1. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭ Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std- 9 Subject - Geometry

2. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ ´ÖÖ ¬µÖ ´Ö ‡µÖ¢ÖÖ 9 ¾Öß ×¾ÖÂÖµÖ - Öם֟ Ö 9.Mensurati on ‡Ó Ö ÏÖ ß ‡µ Ö¢ ÖÖ × ¾ Ö ֵ Ö 9 ¾ Öß Ö×ÖŸÖ (³Öæ×´ÖŸÖß ) ‘Ö™ü ú- 9.Mensurati on

3. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std-9 th Sub - Geometry 9.Mensurati on

4. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION 9.1) AREA OF TRIANGLE 9.2) AREA & PERIMETER OF A QUADRILAERAL 9.3)AREA & PERIMETER OF THE CIRCLE & SEMICIRCLE 9.7) PARTICULAR CASES 1. REGULAR HEXAGON 2. REGULAR OCTAGON 9.7) PARTICULAR CASES 1. REGULAR HEXAGON 2. REGULAR OCTAGON 9.6) TO FIND THE AREA OF THE REGULAR POLYGON IN TERMS OF RADIUS OF CIRCUMSCRIBED CIRCLE 9.6) TO FIND THE AREA OF THE REGULAR POLYGON IN TERMS OF RADIUS OF CIRCUMSCRIBED CIRCLE 9.5) AREA OF n-SIDED REGULAR POLYGON 9.4) AREA OF REGULAR POLYGON  INTRODUCTION Std-9 th Sub - Geometry 9.Mensurati on

5. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION MENSURATION - INTRODUCTION In our daily life we come across many situations like- 1)Fencing a circular garden with barbed wire. We have to consider the total length of wire required,number of turns of wire and supporting poles etc and then we calculate the expenditure. Std-9 th Sub - Geometry 9.Mensurati on

6. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION MENSURATION - INTRODUCTION 2) Painting a walls of drawing room. we must know the area, colour required for unit area and the rate of painting. Then we can calculate the expenditure. Std-9 th Sub - Geometry 9.Mensurati on

7. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION MENSURATION - INTRODUCTION 3)For purchasing land or plot Std-9 th Sub - Geometry 9.Mensurati on it is essential to know the area and rate

8. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION MENSURATION - INTRODUCTION 4)While tiling the floor we must know the area of floor and also the area of each tile. Std-9 th Sub - Geometry 9.Mensurati on

9. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION 5)To decide the number of bricks to built a wall, we must know the volume of wall and a brick. Std-9 th Sub - Geometry 9.Mensurati on

10. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ MENSURATION - INTRODUCTION In above examples we need to measure area, perimeter and volumes. Such a measurements come under the topics mensuration. Std-9 th Sub - Geometry 9.Mensurati on

11. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Triangular region Triangle Observe the adjoining figure. 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. Fig.1 Fig.2 Std-9 th Sub - Geometry 9.Mensurati on

12. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ AREA OF RECTANGULAR REGION. UNITS OF AREA 1 1 Consider a square of side 1 cm. Its area = 1 cm × 1 cm area = 1 sq. cm. Like sq.cm. other units of area are sq.m. and sq.km. Area of square having side 1 cm is 1 sq.cm. area. 6 cm 4 cm 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 Here we get formula, area of rectangle = length× breadth Now, length ×breadth=6 cm ×4cm = 24.sq.cm. = area of rectangle. Thus, we get unit of area sq.cm A BC D Therefore area of rectangle ABCD is 24 sq.cm. Std-9 th Sub - Geometry 9.Mensurati on

13. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ h ca b Triangle Perimeter Area Revision – Some Formulae. Sr.No. P = a+b+c Figure a a a Equilateral Triangle P = 3a 1 2 Std-9 th Sub - Geometry 9.Mensurati on

14. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ h ca b Triangle PerimeterArea Revision – Some Formulae. Sr.No. P = a+b+c Figure a a a Equilateral Triangle P = 3a 1 2 Std-9 th Sub - Geometry 9.Mensurati on

15. v ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ PerimeterArea Revision – Some Formulae. Sr.No. P = 4a Figure 3 4 a a a a Square b l rectangle Std-9 th Sub - Geometry 9.Mensurati on

16. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ PerimeterArea Revision – Some Formulae. Sr.No. P =2(a+b) Figure 5 6 parallelogram a b b a d a P=a+b+c+d h h Trapezium Std-9 th Sub - Geometry 9.Mensurati on

17. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ a PerimeterArea Revision – Some Formulae. Sr.No. P =4a Figure 7 8 Rhombus a a O r rr 9 semicircle circle a Std-9 th Sub - Geometry 9.Mensurati on

18. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 9.1 Area of triangle Area of triangles lies between two parallel lines with common base FED A H B C l m When triangles lies between two parallel lines with common base or congruent bases, their areas are equal. h a 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? 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-9 th Sub - Geometry 9.Mensurati on

19. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Solved examples 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-9 th Sub - Geometry 9.Mensurati on

20. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 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 3050103060 1 2 3 4 5 6 Std-9 th Sub - Geometry 9.Mensurati on

21. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Area of polygonal field ABCDEF 1) A( ∆ APF) = 1/2 × ×.... Formula = 1/2 ×....×......... 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 =.. +... +......+......+ + =...... =..... ×(..+ )×. A B C D E F P Q R S 40 60 50 3050103060 Std-9 th Sub - Geometry 9.Mensurati on

22. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ A B FG D C EH 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. 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 Solution- Std-9 th Sub - Geometry 9.Mensurati on

23. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 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. A B It will be equal to length of seg AB which is equal to the circumference of circle and it is given by Note- circumference of the circle is its perimeter too. Std-9 th Sub - Geometry 9.Mensurati on

24. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 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-9 th Sub - Geometry 9.Mensurati on

25. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ The line segments are called sides of the polygon. 9.4 Area of a regular polygon Polygon: : A simple closed figure formed by line segments is called polygon. Regular polygon : A polygon in which all sides and all angles are congruent is called regular polygon. Triangle Quadrilateral Pentagon Pentagon Hexagon Hexagon Heptagon Heptagon Octagon Octagon They are named according to the number of sides they have. Each common end point of two sides is called vertex of the polygon. Std-9 th Sub - Geometry 9.Mensurati on

26. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 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. An important property of regular polygon is that we can draw a circle passing through its vertices called circumscribed-circle. A circle can drawn touching each side of polygon is called its incribed-circle. A B C DE F R o P r Circumscribed circle of polygon - Incribed-circle 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-9 th Sub - Geometry 9.Mensurati on

27. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 9.5 Area of n-sided regular polygon G AB C P O aa a r R R RR Let OP = r ( In-radius) OG =OA =OB =OC = R.....(Circum-radius) 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  GAB and AABC meet each other in point O. Seg OP  side AB Now, A( ∆ AOB) = ½ x AB xOP Area of regular polygon = n x A(  AOB) = n x ½ x AB x OP = n x ½ x a x r ( ∵ Perimeter of n sided polygon = na) Area of regular polygon(A) = ½ x n a x r sq.units. = ½ x perimeter x In -radius Perimeter = 2Area/r Std-9 th Sub - Geometry 9.Mensurati on

28. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 9.6 Area of regular polygon in terms of circum-radius(R) G AB C P O aa a r R R RR Circum-radius OA = R Here in-radius OP = r  OAP is right angled triangle. Std-9 th Sub - Geometry 9.Mensurati on

29. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ  OAB is equilateral Area of regular hexagon = 6 x A(  OAB) Area of regular hexagon a aa a 9.7 I )Area of Regular Hexagon a a a a Std-9 th Sub - Geometry 9.Mensurati on

30. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ ABCDEFDH is a regular octagon Length of each side is ‘a ’ and its in-radius is ‘r From fig. OP = r = OQ + PQ (PQ = MB... why ?) r = OQ + MB Where BC is diagonal of square MBNC II) Area of regular octagon Area of regular polygon a P a a a a a a A B C D EF G H L M Q O N a r Std-9 th Sub - Geometry 9.Mensurati on

31. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Ex.1) Find the area of a regular hexagon whose side is 5 cm. Side of regular hexagon = =64.95 ∴ Area of regular hexagon is 64.95 sq.cm. Solution : Std-9 th Sub - Geometry 9.Mensurati on

32. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ 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 Ex.3) Find area of regular pentagon whose each of measures 6 cm and the radius of incribed circle is 4 cm. Area of regular polygon = ½ x n x a x r =1/2 x5 x6 x4 =60 Area of regular pentagon = 60 sq.cm. ∴ Area of regular octagon =77.248 sq.cm Std-9 th Sub - Geometry 9.Mensurati on

33. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Ex.4) Find the area of regular polygon of 7 sides whose each side measures 4 cm and the circum-radius is 3 cm( ) Solution : =14 x 2.236 =31.304 The area of regular polygon of 7 sides =31.304 sq.cm. Here, a = 4 cm, R = 3 cm, n =7 Std-9 th Sub - Geometry 9.Mensurati on

34. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Do the activity to verify the formula- Sum of angles of a polygon =180 x (n-2) Std-9 th Sub - Geometry 9.Mensurati on

35. Q-2 ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭ Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std- 9 Subject - Geometry ❀ TEST ❀ 20 marks Q-1.Attempt any two 2 1) Find area of triangle whose height is 3 cm and base is 4 cm? 2) If the side square is 8 cm then find then find the length of its diagonal 3) If radius of circle is 7 cm then find ts area. Q-2.Attempt any two 4 1) If the length of rectangular plot is twice of its breadth and its perimeter is 420 cm then find length and breadth 2) Find the perimeter of semicircle whose radius of is 14 cm 3) If the area of equilateral triangle is sq.cm then find its side. 9.Mensurati on

36. Q-2 ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭ Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std- 9 Subject - Geometry Q-3.Attempt any two 6 1)Redii of two circles are 3m and 4m respectively.Find the radius of circle having area equal to the sum of area these two circles. 2) A regular hexagon is incribed in a circle of radius 6 cm. Find the area of shaded portion. 3) The length of rectangular hall is 5 m more than its breadth. The area of hall is 750 m 2. Find the perimeter of hall. A B CD E F 9.Mensurati on

37. Q-2 ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭ Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std- 9 Subject - Geometry Q-4. Attempt any two 8 1) Find the length of side of an equilateral triangle △ ABC incribed in a circle of radius 4 cm. 2) Two parallel chords AB and CD are drawn on the same side of centre O of a circle. Radius of a circle is 65 m, length of chord are 112 m and 126 m. Find the area of quadrilateral  ABCD. 3) The length of sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of triangle. A B C AB C D O 9.Mensurati on

38. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std-9 th Sub - Geometry 9.Mensurati on prepared by-

39. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ Std-9 th Sub - Geometry 9.Mensurati on

40. ¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓãÖÖ, ú´ÖÔ¾Ö߸ü ×¾ÖªÖ¯ÖϲÖÖê׬֭Öß, ´Ö¬µÖ ×¾Ö³Ö֐Ö, ÃÖÖŸÖÖ¸üÖ ´ÖÖ ¬µÖ ´Ö ‡µÖ¢ÖÖ 9 ¾Öß ×¾ÖÂÖµÖ - Öם֟ Ö ‡Ó Ö ÏÖ ß ‡µ Ö¢ ÖÖ × ¾ Ö ֵ Ö 9 ¾ Öß Ö ם ÖŸÖ 9.Mensuration Geometry