# Quantitative Methods Area & Perimeter of Plane figures

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Quantitative Methods Area & Perimeter of Plane figures
Quantitative Methods Area & Perimeter of Plane figures Pranjoy Arup Das

Plane figure – a figure formed by three or more straight lines or by a circular boundary.
Four basic plane figures: > Triangle – a figure formed by joining three straight lines. > Quadrilateral – a figure formed by joining four straight lines > Polygon – a figure formed by joining more than four lines. > Circle – a figure formed by drawing a line around a point such that the distance from the point to any point on the line is always the same. Area of a plane figure: measurement of the space covered by the figure. Perimeter of a plane figure : length of the outer boundary of the fig. Some important points: Area is always measured in cm2 ( Square centimeter) , m2 ( Square meter), km2 ( Square kilometer), inch2(Square inch), ft2 (Square feet), etc. Perimeter is always measures in cm, m km, inches, feet etc.

A rectangle has four sides, AB, BC, CD & DA
Side AB = CD (Length) and BC = DA (Breadth) AC & BD are the diagonals and AC = BD (Length)2 + (Breadth) 2 = (Diagonal) 2 Perimeter of a rectangle =2 ( Length + Breadth) Area of a rectangle = (Length * Breadth) Area of 4 walls of a room = 2 (Length + Breadth) * Height SQUARE: Four sides. Side AB = Side BC = Side CD = Side DA AC & BD are the diagonals and AC=BD Diagonal = √2 *Side Perimeter of square= 4 * Side Area of a square= (Side) 2 Also Area of a square = ½ (Diagonal) 2 A B D C A D B C

We know that Area of a rectangle = Length * Breadth = 15 * 8 = 120 m2
Page 416 Ex. 1. Solution: Length = AB = CD = 15 m Breadth = AD = BC = 8 m We need to find the Area of the rectangle ABCD and the length of its diagonal AC We know that Area of a rectangle = Length * Breadth = 15 * 8 = 120 m2 (Length of the diagonal ) 2= Length 2 + Breadth2 Length of diagonal = √ (Length 2 + Breadth2 ) AC = √ ( )= √ ( ) = √ 289 m A B D C

Length of the room= AB = CD = 12 m Breadth of the room= AD = BC = 8 m
Page 416 Ex. 3. Solution: Length of the room= AB = CD = 12 m Breadth of the room= AD = BC = 8 m Breadth of carpet = 75 cm = 75 / 100 = 0.75 m As the length of the room is 12 m, if we take length of the carpet as 12 m having breadth of 0.75 m, this bit of carpet will cover an area = (12 * 0.75) = 9 m2 But the area of the whole room = Length * Breadth = 12 * 8 = 96 m2 So if 9 m2 is covered by carpet length of 12 m, 96 m2 will be covered by = 12 / 9 * 96 = 128 m long carpet. Cost of 128 m long Rs. 50 per m = 128 * 50 = Rs. 6400 A B C D

Page 417 Ex. 7. Solution: Let each side of the square ABCD (before the increase) be A. We know that area of a square = ( side) 2 So the area of ABCD = A 2 Now each side is increased by 150%. That means new side = A + 150% of A = A + 150/100 * A = (100 A A) / 100 = 250 A / 100 = 5A/ 2 And the new area = (5A/2) 2 = 25A2 / 4 Increase in Area = 25A2/4 - A2 = 21A2 / 4 Increase % = ____________________________

A triangle has three sides, AB, BC & CA
AB, BC & CA form three angles A, B & C BC is the base of the triangle. AD is the height or altitude of the triangle. A is known as the vertex. Perimeter of a triangle = AB + BC + CA Area of a triangle = If height is not known, A B C D

SOME SPECIAL PROPERTIES AND FORMULAS OF TRIANGLES:
The sum of any two sides is always greater than the 3rd side. For right angled triangles: AB is called the hypotenuse which is = AC is also the height or altitude or perpendicular BC is the Base. For Equilateral triangles: All three sides are equal, i.e., AB = BC = AC All angles are equal to 60O, i.e., <A = <B = <C = 60O Area of an equilateral triangle = Height AD = A B C A B C D

Let the three sides be AB, BC & CA So AB = 15 cm BC = 13 cm CA = 14 cm
Page 417 Ex. 8. Solution: Let the three sides be AB, BC & CA So AB = 15 cm BC = 13 cm CA = 14 cm We know that Area of a triangle = We need to first find the value of s which is = ( ) / 2 = 21 So the area of the triangle = = = = = = ________ cm2 A 15 cm 14 cm B C 13 cm

