ICSE Class 9 Maths Chapter 20 Parimeter and Area of Plane Figures

Unit 6 - Mensuration

Perimeter And Area Of Plane Figures

Points To Remember

1. Perimeter: The perimeter of a plane figure is the length of its boundary, i.e., the sum of its sides. The unit of perimeter is the same as the unit of length.

2. Area: The area of a plane figure is the measure of the surface enclosed by its boundary, i.e., the surface enclosed by its sides.

It is measured in square units such as square centimetres or square metres, written as cm\(^2\) or m\(^2\) respectively.

Perimeter And Area Of Triangles

A. Area of a Triangle = \(\frac{1}{2} \times \text{Base} \times \text{Corresponding Height}\)

Any side of the triangle may be taken as base and the length of perpendicular from the opposite vertex to the base is the corresponding height.

In given figure, Area of \(\triangle ABC = \left(\frac{1}{2} \times BC \times AD\right)\) sq. units

Perimeter = \((a + b + c)\) units

B. Hero's Formula: Let \(a, b, c,\) be the lengths of the sides of a triangle and let \(s = \frac{1}{2}(a + b + c)\), called semi-perimeter of the triangle. Then.

Area of the Triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) sq. units.

C. For a Right-Angled \(\triangle ABC\) in which \(\angle B = 90°\), we have:

\((i) AC^2 = AB^2 + BC^2\) (Pythagoras Theorem)

\((ii)\) Area of \(\triangle ABC = \frac{1}{2} \times\) (Product of sides containing the right angle)

\(= \left(\frac{1}{2} \times BC \times AB\right)\) sq. units

\((iii)\) Perimeter = (Sum of three sides) units

D. For An Equilateral Triangles of Side a, we have:

\((i)\) Height = \(\left(\frac{\sqrt{3}}{2}a\right)\) units.

\(h = \sqrt{a^2 - \left(\frac{a}{2}\right)^2}\)

\((ii)\) Area = \(\left(\frac{\sqrt{3}}{4}a^2\right)\) sq. units

\(\therefore A = \frac{1}{2} \times a \times h\)

\((iii)\) Perimeter = \(3a\) units.

E. For an Isosceles \(\triangle ABC\) in which \(AB = AC = a\) and \(BC = b\), we have:

\((i)\) Height = \(\frac{\sqrt{4a^2 - b^2}}{2}\) units.

\(\therefore h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}\)

\((ii)\) Area = \(\left(\frac{1}{4}b\sqrt{4a^2 - b^2}\right)\) sq. units

\(\therefore A = \frac{1}{2} \times b \times h\)

\((iii)\) Perimeter = \((2a + b)\) units.

Perimeter And Area Of Quadrilaterals

A. Area of Quadrilateral when one diagonal and perpendiculars from remaining vertices to the diagonal are given

\((i)\) Area of quad. ABCD = area (\(\triangle ABD\)) + area (\(\triangle BCD\))

\(= \frac{1}{2} \times BD \times AL + \frac{1}{2} \times BD \times CM\)

\(= \frac{1}{2} \times BD \times (AL + CM)\)

\(= \frac{1}{2} \times \text{one diagonal} \times \text{sum of lengths of perpendiculars on it from remaining vertices}.\)

\((ii)\) Perimeter = sum of four sides units

B. Area of a Quadrilateral whose Diagonals Intersect At Right Angles

Let the diagonals AC and BD of quad. ABCD intersect at O at right angles. Then.

Area of quad. ABCD.

\(= \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD)\)

\(= \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD\)

\(= \frac{1}{2} \times AC \times (BO + OD)\)

\(= \frac{1}{2} \times AC \times BD\)

\(= \frac{1}{2} \times\) (Product of its diagonals) square units.

C. For a Rectangle with length = \(l\) units and Breadth = \(b\) units, we have

\((i)\) Perimeter = 2 (Length + Breadth) = 2 \((l + b)\) units.

\((ii)\) Area = (Length \(\times\) Breadth) = \((l \times b)\) sq. units.

\((iii)\) Diagonal = \(\sqrt{l^2 + b^2}\) units.

D. For a square with side a units, we have:

\((i)\) Perimeter = \((4 \times \text{side}) = 4a\) units.

\((ii)\) Area = \((\text{side})^2 = a^2\) sq. units.

\((iii)\) Area = \(\frac{1}{2} \times (\text{Diagonal})^2\) sq. units.

\((iv)\) Diagonal = \(\sqrt{2}a\) units = \(\sqrt{2 \times \text{Area}}\) units.

E. Area of a Parallelogram = Base \(\times\) Height.

\((i)\) Area of \(\parallel\)gm ABCD = AB \(\times\) DL

\(= \) AD \(\times\) BM.

\((ii)\) Perimeter = 2 (AB + BC) units

F. Area of Rhombus = \(\frac{1}{2} \times\) Product of its diagonals.

\(= \left(\frac{1}{2} \times d_1 \times d_2\right)\) sq. units.

Remark. The diagonals of a rhombus bisect each other at right angles.

G. \((i)\) Area of Trap. ABCD = Area of \(\triangle ABC\) + Area of \(\triangle ACD\)

\(= \frac{1}{2} \times AB \times h + \frac{1}{2} \times CD \times h\)

\(= \frac{1}{2} \times (AB + CD) \times h\)

\(= \frac{1}{2} \times\) (Sum of parallel sides)

\(\times\) (Distance between them) sq. units.

\((ii)\) Perimeter = sum of four sides units

Note: \(\sqrt{2} = 1.414\) or 1.41

\(\sqrt{3} = 1.732\) or 1.73

Exercise 20 (A)

Q. 1. Find the area of a triangle whose base is 15 cm and the corresponding height is 9.6 cm.

Sol. Base BC of triangle ABC = 15 cm and its altitude AD = 9.6 cm

\(\therefore\) Area of \(\triangle ABC = \frac{1}{2}\) base \(\times\) altitude

\(= \frac{1}{2} \times 15 \times 9.6 = 72.0\) cm\(^2\) Ans.

Teacher's Note

Understanding area calculations helps in real-life applications like determining how much paint is needed to cover a triangular wall or calculating the size of a triangular garden.

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