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9.MENSURATION Std.9th Sub- MATHEMATICS 9.Mensuration Sadavarte M.D. 9403347957 9.Mensuration

MENSURATION Std-9th Sub -Geometry 9.Mensuration INTRODUCTION 9.1) AREA OF TRIANGLE 9.2) AREA & PERIMETER OF A QUADRILAERAL 9.3)AREA & PERIMETER OF THE CIRCLE & SEMICIRCLE 9.4) AREA OF REGULAR POLYGON 9.5) AREA OF n-SIDED REGULAR POLYGON 9.6) TO FIND THE AREA OF THE REGULAR POLYGON IN TERMS OF RADIUS OF CIRCUMSCRIBED CIRCLE 9.7) PARTICULAR CASES 1. REGULAR HEXAGON 2. REGULAR OCTAGON Std-9th Sub -Geometry 9.Mensuration

Std-9th 9.Mensuration Sub -Geometry CONCEPT OF AREA. Fig.1 Fig.2 Observe the adjoining figure. Triangular region Triangle Fig .1 represent a region which includes a triangle and its interior. While fig.2 represents only a triangle. The union of a triangle and its interior is called triangular region. It has definite area whereas triangle has no area. Similarly rectangular, circular and polygonal region have area. In this chapter we refer triangular region as area of triangle, circular area as area of circle etc. Std-9th 9.Mensuration Sub -Geometry

Std-9th Sub -Geometry 9.Mensuration AREA OF RECTANGULAR REGION. UNITS OF AREA 1 Consider a square of side 1 cm . 1 Its area = 1 cm × 1 cm 4 cm area = 1 sq. cm. Thus , we get unit of area sq.cm Area of square having side 1 cm is 1 sq.cm. area. B 6 cm C Now we will find out the area of rectangle ABCD whose length is 6 cm and breadth is 4 cm. Rectangle ABCD can be divided into 24 squares with side 1 cm Therefore area of rectangle ABCD is 24 sq.cm. Now, length ×breadth=6 cm ×4cm = 24.sq.cm. = area of rectangle. Here we get formula, area of rectangle = length× breadth Like sq.cm. other units of area are sq.m. and sq.km. Std-9th Sub -Geometry 9.Mensuration

Revision – Some Formulae. Sr.No. Figure Perimeter Area a c P = a+b+c h 1 b Triangle a a P = 3a 2 a Equilateral Triangle Std-9th Sub -Geometry 9.Mensuration

Revision – Some Formulae. Sr.No. Figure Perimeter Area a a a 3 P = 4a a Square b 4 l rectangle Sub -Geometry 9.Mensuration Std-9th

Revision – Some Formulae. Sr.No. Figure Perimeter Area a h 5 P =2(a+b) b parallelogram a P=a+b+c+d a d h 6 b Trapezium Sub -Geometry 9.Mensuration Std-9th

Revision – Some Formulae. Sr.No. Figure Perimeter Area a 7 a P =4a a a Rhombus 8 r O circle r O r 9 semicircle Std-9th Sub -Geometry 9.Mensuration

9.1 Area of triangle Std-9th Sub -Geometry 9.Mensuration Students must do activity given in the book and find the formula for area of triangle. Observe the fig alongside and write the name of triangles. Find the area of each triangle in terms of base b and height h. what do you find? H F E D A l h m B a C When triangles lies between two parallel lines with common base or congruent bases, their areas are equal. Find the parallelograms in the fig.Whether they have same area? Find the relation between area of parallelogram and area of the triangle with same or congruent base and lies between two parallel lines. Area of triangle = ½ area of parallelogram. (When their bases and heights are same.) Std-9th Sub -Geometry 9.Mensuration

Solved examples Std-9th Sub -Geometry 9.Mensuration Ex-1 Area of right angled triangle is 240 sq.m. and its base is one and half of its altitude. Find the base and altitude. Solution- Std-9th Sub -Geometry 9.Mensuration

Ex-2) The polygonal field is as shown in figure. All Ex-2) The polygonal field is as shown in figure. All measurements are given in meters. Find the area of the polygonal field. A B C D E F P Q R S 40 60 50 30 10 5 6 4 1 3 2 Sub -Geometry 9.Mensuration Std-9th

Area of polygonal field ABCDEF 1) A(∆APF) = 1/2× . . . . × . . . . Formula = 1/2× . . . .×. . . . . . . . . A B C D E F P Q R S 40 60 50 30 10 = 2)A(□ABSP) = . . . . × . . . . . . . . = . . . . . × … . = . . . . . . . . . . . . 3) A(∆ BSC) =1/2× . × . . . .. . . = 4) A(∆ DRC) = 1/2× . . .× . . . . . . 5)A(□EQRD)= . .× ( . . . . .. +. . . . .)× (Trapezium) = =. . . ..×( . .+ )× . 6) A(∆ FQE) =1/2× . . . × . . . . . . = . . .. . = . . . . . AREA OF POLOGONAL FIELD = . . + . . . +. . . . . .+ . . . . . .+ + = . . . . . . Std-9th Sub -Geometry 9.Mensuration

