Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1

Question 1.

A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Solution:

(a) Side of the square = 60 m

ā“ Its perimeter = 4 Ć Side

= 4 v 60 m = 240 m

Area of the square = Side Ć Side

= 60 m Ć 60 m

= 3600 m

(b) āµ Perimeter of the rectangle = Perimeter of the given square

ā“ Perimeter of the rectangle = 240 m

or 2 Ć [Length + Breadth] = 240 m

or 2 Ć [80 m + Breadth] = 240 m

or 80 m + Breadth = \(\frac{240}{2}\)m = 120 m

ā“ Breadth = (120 – 80)m = 40 m

Now, Area of the rectangle = Length Ć Breadth

= 80 m Ć 40 m

= 3200 m2

Since, 3600 m^{2} Ć 3200 m^{2}

ā“ Area of the square field (a) is greater.

Question 2.

Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ā¹ 55 per m^{2}.

Solution:

āµ The given plot is a square with side as 25 m.

ā“ Area of the plot = Side Ć Side

= 25 m Ć 25 m

= 625 m^{2}

āµ The constructed portion is a rectangle having length = 20 m and breadth = 15 m.

ā“ Area of the constructed portion

= 20 m Ć 15 m

= 300 m^{2}

Now area of the garden

= [Total plot area] – [Total constructed area]

= (625 – 300) m^{2}

= 325 m^{2}

ā“ Cost of developing the garden

= ā¹ 55 Ć 325 = ā¹17,875

Question 3.

The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden? [Lenght of rectangle is 20 – (3.5 + 3.5) metres.]

Solution:

For the semi-circular part:

Diameter of the semi-circle = 7m

ā“ Radius of the semi-circle = \(\frac{7}{2}\) = 3.5 m

Area of the 2 semi – circles = 2[\(\frac{1}{2}\) Ļr^{2}]

Also, perimeter of the 2 semicircles

= 2(\(\frac{2Ļr}{2}\)) = 2Ļr = 2 Ć \(\frac{22}{7}\) Ć \(\frac{35}{10}\)m = 22m

For rectangular part:

Length of the rectangle = 20 – (3.5 + 3.5)m

= 13 m

Breadth of the rectangle = 7 m

ā“ Area = Length Ć Breadth = 13 m Ć 7 m

= 91 m^{2}

Perimeter = 2 Ć [Length + Breadth]

= 2 Ć [13 m + 0m] = 2 Ć 13 m = 26 m

Now, Area of the garden = (38.5 + 91)m^{2}

= 129.5 m^{2}

Perimeter of the garden = (38.5 + 91)m^{2} = 129.5 m^{2}

Perimeter of the garden = 22 m + 26 m = 48 m

Question 4.

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m^{2}? (If required you can split the tiles in whatever way want to fill up the comers.)

Solution:

Area of a parallelogram

= Base Ć Corresponding height

Area of a tile = (\(\frac{24}{100}\) Ć \(\frac{10}{100}\))m^{2} = \(\frac{240}{10000}\)

m^{2} = 0.024 m^{2}

ā“ Area of the floor = 1080 m^{2
}Now, number of tiles =

= \(\frac{1080}{0.024}\) = 45000 tiles

Question 5.

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2Ļr, where r is the radius of the circle?

Solution:

(a) Diameter = 2.8 cm

ā Radius = \(\frac{2.8}{2}\)

= 1.4 cm

Perimeter of a circle = 2Ļr

ā Perimeter of a semicircle = \(\frac{2Ļr}{2}\) = Ļr

ā“ Perimeter of the figure = Ļr + Diameter

= \(\frac{22}{7}\) Ć 1.4 cm + 2.8 cm

= \(\frac{22}{7}\) Ć \(\frac{14}{10}\) cm + 2.8 cm

= 4.4 cm + 2.8 cm = 7.2 cm

Note: In this fugure, the diameter is also a part of the figure.

(b) Perimeter of the semi-circular part = Ļr

= \(\frac{22}{7}\) Ć 1.4 cm = 4.4 cm

Perimeter of the remaining part

= 1.5 cm + 28 cm + 1.5 cm = 5.8 cm

ā“ Perimeter of the figure

= 4.4 cm + 5.8 cm = 10.2 cm

(c) Perimeter of the semi – circular part

ā“ Perimeter of the figure

= 4.4 cm + 2 cm + 2 cm = 8.4 cm

Since, 7.2 cm < 8.4 cm < 10.2 cm

ā“ Perimeter of figure ‘b’ has the longest round.

Note: In figures ‘b’ and ‘c’, the diameters are not part of the figures.