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Class 10 Class 12

(a) Match the following figures with their respective areas in the box.

(b) Write the perimeter of each shape.

(a)
(a)(b) (i) The given figure is a rectangle in which Length =
(b) (i) The given figure is a rectangle in which
Length = 14 cm
Breadth = 7 cm
∵ Perimeter of a rectangle = 2 x [Length + Breadth]
∴ Perimeter of the given figure = 2 x [14 cm + 7 cm]
= 2 x 21 cm
= 42 cm
(a)(b) (i) The given figure is a rectangle in which Length =
(ii) The figure is a square housing its side as 7 cm.
∵ Perimeter of a square = 4 x side
∴ Perimeter of the given figure = 4 x 7 cm
= 28 cm.
   (a)(b) (i) The given figure is a rectangle in which Length =

2482 Views

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

(a) Side of the square = 60 m
∴ Its perimeter = 4 x side
= 4 x 60 m
= 240 m
Area of the square = Side x Side
= 60 m x 60 m
= 3600 m²
(b) ∵ Perimeter of the rectangle = Perimeter of the given square
∴ Perimeter of the rectangle = 240 m
or 2 x [Length + Breadth] = 240 m
or 2 x [80 m + Breadth] = 240 m
or 80 m + Breadth = $240 over 2 straight m space equals space 120 space straight m$
∴ Breadth = (120 - 80) m = 40 m
Now, Area of the rectangle = Length x Breadth
= 80 m x 40 m
= 3200 m²
Since, 3600 m² > 3200 m²
∴ Area of the square field (a) is greater.

940 Views

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 Rs 55 per m².

∵ The given plot is a square with side as 25 m.
∴ Area of the plot = Side x Side
= 25 m x 25 m = 625 m²
∵ The constructed portion is a rectangle having length = 20 m and breadth = 15 m.
∴ Area of the constructed portion = 20 m x 15 m
= 300 m²
Now area of the garden = [Total plot area] - [Total constructed area]
= (625 - 300)m²
= 325 m²
∴ Cost of developing the garden = Rs 55 x 325
= Rs 17, 875

1121 Views

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.]

For the semi-circular part:
Diameter of the semi-circle = 7 m
∴ Radius of the semi-circle =

7 over 2 equals 3.5 space straight m

Area of the 2 semi-circles =

space space space space 2 open square brackets 1 half πr squared close square brackets

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equals 22 over 7 cross times 35 over 10 cross times 35 over 10 space straight m squared

equals space fraction numerator 11 cross times 35 over denominator 10 end fraction straight m squared space equals space 38.5 space straight m squared

Also, perimeter of the 2 semicircles =

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equals space 2 cross times 22 over 7 cross times 35 over 10 space straight m

= 22 m
For rectangular part:
Length of the rectangle = 20 - (3.5 + 3.5) m
= 13 m
Breadth of the rectangle = 7 m
∴ Area = Length x Breadth
= 13 m x 7 m
= 91 m²
Perimeter = 2 x [Length + Breadth]
= 2 x [13 m + 0 m]
= 2 x 13 m
= 26 m
Now, Area of the garden = (38.5 + 91)m²
= 129.5 m²
Perimeter of the garden = 22 m + 26 m
= 48 m

2143 Views

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 area required to cover a floor of area 1080 m²?
(If required you can split the tiles in whatever way want to fill up the corners.)

Area of a parallelograms = Base x Corresponding height
Area of a tile = $space space space space open parentheses 24 over 100 cross times 10 over 100 close parentheses straight m squared space equals space 240 over 10000 straight m squared space equals space 0.024 space straight m squared$
∴ Area of the floor = 1080 m²
Now, number of tiles = $fraction numerator Total space area over denominator Area space of space one space tiles end fraction$

= $fraction numerator 1080 over denominator 0.024 end fraction$
= 45000 tiles