RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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Question 1:

Find the area, in square metres, of a rectangle whose
(i) Length = 5.5 m, breadth = 2.4 m
(ii) Length = 180 cm, breadth = 150 cm

Answer 1:

We have,
(i) Length = 5.5 m, Breadth = 2.4 m
    Therefore,
    Area of rectangle = Length x Breadth
                                 = 5.5 m x 2.4 m
                                 = 13.2 m2 
              
  (ii) Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m     [ Since 100 cm = 1 m]
       Therefore,
       Area of rectangle = Length x Breadth
                                    = 1.8 m x 1.5 m
                                    = 2.7 m2

Question 2:

Find the area, in square centimetres, of a square whose side is
(i) 2.6 cm
(ii) 1.2 dm

Answer 2:

We have,
(i) Side of the square = 2.6 cm
     Therefore, area of the square = (Side)2
                                             = (2.6 cm)= 6.76 cm2


(ii) Side of the square = 1.2 dm = 1.2 x 10 cm = 12 cm        [ Since 1 dm = 10 cm]
      Therefore, area of the square = (Side)2
                                             = (12 cm)= 144 cm2


Question 3:

Find in square metres, the area of a square of side 16.5 dam.

Answer 3:

We have,
Side of the square = 16.5 dam = 16.5 x 10 m = 165 m     [ Since 1 dam = 10 m ]
Area of the square = (Side)2 = (165 m)2
                                         = 27225 m2


Question 4:

Find the area of a rectangular feild in ares whose sides are:
(i) 200 m and 125 m
(ii) 75 m 5 dm and 125 m

Answer 4:

We have,
(i) Length of the rectangular field = 200 m
    Breadth of the rectangular field = 125 m
    Therefore,
    Area of the rectangular field =  Length x Breadth
                                                   = 200 m x 125 m
                                                   = 25000 m2 = 250 ares     [Since 100 m2 = 1 are]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m = 75.5 m     [Since 1 dm = 10 cm = 0.1 m]
    Breadth of the rectangular field = 120 m
    Therefore,
    Area of the rectangular field =  Length x Breadth
                                                   = 75.5 m x 120 m
                                                   = 9060 m2 = 90.6 ares     [Since 100 m2 = 1 are]


Question 5:

Find the area of a rectangular field in hectares whose sides are:
(i) 125 m and 400 m
(ii) 75 m 5 dm and 120 m

Answer 5:

We have,
(i) Length of the rectangular field = 125 m
     Breadth of the rectangular field = 400 m
     Therefore,
     Area of the rectangular field =  Length x Breadth
                                                   = 125 m x 400 m
                                                   = 50000 m2 = 5 hectares     [Since 10000 m2 = 1 hectare]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m = 75.5 m     [Since 1 dm = 10 cm = 0.1 m]
     Breadth of the rectangular field = 120 m
     Therefore,
     Area of the rectangular field =  Length x Breadth
                                                   = 75.5 m x 120 m
                                                   = 9060 m2 = 0.906 hectares     [Since 10000 m2 = 1 hectare]


Question 6:

A door of dimensions 3 m × 2m is on the wall of dimension 10 m × 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.

Answer 6:

We have,
Length of the door = 3 m
Breadth of the door = 2 m
Side of the wall = 10 m
Area of the wall = Side x Side = 10 m x 10 m = 100 m2
Area of the door = Length x Breadth = 3 m x 2 m = 6 m2
Thus,
Required area of the wall for painting = Area of the wall − Area of the door = (100 − 6 ) m2 = 94 m2
Rate of painting per square metre = Rs. 2.50
Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235


Question 7:

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?

Answer 7:

We have,
Perimeter the of rectangle = 2(Length + Breadth)
                                     = 2(40 cm + 22 cm) = 124 cm
It is given that the wire which was in the shape of a rectangle is now bent into a square.
Therefore, the perimeter of the square = Perimeter of the rectangle
            =>   Perimeter of the square = 124 cm
⇒ 4 x side = 124 cm
∴ Side = 124/4=31 cm
Now,
Area of the rectangle = 40 cm x 22 cm = 880 cm2
Area of the square = (Side)2 = (31 cm)2 = 961 cm2
Therefore, the square-shaped wire encloses more area.


Question 8:

How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?

Answer 8:

We have,
Length of the glass pane = 25 cm
Breadth of the glass pane = 16 cm
Area of one glass pane = 25 cm x 16 cm = 400 cm2 = 0.04 m2   [ Since 1 m2 = 10000 cm2 ]
Thus,
Area of 12 such panes = 12 x 0.04 = 0.48 m2


Question 9:

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 9:

We have,
Area of the wall = 3 m x 4 m = 12 m2
Area of one marble tile = 10 cm x 12 cm = 120 cm2 = 0.012 m2    [ Since 1 m2 = 10000 cm2 ]
Thus,

Number of tiles =  RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math Class 7 Notes | EduRev

Cost of one tile = Rs. 2
Total cost = Number of tiles x Cost of one tile
                = Rs. (1000 x 2) = Rs. 2000


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