m e t e r}{2} -$ Length
= $\frac{80}{2} -$ 24 = 16 â€‹m

Page No 20.23:

Question 4:

The cost of putting a fence around a square field at Rs 2.50 per metre is Rs 200. 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

Cost of fencing the square field = Rs. 200

Rate of fencing the field = Rs. 2.50

Now, perimeter of the square field = $\frac{C o s t o f f e n c i n g}{R a t e o f f e n c i n g}$ = $\frac{200}{2.50}$ = 80 m

Perimeter of square = 4 $\times$ Side of the square

Therefore, side of the square = $\frac{P e r i m e t e r}{4}$ = $\frac{80}{4}$ = 20 m

Page No 20.23:

Question 5:

The length of a rectangle is three times of its width. If the length of the diagonal is $8 \sqrt{10}$ m, then the perimeter of the rectangle is

(a) $15 \sqrt{10 m}$
(b) $16 \sqrt{10 m}$
(c) $24 \sqrt{10 m}$
(d) 64 m

Answer:

(d) 64 m
Let us consider a rectangle ABCD.
Also, let us assume that the width of the rectangle, i.e., BC be x m.
It is given that the length is three times width of the rectangle.
Therefore, length of the rectangle, i.e., AB = 3x m

Now, AC is the diagonal of rectangle.
In right angled triangle ABC.
   AC² = AB² + BC²                                 {Using Pythagoras theorem}
   ${(8 \sqrt{10})}^{2} = {(3 x)}^{2} + {(x)}^{2}$
   640 = 9x² + x²
   640 = 10x²
     x² = $\frac{640}{10}$ = 64
     x = $\sqrt{64}$ = 8 m

Thus, breadth of the rectangle = x = 8 m
Similarly, length of the rectangle = 3x = 3 $\times$ 8 = 24 m
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (24 + 8)
= 2 $\times$ 32 = 64 m

Page No 20.23:

Question 6:

If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio

(a) 3 : 1
(b) $\sqrt{3} : 1$
(c) $\sqrt{2} : 1$
(d) $2 \sqrt{2} : 1$

Answer:

(d) $2 \sqrt{2}$ : 1
Let us assume that the length of the smaller side of the rectangle, i.e., BC be x and length of the larger side , i.e., AB be y.
It is given that the length of the diagonal is three times that of the smaller side.
Therefore, diagonal = 3x = AC
Now, applying Pythagoras theorem, we get:
(Diagonal)² = (Smaller side)² + (Larger side)²

(AC)² = (AB)²+ (BC)²
(3x)² = (x)² + (y)²
9x² = x² + y²
8x² = y²
Now, taking square roots of both sides, we get:
$2 \sqrt{2}$ x = y
or, $\frac{y}{x} = \frac{2 \sqrt{2}}{1}$

Thus, the ratio of the larger side to the smaller side = $2 \sqrt{2}$ : 1

Page No 20.23:

Question 7:

The ratio of the areas of two squares, one having its diagonal double than the other, is

(a) 1 : 2
(b) 2 : 3
(c) 3 : 1
(d) 4 : 1

Answer:

(d) 4 : 1

Let the two squares be ABCD and PQRS. Further, the diagonal of square PQRS is twice the diagonal of square ABCD.

PR = 2 AC

Now, area of the square = $\frac{{(D i a g o n a l)}^{2}}{2}$

Area of PQRS = $\frac{{(P R)}^{2}}{2}$

Similarly, area of ABCD = $\frac{{(A C)}^{2}}{2}$

According to the question:

If AC = x units, then, PR = 2x units

Therefore, $\frac{A r e a o f P Q R S}{A r e a o f A B C D}$ = $\frac{{(P R)}^{2} \times 2}{2 \times {(A C)}^{2}}$ = $\frac{{(2 x)}^{2} \times 2}{2 \times {(x)}^{2}}$ = $\frac{4}{1}$

Thus, the ratio of the areas of squares PQRS and ABCD = 4 : 1

Page No 20.23:

Question 8:

If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is

(a) 225 : 256
(b) 256 : 225
(c) 15 : 16
(d) 16 : 15

Answer:

(c) 15 : 16

Let the two squares be ABCD and PQRS.

Further, let the lengths of each side of ABCD and PQRS be x and y, respectively.

