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Selina Concise Mathematics Class 9 ICSE Solutions Area and Perimeter of Plane Figures

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Selina Concise Mathematics Class 9 ICSE Solutions Area and Perimeter of Plane Figures

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APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 20 Area and Perimeter of Plane Figures. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Maths Chapter 20 Area and Perimeter of Plane Figures

Exercise 20(A)

Solution 1:
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Solution 2:
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Solution 3:
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Solution 4:
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Solution 5:
Since the perimeter of the isosceles triangle is 36cm and base is 16cm.
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Solution 11:
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Exercise 20(B)

Solution 1:
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Solution 2:
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Solution 9:
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Solution 10:
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Solution 12:
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Solution 13:
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We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.
Then
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Solution 20:
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Solution 27:
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In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.
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Solution 36:
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Exercise 20(C)

Solution 1:
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Solution 2:
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Solution 9:
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Solution 19:
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Solution 20:
From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.
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More Resources for Selina Concise Class 9 ICSE Solutions

Area of Polygons and Circles

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Area of Polygons and Circles

Area formulas can be found at “Reference Table for Areas”
Let’s pick up some hints for those more challenging problems involving area.

Regular polygons have a center and a radius (coinciding with their circumscribed circle), and the distance from the center perpendicular to any side is called its apothem.

Area of Polygons 1

The apothem of a regular polygon is a line segment from the center of the polygon perpendicular to any side of the polygon. Triangle DOC is an isosceles triangle, making the apothem the altitude of this triangle and the median of this triangle (going to the midpoint P.) The apothem is also the radius of the inscribed circle.
The apothem can be used to determine area:

Area of Polygons 2

Area of Polygons 3

Area of Sector of a Circle

Area of Polygons 4

How To Find The Area Of A Plane Figure

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How To Find The Area Of A Plane Figure

The area of the portion of the plane or shape can be defined as the amount of stuff required to cover it.

For finding the area of a polygon, we consider the enclosed region of the polygon. Let us consider an illustration to clarify the idea.

A news bulletin board in a class is to be covered with squared colour paper sheets. To know that how many colour paper sheets are required for the said purpose, the best way is to lay the board on the floor and cover it with a coloured paper sheet without any gap as shown in the figure.

Area-Of-Plane-Figure

Count the number of sheets used for this purpose. Suppose 21 sheets are used to cover the complete board, then, we can measure the region enclosed by a polygon. This measurement is called the area of the polygon.
∴ Area = 21 × Area of 1 sheet
The amount of surface enclosed by a plane figure is called its area. It is measured in square units of lengths.

To find the area of a figure using a graph

We can find the area of regular and irregular figures by using a graph or squared paper. To find the area, first we draw the figure on the graph paper covering as many squares as possible.

area-of-figure-using-graphFor finding the area by using squared paper, we have to follow the following steps:

  1. Count the number of complete squares of unit length enclosed by the figure.
  2. Count the number of those squares whose more than half parts are enclosed by the figure.
  3. Count the number of those squares whose exact half parts are enclosed by the figure and divide the number by 2.
  4. Leave those squares whose less than half parts are enclosed by the figure.
  5. The total of steps (i), (ii), and (iii) are added to obtain the area of the figure.

∴ Area of the figure = Number of complete squares + Number of squares having more than half parts enclosed + 1/2 (Number of squares having half parts enclosed)

Consider the figures given in figure.
(i) Area of rectangle ABCD
= 8 × Area of one square
= 8 × 1 cm2 = 8 cm2
Thus, if the length and breadth of a rectangle are 4 cm and 2 cm respectively, then
Area = 8 cm2
= (4 × 2) sq cm
∴ Area = length × breadth

(ii) Area of ΔPQR = Number of complete square + Number of squares having more than half parts + 1/2 × Number of halved squares
Number of complete squares enclosed in the figure = 2
Number of squares whose more than half parts are enclosed = 4
Neglect the number of squares whose less than half part is enclosed.
Area of ΔPQR = 2 + 4 + 1/2 × 0
= 6 squares
= 6 × 1 cm2
= 6 cm2
Thus, if the base and height of the triangle are 4 cm and 3 cm respectively, then
Area = 6 cm2
= 1/2 × 4 × 3
= 1/2 × base × height

(iii) Area of square WXYZ
= 9 × Area of one square
= 9 × 1 cm2
= 9 cm2
Thus, if the side of the square is 3 cm, then
Area = 9 cm2
= (3 × 3) cm2
= side × side

(iv) Similarly, area of figure ABCDEFGH
= 7 × Area of one square
= 7 × 1 cm2
= 7 cm2

The Area Of Plane Figures Example Problems with Solutions

Example 1: Find the area of a square whose side is 8 cm.
Solution: Area of square
= side × side
= 8 cm × 8 cm
= 64 cm2

Example 2: Find the area of a rectangle whose length is 21 cm and breadth is 5 cm.
Solution: Area of a rectangle
= length × breadth
= 21 cm × 5 cm
= 105 cm2

Example 3: The side of a square plot of land is 30 m. Find the cost of levelling the plot, if the rate is Rs 2 per square metre.
Solution: Area of square
= side × side
= 30 m × 30 m
= 900 m2
Levelling cost of 1 m2
= Rs. 2
Levelling cost of 900 m2
= Rs. (2× 900)
= Rs. 1800

Example 4: Find the area of the given figure.
find-area
Solution: Area of the rectangle DEFG = l × b
= 7 cm × 3 cm = 21 cm2
Area of the rectangle ABCG  l × b
= 7 cm × 3 cm = 21 cm2
Area of the complete figure
= Area of rectangle DEFG
= 21 cm2 + 21 cm2
= 42 cm2

Example 5: How many tiles, each measuring 2 m by 1 m, are required to cover a rectangular hall 12 m long and 8 m broad? Find the cost of tiles at Rs. 25 per tile.
Solution: Length of a tile = 2 m
Breadth of a tile = 1 m
∴ Area of each tile = 2 m × 1 m
= 2 m2
Length of rectangular hall = 12 m
Breadth of rectangular hall = 8 m
∴ Area of the hall = 12 m × 8 m
= 96 m2
Hence, number of tiles required
= Area of the hail/Area of one tile
= 96/2 = 48
∴ Cost of tiles 48 × 25
= Rs. 1200