Class 8 - Mathematics
Algebraic Expressions and Identities - Exercise 11.1
Top Block 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?
Answer :
Given: The side of a square = 60 m
And the length of rectangular field = 80 m
According to question,
Perimeter of rectangular field
= Perimeter of square field
⇒ 2(l + b) = 4 x side
⇒ 2(80 + b) = 4 x 60
⇒ 160 + 2b = 240
⇒ 2b = 240 -160 =80
⇒ b = 40m
Now Area of Square field
= (side)²
= 60² = 3600 m²
And Area of Rectangular field
= length x breadth = 80 x 40
= 3200 m²
Hence, area of square field is larger.
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 ₹55 per m2.
Mddle block 1
Answer :
Side of a square plot = 25 m
∴ Area of square plot = (side)²
= 25² = 625 m²
Length of the house = 20 m and
Breadth of the house = 15 m
∴ Area of the house = length x breadth
= 20 x 15 = 300 m²
Area of garden = Area of square plot – Area of hou
se
= 625 – 300 = 325 m²
Cost of developing the garden per sq. m = 荷55
Cost of developing the garden 325 sq. m = 荷55 x 325
= 荷 17,875
Hence total cost of developing a garden around is 荷17,875.
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 [Length of rectangle is 20 – (3.5 + 3.5 meters]
Answer :
Given: Total length = 20 m
Diameter of semi circle = 7 m
∴ Radius of semi circle = 7⁄2 = 3.5 m
Length of rectangular field
= 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = l x b
= 13 x 7 = 91 m²
Area of two semi circles = 2 x (½πr²)
= 2 x ½ x 3.14 x 3.5 x 3.5 = 38.5 m2
Area of garden = 91 + 38.5 = 129.5 m2
Now Perimeter of two semi circles = 2 x πr = 2 x 3.14 x 3.5 = 22 m
And Perimeter of garden
= 22 + 13 + 13
= 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 1080m²? [If required you can split the tiles in whatever way you want to fill up the corners]
Answer :
Given: Base of flooring tile = 24 cm
= 0.24 m
Corresponding height of a flooring tile
= 10 cm = 0.10 m
Now Area of flooring tile
= Base Altitude
= 0.24 0.10
= 0.024 m²
∴ Number of tiles required to cover the floor
= Area of floor⁄Area of one tile
= 1080⁄0.024
= 45000 tiles
Hence 45000 tiles are required to cover the floor.
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 2πr where r is the radius of the circle.
Answer :
(a) Radius = diameter⁄2 =2.8⁄2
= 1.4 cm
Circumference of semi circle = πr
= 3.14 x 1.4 = 4.4 cm
= Circumference of semi circle + Diameter
= 4.4 + 2.8 = 7.2 cm
(b) Diameter of semi circle = 2.8 cm
∴ Radius = r = 1.4 cm
Circumference of semi circle = πr
= 3.14 x 1.4 = 4.4 cm
= 1.5 + 2.8 + 1.5 + 4.4 = 10.2 cm
(c) Diameter of semi circle = 2.8 cm
∴ Radius = diameter⁄2 =2.8⁄2
= 1.4 cm
Circumference of semi circle = πr
= 3.14 x 1.4 = 4.4 cm
= 2 + 2 + 4.4 = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.