Therefore

$\frac{A r e a o f s q . A B C D}{A r e a o f s q . P Q R S}$ = $\frac{x^{2}}{y^{2}}$

⇒ $\frac{x^{2}}{y^{2}}$ = $\frac{225}{256}$

Taking square roots on both sides, we get:

$\frac{x}{y}$ = $\frac{15}{16}$

Now, the ratio of their perimeters:

$\frac{P e r i m e t e r o f s q . A B C D}{P r i m e t e r o f s q . P Q R S}$ = $\frac{4 \times s i d e o f s q . A B C D}{4 \times S i d e o f s q . P Q R S}$ = $\frac{4 x}{4 y}$

⇒ $\frac{P e r i m e t e r o f s q . A B C D}{P r i m e t e r o f s q . P Q R S}$ = $\frac{x}{y}$

⇒ â€‹ $\frac{P e r i m e t e r o f s q . A B C D}{P r i m e t e r o f s q . P Q R S}$ = $\frac{15}{16}$

Thus, the ratio of their perimeters = 15 : 16

Page No 20.23:

Question 9:

If the sides of a square are halved, the its area

(a) remains same
(b) becomes half
(c) becomes one fourth
(d) becomes double

Answer:

(c) becomes one fourth

Let the side of the square be x.

Then, area = (Side $\times$ Side) = (x $\times$ x) = x²

If the sides are halved, new side = $\frac{x}{2}$

Now, new area = ${(\frac{x}{2})}^{2}$ = $\frac{x^{2}}{4}$

It is clearly visible that the area has become one-fourth of its previous value.

Page No 20.23:

Question 10:

A rectangular carpet has area 120 m² and perimeter 46 metres. The length of its diagonal is

(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m

Answer:

(c) 17 m
Area of the rectangle = 120 m²
Perimeter = 46 m
Let the sides of the rectangle be l and b.
Therefore
Area = lb = 120 m² ...(1)
Perimeter = 2 (l + b) = 46
or, ( l + b ) = $\frac{46}{2}$ = 23 m ...(2)
Now, length of the diagonal of the rectangle = $\sqrt{(l^{2} + b^{2})}$
So, we first find the value of (l²+ b²)
Using identity:
(l² + b²) = (l + b)² $-$ 2 (lb) [From (1) and (2)]
Therefore
(l² + b²) = (23)² $-$ 2 (120)
= 529 $-$ 240 = 289
Thus, length of the diagonal of the rectangle = $\sqrt{(l^{2} + b^{2})}$ = $\sqrt{289}$ = 17 m

Page No 20.23:

Question 11:

If the ratio between the length and the 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
It is given that $\frac{L e n g t h o f r e c \tan g l e}{P e r i m e t e r o f r e c \tan g l e}$ = $\frac{1}{3}$
⇒ $\frac{l}{2 l + 2 b}$ = $\frac{1}{3}$
After cross multiplying, we get:
3l = 2l + 2b
⇒ l = 2b
⇒ $\frac{l}{b}$ = $\frac{2}{1}$
Thus, the ratio of the length and the breadth is 2 : 1.

Page No 20.23:

Question 12:

If the length of the diagonal of a square is 20 cm, then its perimeter is

(a) $10 \sqrt{2} cm$
(b) 40 cm
(c) $40 \sqrt{2} cm$
(d) 200 cm

Answer:

(c) 40 $\sqrt{2}$ cm

Length of the diagonal = 20 cm

Length of the side of a square = $(\frac{L e n g t h o f D i a g o n a l}{\sqrt{2}})$ = $\frac{20}{\sqrt{2}}$ = $\frac{2 \times 10}{\sqrt{2}}$
= $\frac{\sqrt{2} \times \sqrt{2} \times 10}{\sqrt{2}}$ = 10 $\sqrt{2}$ cm

Therefore, perimeter of the square = 4 $\times$ Side = 4 $\times$ 10 $\sqrt{2}$ = 40 $\sqrt{2}$ cm

Page No 20.24:

Question 1:

Mark the correct alternative in the following question:

If the perimeter of a square is 40 cm, then the length of its each side is

(a) 20 cm (b) 10 cm (c) 5 cm (d) 40 cm

Answer:

$The length of each side of the square = \frac{Perimeter of the square}{4} = \frac{40}{4} = 10 cm$

Hence, the correct alternative is option (b).

Page No 20.24:

Question 2:

Mark the correct alternative in the following question:

The area of a square of side 14 cm is

(a) 49 cm² (b) 156 cm² (c) 56 cm² (d) 196 cm²

Answer:

The area of the square = Side $\times$ Side
= 14 $\times$ 14
= 196 cm²

Hence, the correct alternative is option (d).

