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Quantitative Methods Area & Perimeter of Plane figures Pranjoy Arup Das

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.

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Quantitative Methods Area & Perimeter of Plane figures Pranjoy Arup Das

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  1. Quantitative MethodsArea & Perimeter of Plane figuresPranjoy Arup Das

  2. 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.

  3. A B RECTANGLE : • 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 D C D A B C

  4. A B • 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 = √ (152 + 82 )= √ (225 + 64) = √ 289 m D C

  5. A B • 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 carpet @ Rs. 50 per m = 128 * 50 = Rs. 6400 C D

  6. 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 + 150 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 % = ____________________________

  7. A TRIANGLE: • 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, Area of a triangle = B C D

  8. A 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 = B C A B C D

  9. A 15 cm 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 = (15 + 13+ 14) / 2 = 21 So the area of the triangle = = = = = = = ________ cm2 B C 13 cm

  10. A 8 cm 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 = (8 + 8+ 8) / 2 = 12 So the area of triangle ABC= = = = C B 8 cm

  11. B D 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 A C

  12. 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 * 10.5 = _________cm

  13. 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 = _____________ %

  14. 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 = 0.875 m

  15. A F B PARALLELOGRAM: • 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) C D E A B D C

  16. 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

  17. 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

  18. 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 .

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