Find the areas of the rectangles whose sides are:
EXERCISE 10.3
1.
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
ad its
(a) 10 cm
(a) 9 m and 6 m
rate of 78
sq m.
Find the areas of the squares whose sides are :
(d) 2 m and 70 cm
(b) 14 cm
(c) 5 m
The length and breadth of three rectangles are as given below:
(b) 17 m and 3 m (c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
The area of a rectangular garden 50 m long is 300 sqm. Find the width of the garden.
5 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the
per hundred
6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
1. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is
needed to cover the floor of the room?
8. Afloor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find
the area of the floor that is not carpeted.
9. Five square flower beds each of sides I m are dug on a piece of land 5 m long and 4
m wide. What is the area of the remaining part of the land?
10. By splitting the following figures into rectangles, find their areas
(The measures are given in centimetres),
(b)
(a)
of
3
4
219
3
1
Answers
Answer:
Periodic Classification of Elements Class 10 Notes Science Chapter 5
October 24, 2019 by Bhagya
CBSE NotesCBSE Notes ScienceNCERT Solutions Science
CBSE Class 10 Science Notes Chapter 5 Periodic Classification of Elements Pdf free download is part of Class 10 Science Notes for Quick Revision. Here we have given NCERT Class 10 Science Notes Chapter 5 Periodic Classification of Elements.
According to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks.
CBSE Class 10 Science Notes Chapter 5 Periodic Classification of Elements
“The periodic table is a tabular method of displaying the elements in such a way, that the elements having similar properties occur in the same vertical column or group”.
Earlier attempts of the classification of elements: Dobereiner’s Triads, Newland’s law of octaves.
Dobereiner’s Triads: This classification is based on the atomic mass. According to this, when elements are arranged in order of increasing atomic masses, groups of three elements, having similar properties are obtained. The atomic mass of middle element of the triad being nearly equal to the average of the atomic masses of the other two elements.
For Example Li (6.9), Na (23), K (39).
Limitation: It fails to arrange all the known elements in the form of triads, even having similar properties.