Page No 20.24:

Question 3:

Mark the correct alternative in the following question:

The length of the diagonal of a square is 20 cm. Its area is

(a) 400 cm² (b) 200 cm² (c) 300 cm² (d) $100 \sqrt{2}$ cm²

Answer:

$The area of the square = \frac{1}{2} \times Diagonal \times Diagonal = \frac{1}{2} \times 20 \times 20 = \frac{400}{2} = 200 {cm}^{2}$

Hence, the correct alternative is option (b).

Page No 20.24:

Question 4:

Mark the correct alternative in the following question:

The length and breadth of a rectangle of area A are doubled. The area of the new rectangle is

(a) 2A (b) A² (c) 4A (d) None of these

Answer:

Let the length and breadth of the given rectangle be l and b, respectively.

We have,
A = lb .....(i)

Also,
the length of the new rectangle, l' = 2l and
the breadth of the new rectangle, b' = 2b

Now, the area of the new rectangle = l' $\times$ b'
= (2l) $\times$ (2b)
= 4lb
= 4A [Using (i)]

Hence, the correct alternative is option (c).

Page No 20.24:

Question 5:

Mark the correct alternative in the following question:

The maximum length of the side of a square sheet that can be cut off from a rectangular sheet of size 8 m $\times$ 3 m is

(a) 3 m (b) 4 m (c) 6 m (d) 4 m

Answer:

The maximum length of the side of a square sheet that can be cut off from a rectangular sheet of size 8 m $\times$ 3 m is 3 m.

Hence, the correct alternative is option (a).

Page No 20.24:

Question 6:

Mark the correct alternative in the following question:

The perimeter of a square whose area is 225 m² is

(a) 15 m (b) 60 m (c) 225 m (d) 30 m

Answer:

$We have, Area of the square = 225 m^{2} As, the side of the square = \sqrt{Area} = \sqrt{225} = 15 m So, the perimeter of the square = 4 \times side = 4 \times 15 = 60 m$

Hence, the correct alternative is option (b).

Page No 20.24:

Question 7:

Mark the correct alternative in the following question:

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:

$We have, Area of the rectangle = 650 {cm}^{2} and Breadth of the rectangle = 13 cm As, length of the rectangle = \frac{Area}{Breadth} = \frac{650}{13} = 50 cm So, the perimeter of the rectangle = 2 \times (length + breadth) = 2 \times (13 + 50) = 2 \times 63 = 126 cm$

Hence, the correct alternative is option (d).

Page No 20.24:

Question 8:

Mark the correct alternative in the following question:

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:

$We have, length of the sheet of the paper = 72 cm, breadth of the sheet of the paper = 48 cm, length of the envelope = 18 cm and breadth of the envelope = 12 cm The area of the sheet of the paper = length \times breadth = (72 \times 48) {cm}^{2} And, the area of the envelope = length \times breadth = (18 \times 12) {cm}^{2} Now, the number of envelope that can be made out = \frac{Area of the sheet of the paper}{Area of the envelope} = \frac{(72 \times 48)}{(18 \times 12)} = 4 \times 4 = 16$

Hence, the correct alternative is option (d).

Page No 20.24:

Question 9:

Mark the correct alternative in the following question:

If the diagonal of a square is $\sqrt{8}$ metre, then its area is

(a) 8 m² (b) 4 m² (c) 16 m² (d) 6 m²

Answer:

$We have, length of the diagonal of the square = \sqrt{8} m Now, the area of the square = \frac{1}{2} \times diagonal \times diagonal = \frac{1}{2} \times \sqrt{8} \times \sqrt{8} = \frac{8}{2} = 4 m^{2}$

Hence, the correct alternative is option (b).

Page No 20.24:

Question 10:

Mark the correct alternative in the following question:

The area of the shaded path in the following figure is

(a) 16 m² (b) 18 m² (c) 14 m² (d) 20 m²

Answer:

$Area of the shaded region = Area of the rectangle + Area of the isosceles right angled triangl = length \times breadth + \frac{1}{2} \times base \times height = 8 \times 2 + \frac{1}{2} \times 2 \times 2 = 16 + 2 = 18 m^{2}$

Hence, the correct alternative is option (b).

Page No 20.24:

Question 11:

The area of a rectangular ground is 120 m² and its length is 12 m. Find the cost of fencing the ground at the rate of â‚¹125 per metre.

