Parimeter and Area of Plane Figures Class 9 ICSE — Arun 20
Parimeter and area of plane figures class 9 ICSE teaches you how to measure the boundary and surface of triangles, quadrilaterals and circles. This Arun Deep chapter builds the formula kit you will reuse through the rest of the Class 9 ICSE syllabus. You get Heron’s rule for triangles, the area of rectangles, squares, rhombus and trapezium, then circles and rings. Each formula appears with its condition of use.
Download Parimeter and Area of Plane Figures Class 9 ICSE Mathematics Chapter 20 PDF
Download the Parimeter and Area of Plane Figures Class 9 ICSE Mathematics chapter PDF from the Arun Deep textbook to revise offline or print for practice. It is the full chapter as printed in the book — 28 pages, 1 sections, 74 questions.
What is inside Chapter 20
28 pages · 1 sections · 74 exercise questions
| Section | Title |
|---|---|
| 4 | PERIMETER AND AREA OF QUADRILATERALS |
- EXERCISE 20 (A): 19 questions
- EXERCISE 20 (B): 35 questions
- EXERCISE 20 (C): 20 questions
4 PERIMETER AND AREA OF QUADRILATERALS
A. Area of a Quadrilateral when one diagonal and perpendiculars from remaining vertices to the diagonal are given
This first rule handles any four-sided figure split by a single diagonal. In quadrilateral ABCD, draw diagonal BD. It divides the shape into two triangles, ABD and BCD. Drop perpendiculars AL and CM from the two remaining vertices onto BD. The quadrilateral’s area is the sum of the two triangle areas.
\text{Area} = \frac{1}{2} \times BD \times AL + \frac{1}{2} \times BD \times CM
Factor out the common base BD:
= \frac{1}{2} \times BD \times (AL + CM)
So the rule reads: half of one diagonal, times the sum of the perpendiculars on it from the remaining vertices. The perimeter is simply the sum of the four sides. Exercise 20(B) questions 1 and 3 rely on this formula.
Worked pattern. Let diagonal BD = 12 cm, with perpendiculars AL = 5 cm and CM = 7 cm.
\frac{1}{2} \times 12 \times (5 + 7) = \frac{1}{2} \times 12 \times 12 = 72 \text{ cm}^2
Why this works: both triangles share the same base BD. Only their heights differ, so you add the heights, multiply by the base, then halve.
B. Area of a Quadrilateral whose Diagonals Intersect At Right Angles
When the two diagonals of a quadrilateral cross at a right angle, the area takes a compact form. Let diagonals AC and BD meet at O with \angle AOB = 90^\circ. The diagonal AC splits the figure into triangles ABC and ACD. Each uses AC as its base, with BO and OD as the two heights.
\text{Area} = \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD
Combine the two terms:
= \frac{1}{2} \times AC \times (BO + OD) = \frac{1}{2} \times AC \times BD
So the area equals half the product of the two diagonals. You need no side length at all. Exercise 20(B) question 2 applies this directly.
Worked pattern. Diagonals 10 cm and 15 cm meeting at right angles:
\frac{1}{2} \times 10 \times 15 = 75 \text{ cm}^2
C. For a Rectangle
A rectangle is the shape students meet most often. With length l and breadth b, the chapter gives three results:
- Perimeter: P = 2(l + b)
- Area: A = l \times b
- Diagonal: d = \sqrt{l^2 + b^2}
The diagonal formula comes from Pythagoras. The diagonal is the hypotenuse of a right triangle whose legs are length and breadth. You will use these three in the path and carpet problems of Exercise 20(B), questions 8 to 17.
Worked pattern. Length 8 m, breadth 5 m:
P = 2(8 + 5) = 26 \text{ m}, \quad A = 8 \times 5 = 40 \text{ m}^2, \quad d = \sqrt{64 + 25} = \sqrt{89} \text{ m}
D. For a Square
A square is a rectangle with all four sides equal. If each side is a, the chapter gives four results:
- Perimeter: P = 4a
- Area from side: A = a^2
- Area from diagonal: A = \frac{1}{2} d^2
- Diagonal: d = \sqrt{2} a
The second area formula helps when a problem gives the diagonal instead of the side. The diagonal formula is Pythagoras on a right isosceles triangle with legs a and a. Exercise 20(B) questions 20 to 22 test the diagonal forms, including hectare-to-square-metre conversions.
Worked pattern. Side 7 cm:
P = 28 \text{ cm}, \quad A = 49 \text{ cm}^2, \quad d = 7\sqrt{2} \approx 9.9 \text{ cm}
E. Area of a Parallelogram
A parallelogram has two pairs of parallel sides. Its area is base times height, where the height is the perpendicular distance between the chosen base and its opposite side.
\text{Area} = \text{Base} \times \text{Height}, \quad \text{Perimeter} = 2(AB + BC)
The height must be perpendicular to the base — never a slanted side. Mixing up base and height pairs is the common mistake here. Exercise 20(B) question 28 gives two adjacent sides and one height, then asks for the other height. Because both base-height pairs give the same area, you can equate them and solve.
Worked pattern. Base 10 cm, height 6 cm:
A = 10 \times 6 = 60 \text{ cm}^2
F. Area of a Rhombus
A rhombus is a parallelogram with all four sides equal. Its area uses the two diagonals:
A = \frac{1}{2} \times d_1 \times d_2
The chapter states the key remark: the diagonals of a rhombus bisect each other at right angles. That property splits the rhombus into four right triangles. If you know one diagonal and one side, apply Pythagoras to find half of the other diagonal, then double it. Exercise 20(B) questions 24, 26 and 27 follow this path.
Worked pattern. Diagonals 8 cm and 6 cm:
A = \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^2
G. For a Trapezium
A trapezium has one pair of parallel sides. Its area is the average of the parallel sides, times the distance between them.
A = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})
The derivation splits the trapezium along a diagonal into two triangles sharing the same height h. The perimeter is the sum of the four sides. When the non-parallel sides are equal, as in question 25 of Exercise 20(B), drop a perpendicular to form a right triangle, then use Pythagoras to find the height.
Worked pattern. Parallel sides 12 cm and 8 cm, distance 5 cm:
A = \frac{1}{2} \times (12 + 8) \times 5 = 50 \text{ cm}^2
Square roots the chapter expects you to know. Two surd values appear throughout the exercises: \sqrt{2} = 1.414 or 1.41, and \sqrt{3} = 1.732 or 1.73. Exercise 20(A) question 10 asks for answers correct to two decimal places, so keep the three-decimal forms ready. Equilateral triangle problems need \sqrt{3}; square diagonal problems need \sqrt{2}.
Two related chapters sit on either side of this one. Mean and Median covers statistics. The surface area and volume of 3D solids chapter extends area into three dimensions.
What to practise from Parimeter and Area of Plane Figures Class 9 ICSE
This ICSE chapter drills three formula families across three exercise blocks. Each block targets a different figure type.
- EXERCISE 20 (A) — 19 questions on triangles. Base-times-height area, Heron’s formula with all three sides, the equilateral triangle formulas \frac{\sqrt{3}}{4} a^2 for area and \frac{\sqrt{3}}{2} a for height, isosceles triangles solved via Pythagoras, right triangles, ratio-of-sides problems (3:4:5 and 25:17:12), and cost-of-cultivation problems.
- EXERCISE 20 (B) — 35 questions on quadrilaterals and rectilinear figures. One-diagonal quadrilateral area, rectangle perimeter and area with path and carpet problems, square diagonal and hectare problems, parallelogram base-altitude pairs, rhombus from diagonals and side, trapezium with equal legs.
- EXERCISE 20 (C) — 20 questions on circles. Area and circumference, rings of concentric circles, wire bent into other shapes, wheel revolutions, and a square inscribed in a circle.
Cost problems follow one pattern: find the area first, then multiply by the rate per unit area. Question 5 of Exercise 20(A) multiplies an area of 9000 m² by Rs. 24.60 per 100 m². Question 18 of Exercise 20(B) reverses the logic, dividing the total cost by the rate to recover the area.
In triangles, each side can serve as the base, and the height must be perpendicular to that chosen base. Question 2 of Exercise 20(A) finds the height on the longest side — a different base gives a different height.
The comparison table below maps each figure to its formula and its key condition.
| Figure | Area formula | Perimeter formula | Key condition |
|---|---|---|---|
| Quadrilateral, one diagonal | \frac{1}{2} \times d \times (p_1 + p_2) | sum of 4 sides | need diagonal + two perpendiculars |
| Quadrilateral, right diagonals | \frac{1}{2} \times d_1 \times d_2 | sum of 4 sides | diagonals meet at 90° |
| Rectangle | l \times b | 2(l+b) | need length and breadth |
| Square | a^2 or \frac{1}{2} d^2 | 4a | need side or diagonal |
| Parallelogram | base × height | 2(AB+BC) | height ⊥ base |
| Rhombus | \frac{1}{2} d_1 d_2 | 4a | diagonals bisect at 90° |
| Trapezium | \frac{1}{2}(a+b)h | sum of 4 sides | parallel sides + distance |
You will find every chapter of this book on the Class 9 ICSE Maths book hub, and the complete Class 9 books list on the books page. The official CISCE website publishes the Class 9 syllabus for reference.
Parimeter and Area of Plane Figures Class 9 ICSE frequently asked questions
When do I use Heron’s formula instead of half base times height in perimeter and area of plane figures class 9 ICSE?
Use Heron’s formula when you know all three sides but not the height. The half-base-times-height rule needs a base and its perpendicular height. If a problem gives only sides, apply Heron’s rule first, then recover the height from \text{Area} = \frac{1}{2} \times \text{base} \times h.
For example, sides 9 cm, 12 cm and 15 cm give semi-perimeter s = 18. The area is \sqrt{18 \times 9 \times 6 \times 3} = 54 cm². The height on the 15 cm base comes from 54 = \frac{1}{2} \times 15 \times h, so h = 7.2 cm.
How do I find the height of a triangle in Class 9 ICSE when only its three sides are given?
Compute the semi-perimeter s, then Heron’s area. Equate that area to \frac{1}{2} \times \text{base} \times h and solve for h. Exercise 20(A) questions 2 and 3 do exactly this, working back from the area to the altitude.
Why do the diagonals of a rhombus intersect at right angles in this ICSE chapter?
It is a property of the rhombus: the diagonals bisect each other at right angles. Each diagonal splits the rhombus into two congruent right triangles. If you know one diagonal and the side, Pythagoras gives half of the other diagonal. Double it for the full length. Questions 24 and 26 of Exercise 20(B) follow this path.
How do I solve wheel revolution problems in Exercise 20(C) of Class 9 ICSE?
One revolution covers one circumference, 2\pi r. Divide the total distance by the circumference. Question 17 uses a wheel of radius 0.25 m travelling 11 km. Convert 11 km to 11000 m, then divide by \frac{11}{7} m to get 7000 revolutions.
How do I reject an impossible answer in a Class 9 ICSE geometry quadratic?
A length, side or radius can never be negative. When a quadratic gives two roots, discard the negative one. Exercise 20(A) question 11 produces x = -8 and x = 6. The side is 6 cm because a side of -8 cm has no meaning.
Reference: ICSE Class 9 Mathematics textbook material, chapter 20: Parimeter and Area of Plane Figures — CISCE-aligned.