Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math

RD Sharma Solutions (Part - 1) - Ex-20.1, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

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 | RD Sharma Solutions for Class 7 Mathematics

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


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

1. What is RD Sharma Solutions?
Ans. RD Sharma Solutions is a comprehensive guidebook that provides step-by-step solutions to the questions present in the RD Sharma textbook. It is specifically designed for students of Class 7 studying Mathematics, and it covers various topics including Mensuration - I.
2. What is the significance of studying Mensuration - I in Class 7?
Ans. Mensuration - I is a crucial topic in Class 7 Mathematics as it introduces students to the concept of measuring and calculating various geometric shapes and figures. It helps students develop their skills in understanding perimeter, area, and volume, which are fundamental concepts required in higher classes and real-life applications.
3. How can RD Sharma Solutions help in understanding Mensuration - I?
Ans. RD Sharma Solutions provides detailed step-by-step explanations and solutions to the questions present in the Mensuration - I chapter of Class 7 Mathematics. By referring to these solutions, students can understand the concepts, formulas, and methods used to solve problems related to measuring and calculating different shapes and figures.
4. Can RD Sharma Solutions be used as a standalone study material for Mensuration - I?
Ans. RD Sharma Solutions can be used as a supplementary study material for Mensuration - I, but it is recommended to study the corresponding chapter in the RD Sharma textbook as well. The solutions provide clarity and guidance on solving problems, but the textbook provides a comprehensive understanding of the topic with additional examples and exercises.
5. Are RD Sharma Solutions helpful for exam preparation in Class 7?
Ans. Yes, RD Sharma Solutions are extremely helpful for exam preparation in Class 7. By practicing the solved examples and exercises provided in the solutions, students can strengthen their problem-solving skills and gain confidence in answering questions related to Mensuration - I. Additionally, the solutions serve as a valuable resource for revision and self-assessment before exams.
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