Answer:

$We have, Area of the rectangular ground = 120 m^{2} and Length of the ground = 12 m Now, Breadth of the ground = \frac{Area of the ground}{Length of the ground} = \frac{120}{12} = 10 m Also, Perimeter of the ground = 2 \times (Length + Breadth) = 2 \times (12 + 10) = 2 \times 22 = 44 m Since, the rate of fencing the ground = ₹ 125 per metre So, the cost of fencing the ground = 125 \times 44 = ₹ 5500$

Page No 20.24:

Question 12:

Find the perimeter of the following figure:

Answer:

$Perimeter of the figure = AB + BC + CD + DE + EF + FG + GH + HA = 10 + 10 + 20 + 30 + 20 + 20 + 10 + 20 = 140 m$

Page No 20.25:

Question 13:

A room is 8 m long and 6 m wide. Its floor is to be covered with rectangular tiles of size 25 cm by 20 cm. How many tiles will be required? Find the cost of these tiles at â‚¹25 per tile.

Answer:

We have,
Length of the room = 8 m = 800 cm,
Breadth of the room = 6 m = 600 cm,
Length of the tile = 25 cm and
Breadth of the tile = 20 cm

$Now, The number of the tiles required = \frac{Area of the floor of the room}{Area of a tile} = \frac{Length of the room \times Breadth of the room}{Length of the tile \times Breadth of the tile} = \frac{800 \times 600}{25 \times 20} = 960 So, the number of tiles required to cover the floor of the room is 960 . Also, Since, the cost of 1 tile = ₹ 25 So, the cost of the 960 tiles such tiles = 25 \times 960 = ₹ 24, 000 .$

Page No 20.25:

Question 14:

The length and breadth of a rectangular park are in the ratio 5 : 3 and its perimeter is 128 m. Find the area of the park.

Answer:

$Let the length of the park be 5 x m and its breadth be 3 x m . As, perimeter of the park = 2 \times (Length of the park + Breadth of the park) = 2 \times (5 x + 3 x) = 2 \times 8 x = 16 x m But the perimeter of the park = 128 m \Rightarrow 16 x = 128 \Rightarrow x = \frac{128}{16} \Rightarrow x = 8 ∴ The length of park = 5 x = 5 \times 8 = 40 m and the breadth of the park = 3 x = 3 \times 8 = 24 m Now, the area of the park = Length \times Breadth = 40 \times 24 = 960 m^{2}$

So, the area of the park is 960 m².

Page No 20.25:

Question 15:

The total cost of flooring a room at â‚¹85 per m² is â‚¹5100. If the length of the room is 8 metres, then find its breadth.

Answer:

$As, the total cost of flooring = ₹ 5100 and the rate of flooring = ₹ 85 per m^{2} So, the area of the floor = \frac{Total cost of the flooring}{Rate of the flooring} = \frac{5100}{85} = 60 m^{2} Also, the length of the room = 8 m Now, the breadth of the room = \frac{Area of the floor}{Length of the room} = \frac{60}{8} = \frac{15}{2} = 7.5 m$

So, the breadth of the room is 7.5 m.

Page No 20.25:

Question 16:

A square sheet of side 5 cm is cut out from a rectangular piece of an aluminimum sheet of length 9 cm and breadth 6 cm. What is the area of the aluminium sheet left over?

Answer:

We have,
Side of the square sheet cut out = 5 cm,
Length of the rectangular sheet = 9 cm and
Breadth of the rectangular sheet = 6 cm

$Now, the area of the square sheet cut out = Side \times Side = 5 \times 5 = 25 {cm}^{2} Also, the area of the recatngular sheet = Length \times Breadth = 9 \times 6 = 54 {cm}^{2} So, the area of the sheet left over = Area of the rectangular sheet - Area of the square sheet cut out = 54 - 25 = 29 {cm}^{2}$

Hence, the area of the aluminium sheet left over is 29 cm^2.

Page No 20.25:

Question 17:

Find the total cost of levelling the shaded path of uniform width 2 metres, laid in the rectangular field shown below, if the rate per m² is â‚¹100.

Answer:

The shaded path can be formed as two rectangles as shown below:

$Now, the area of the shaded path = 38 \times 2 + 50 \times 2 = 76 + 100 = 176 m^{2} As, the rate of levelling the shaded path = ₹ 100 per m^{2} So, the total cost of levelling the shaded path = 176 \times 100 = ₹ 17, 600$

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Page No 20.25:

Question 18:

A square sheet of 2 cm is cut from each corner of a rectangular piece of aluminium sheet of length 12 cm and breadth 8 cm. Find the area of the left over aluminium sheet.