S P M Q R 9.2 AREA AND PERIMETER OF QUADRILATERAL Std-9th EXAMPLES . EX-1 In rhombus PQRS, the perimeter is 20 cm and diagonal QS has length 8 cm. Find the length of the second diagonal. Solution of example Perimeter of rhombus = 4 x side 20 = 4 x side side = 5 cm Diagonals PR and QS bisect each other at right angles in point M. QS=8 cm ….. given 25 = PM2 + 16 Area of rhombus = ½ x product of diagonal PM2 = 25 - 16 PM2 = 9 =1/2 x QS x PR PM = 3 cm =1/2 x 8 x 6 In right angled PMQ = 24 cm2 PQ2= PM2 +QM2 (by Pythagoras thm) diagonal PR =2x3 = 6 cm Area of rhombus is 24 cm2 . 52 = PM2 + 42 Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration A B F G D C E H EX-2) A rectangular lawn 75 m by 69 m wide , has two roads each 4 m wide running through the middle of the lawn, one parallel to the length and other parallel to breadth as shown in figure. Find the cost of gravelling the roads at the rate of Rs. 450 per sq.m. Solution- A B F G D C E H 1)Area of rectangular Road ABCD= length x breadth = 75mx 4m = 300 sq.m. 2)Area of rectangular Road EFGH= length x breadth = 60 m x 4 m = 240 sq. m 3)Area of the common part of road i.e. square ∴Area of the road to be gravelled = 300 sq.m+ 240sq.m.- 16 sq.m. = 524 sq. m. Cost of the road to be gravelled = rate x area. = 4.50 x 524 Cost of the road to be gravelled = Rs.2358 Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration Circumference – 9.3 AREA AND PERIMETER OF THE CIRCLE AND SEMICIRCLE We know formulae .. Circumference – Consider a circular wire of the copper as shown in the fig. If we cut this circular wire at point A then what will be the length of wire ? Observe the activity carefully. .. It will be equal to length of seg AB which is equal to the circumference of circle and it is given by B A Note- circumference of the circle is its perimeter too. Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration Thus, circumference is the total length of the path of the circle which is given by the formula But why is circumference is given by above formula. Why is the above formula? For this ,see the proof carefully given in the book. In this proof, It is shown that the ratio of the circumference of the circle and its diameter is always constant. This constant is denoted by Greek letter pi ( ) The perimeter of a semicircle = ½ x circumference + diameter Std-9th Sub -Geometry 9.Mensuration

9.4 Area of a regular polygon Polygon:: A simple closed figure formed by line segments is called polygon. They are named according to the number of sides they have. Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon The line segments are called sides of the polygon . Each common end point of two sides is called vertex of the polygon. Regular polygon : A polygon in which all sides and all angles are congruent is called regular polygon. Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Std-9th Sub -Geometry 9.Mensuration

Circumscribed circle of polygon - E D An important property of regular polygon is that we can draw a circle passing through its vertices called circumscribed-circle. o F C R Incribed-circle r A circle can drawn touching each side of polygon is called its incribed-circle . A P B Central point of a polygon : The inscribed and circumscribed circles of a regular polygon have the same centre. This centre is called central point of a regular polygon. In fig point o is the central point of regular polygon ABCDEG. In-radius : The length of the perpendicular drawn from central point of a polygon to any of its side is called the radius of inscribed circle of the polygon it is denoted by r In fig OP = r Circum-radius: The line segment joining central point of a polygon to any vertex is called the radius of the circumscribed circle. OA = R Std-9th Sub -Geometry 9.Mensuration

9.5 Area of n-sided regular polygon B C P O a r R In terms of in-radius Let G,A,B,C be any four vertices of regular polygon of n-sides with central point o and side a Let bisectors of ÐGAB and Ð ABC meet each other in point O. Seg OP^ side AB Let OP = r ( In-radius) OG =OA =OB =OC = R . . . . .(Circum-radius) Now, A( ∆ AOB) = ½ x AB xOP Area of regular polygon = n x A(DAOB) = n x ½ x AB x OP = n x ½ x a x r Area of regular polygon(A) = ½ x n a x r sq.units. = ½ x perimeter x In -radius ∵Perimeter of n sided polygon = na Perimeter = 2A/r Std-9th Sub -Geometry 9.Mensuration

G A B C P O Std-9th Sub -Geometry 9.Mensuration 9.6 Area of regular polygon in terms of circum-radius(R) Here in-radius OP = r Circum-radius OA = R G A B C P O a r R DOAP is right angled triangle. Std-9th Sub -Geometry 9.Mensuration

9.7 I )Area of Regular Hexagon D B C O a a Area of regular hexagon = 6 x A(DOAB) a a a a a DOAB is equilateral Area of regular hexagon sq.unit Std-9th Sub -Geometry 9.Mensuration

II) Area of regular octagon ABCDEFDH is a regular octagon G D Length of each side is ‘a ’ and its in-radius is ‘r From fig. O a a OP = r = OQ + PQ (PQ = MB . . . why ?) M Q r = OQ + MB . . . why ? H C r Where BC is diagonal of square MBNC a a P N A B Area of regular polygon Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration Ex.1) Find the area of a regular hexagon whose side is 5 cm. ( = 1.732 ) Solution : Side of regular hexagon = =64.95 ∴Area of regular hexagon is 64.95 sq.cm. Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration Ex.2) Find area of regular octagon each of its side measures 4 cm. Solution : Area of regular octagon = 32 x 2.414 = 77.248 ∴Area of regular octagon =77.248 sq.cm Ex.3) Find area of regular pentagon whose each of measures 6 cm and the radius of incribed circle is 4 cm. Solution : Area of regular polygon = ½ x n x a x r =1/2 x5 x6 x4 =60 Area of regular pentagon = 60 sq.cm. Std-9th Sub -Geometry 9.Mensuration

Std-9th Sub -Geometry 9.Mensuration Ex.4) Find the area of regular polygon of 7 sides whose each side measures 4 cm and the circum-radius is 3 cm( ) Here, a = 4 cm, R = 3 cm, n =7 Solution : =14 x 2.236 =31.304 The area of regular polygon of 7 sides =31.304 sq.cm. Std-9th Sub -Geometry 9.Mensuration

PRESENTED BY – NEHRU VIDYALAYA HINGANGAON BK. MR. RASKAR .K.B B.Sc.B.Ed. NEHRU VIDYALAYA HINGANGAON BK.

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