Let the three sides be AB, BC & CA So AB = 8 cm BC = 8 cm CA = 8 cm
Page 417 Ex. 10. Solution: Let the three sides be AB, BC & CA So AB = 8 cm BC = 8 cm CA = 8 cm We know that Area of a triangle = We need to first find the value of s which is = ( ) / 2 = 12 So the area of triangle ABC= = A 8 cm 8 cm B C 8 cm

CIRCLE: The boundary of a circle is called the circumference
The point A is the centre of the circle. BC passing through point A is the diameter (D) AD is the radius(R). So are AB & AC. Radius (R)= ½ of diameter, i.e., AB = AC = AD = ½ * BC Perimeter = Circumference = 2𝞹 R , Where 𝞹 (pi) = 22/7 Area = 𝞹 R 2 B D A C

Page 418, Ex 14. Area of a circle = 𝞹 R 2 It is given that radius R = 10.5 cm and it is known that 𝞹 = 22/7 So, Area of the circle = 22/7 * (10.5) 2 = 22/7 * 10.5 * 10.5 = 22/7 * (105/10) * (105/10) = _______cm 2 And circumference = 2𝞹 R = 2 * 22/7 * = _________cm

We know that Area of a circle = 𝞹 R 2 => 154 = 22/7 * R 2
Page 418, Ex 15. We know that Area of a circle = 𝞹 R 2 => 154 = 22/7 * R 2 => R2 = (154 * 7)/22 => R = √49 = 7 cm And Circumference = 2𝞹 R = 2 * 22/7 * 7 = _________cm Page 418, Ex. 16 Area of a circle = (𝞹 R 2 ) cm 2 If the radius is decreased by 20%, the new radius = R – 20% of R = 80/100 * R = 4R/5cm And the new Area will be : 𝞹 (4R/5) 2 = 16 𝞹 R 2 /25 cm 2 The area decreases by = (𝞹 R 2 ) – (16 𝞹 R 2 /25) = 9 𝞹 R 2 /25 cm 2 Decrease % = (9 𝞹 R 2 /25) / 𝞹 R 2 * 100 = _____________ %

Page 419, Ex 17. In one revolution, a wheel covers a distance equal to its circumference. So if in 2000 revolutions the wheel covers a distance of 11 Km, In one revolution it will cover : 11/ 2000 * 1 = 11/2000 Km Converting km to m, we get : 11/2000 * 1000 = 11/2 m This means that the circumference of the wheel is 11/2 m And we also know that Circumference = 2𝞹 R => 11/2 = 2 * 22/7 * R => R = (11/2) * (7 /44) = 7 / 8 m So the radius of the wheel = 7/8 m = m

A Parallelogram has four sides, AB, BC, CD & DA
Side AB is equal and parallel to side CD Side BC is equal and parallel to side DA Side CD is the Base. AE or CF is the height or altitude of ABCD Perimeter of a parallelogram = 2 (Long side + Short side) = 2 ( CD + BC) Area of a parallelogram = Base * Height = CD * AE RHOMBUS: All sides equal. Opposite sides are parallel Perimeter of a rhombus= 4 * Side AC & BD are the diagonals. AC perpendicular to BD Side = ½ √(AC 2 + BD 2) ALSO Perimeter = 2 √(AC 2 + BD 2) Area of a rhombus= ½ * (Product of the diagonals) = ½ * ( AC * BD) A F B D C E A B D C

Page 418, Ex 12. Since the base is twice that of the height: If the height is assumed as x cm, the base will be 2x cm. Area of a parallelogram = Base * Height It is given that the Area = 128 cm2 128 = x * 2x 128 /2 = x2 64 = x2 x = √64 = 8 cm So the height = 8 cm And the Base = ____________cm

Page 418, Ex 13. Area of a rhombus= ½ * (Product of the diagonals) Here the diagonals are 24 cm & 32 cm So the Area = ½ * ( 24 * 32) = ___________cm 2

PRACTICE ASSIGNMENT : Arithmetic, Dr. R.S Aggarwal Page 419 Exercise 23A Problem nos. 1, 3, 15, 24, 25, 43, 46, 54, 64, 68, 72, 87, 88, 91, 102, 103, 113 & 115 .