Answer:

We have,
Side of each square sheet cut from each corner = 2 cm,
Length of the rectangular piece of sheet = 12 cm and
Breadth of the rectangular piece of sheet = 8 cm

$As, the area of each square = Side \times Side = 2 \times 2 = 4 {cm}^{2} So, the area of four square sheets = 4 \times 4 = 16 {cm}^{2} Also, the area of the rectangular sheet = Length \times Breadth = 12 \times 8 = 96 {cm}^{2} Now, the area of the sheet left over = Area of the rectangular sheet - Area of the square sheet = 96 - 16 = 80 {cm}^{2}$

So, the area of the left over aluminium sheet is 80 cm².

Page No 20.25:

Question 19:

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:

We have,
Length of the room = 12.5 m,
Breadth of the room = 8 m and
Side of the square carpet = 8 m

$The area of the room = Length \times Breadth = 12.5 \times 8 = 100 m^{2} Also, the area of the square carpet = Side \times Side = 8 \times 8 = 64 m^{2} Now, the area of the floor which is not carpeted = Area of the room - Area of the carpet = 100 - 64 = 36 m^{2}$

Page No 20.25:

Question 20:

The area of a square field is 36 m². A path of uniform width is laid around and outside of it. If the area of the path is 28 m², then find the width of the path.

Answer:

$As, the area of the square field = 36 m^{2} So, the side of the square field = \sqrt{36} = 6 m Also, the area of the outer square = Area of the square + Area of the path = 36 + 28 = 64 m^{2} So, the side of the outer square = \sqrt{64} = 8 m Now, the width of the path = \frac{Side of the outer square - Side of the inner square}{2} = \frac{8 - 6}{2} = \frac{2}{2} = 1 m$

So, the width of the path is 1 m.

Page No 20.25:

Question 21:

The perimeter of a square 16 cm, then its area is __________ cm².

Answer:

$As, the perimeter of the square = 16 cm So, the side of the square = \frac{Perimeter}{4} = \frac{16}{4} = 4 cm Now, the area of the square = Side \times Side = 4 \times 4 = 16 {cm}^{2}$

Hence, the area of the square is 16 cm².

Page No 20.25:

Question 22:

The length and breadth of a rectangle are 12 cm and 5 cm, respectively, then its diagonal is __________ cm.

Answer:

$We have, Length of the rectangle = 12 cm and Breadth of the rectangle = 5 cm Now, the diagonal of the rectangle = \sqrt{{(Length)}^{2} + {(Breadth)}^{2}} = \sqrt{12^{2} + 5^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 cm$

Hence, the diagonal of the rectangle is 13 cm.

Page No 20.25:

Question 23:

The perimeter of a square whose area is 225 m² is _________.

Answer:

$As, the area of the square = 225 m^{2} So, the side of the square = \sqrt{Area} = \sqrt{225} = 15 m Now, the perimeter of the square = 4 \times Side = 4 \times 15 = 60 m$

Hence, the perimeter of the square is 60 m .

Page No 20.25:

Question 24:

The area of a rectangle is 120 cm², If the breadth is 6 cm, then its length is _________.

Answer:

$We have, Area of the rectangle = 120 {cm}^{2} and Breadth of the rectangle = 6 cm As, the length of the rectangle = \frac{Area}{Breadth} = \frac{120}{6} = 20 cm$

Hence, the length of the rectangle is 20 cm.

Page No 20.25:

Question 25:

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 _______.

Answer:

$Let the length of the rectangular plot be x and its perimeter be 3 x . As, the breadth of the rectangular plot = (\frac{Perimeter}{2}) - Length = \frac{3 x}{2} - x = \frac{3 x - 2 x}{2} = \frac{x}{2} Now, the ratio between the length and breadth of the plot = \frac{Length}{Breadth} = \frac{x}{(\frac{x}{2})} = \frac{2 x}{x} = \frac{2}{1} = 2 : 1$

Hence, the ratio between the length and breadth of the rectangular plot is 2 : 1.

Page No 20.5:

Question 1:

Which of the following are closed curves? Which of them are simple?

Answer:

Fig. (ii), (iii), (iv), (vi) and (vii) are closed curves, whereas fig. (ii), (iii), (iv) and (vi) are simple closed curves.

Page No 20.5:

Question 2:

Define perimeter of a closed figure.

Answer:

The length of the boundary of a closed figure is known as its perimeter.

Page No 20.5:

Question 3:

Find the perimeter of each of the following shapes:

Answer:

Perimeter = Sum of lengths of all sides of a closed figure
(i)

Perimeter = (4 + 2 + 1 + 5) cm = 12 cm

(ii)

Perimeter = (23 + 35 + 40 + 35) cm = 133 cm

(iii)

Perimeter = (15 + 15 + 15 + 15) cm = 60 cm

(iv)

Perimeter = (3 + 3 + 3 + 3 + 3) cm = 15 cm

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