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16. Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each. It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors. Area of the sector is:-                                                                                   And, area of the triangle:-                                         Hence the area of the designed region is :                                                              ...

15. In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. Consider  ABC,                                                                                                                                           Area of triangle is :                                                                                             Now, area of sector is :                                                                                         And area of...

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If , find the area of the shaded region. Area of the shaded region is   =   Area of larger sector   -   Area of smaller sector                                                                                                                                   Hence area of shaded region is

13. In Fig, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use ) In the given figure we need to find the radius of the circle:-    Consider OAB,                                                                                                                            Thus area of quadrant:-                                                                                                     Also, area of the square is  :      Area of shaded region is :

12.  In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the For area of shaded region we need to find area of the triangle. Area of triangle is :-                                                                           Hence area of shaded region is  =  Area of quadrant   -  Area of triangle

12. In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the The quadrant OACB is a sector with angle 90o and radius 3.5 cm. Thus the area of the quadrant is:-                                                                                                    Hence the area of the quadrant is       .

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief. Since one side of the square has 3 circles, thus the side of the square is 42 cm.          Area of the square :     And, area of a circle  :                                                                                              Hence the area of the remaining portion is  :

10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use and ) Area of an equilateral triangle is:-                                                                                                                                                                                 Now, consider the sector  :- Angle of sector is 60o and radius is 100 cm.  Thus the area of the sector:-                                                                              ...

9.  In Fig, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. Firstly, area of the smaller circle is :                                                                                                                                                                                             Now, area of :-                                                    or                                              And, area of the bigger semicircle is :         ...

8. Fig. depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(ii) the area of the track.

The area of track  =   Area of outer structure   -   Area of inner structure. Area of outer structure is: =   Area of square  + Area of 2 semicircles                                               And, area of inner structure: =  Area of inner square  +   Area of 2 inner semicircles                                                 Thus the area of the track is  :                                 ...

8.  Fig. depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(i) the distance around the track along its inner edge

The distance around the track is Length of two straight lines +  Length of two arcs. Length of the arc is  -                                                                      Thus the length of the inner track is :

7. In Fig, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region. It is clear from the figure that area of all sectors are equal (due to symmetry). Also, the angle of sector is 90o and radius is 7 cm. Thus area of sector is :-                                                         or                                                   And, the area of square is :        Hence area of shaded region is  :                                                          ...

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design. Assume the centre of the circle to be point C and AD as the median of the equilateral triangle. Then we can write:-                                                                              or                                                       Thus                                                  Consider  ABD,                                                      or                       ...

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig.  Find the area of the remaining portion of the square. Consider the quadrant in the given figure:-  We have an angle of the sector as 90o and radius 1 cm.     Thus the area of the quadrant is:-                                                       or                                                    And the area of the square is :                                                                                                             And,...

4. Find the area of the shaded region in Fig, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. Area of the shaded region is given by  =   Area of triangle  +  Area of the circle  -  Area of the sector  Area of the sector is : -                                                     or                                                And, the area of the triangle is :                                                And, the area of the circle is : or                                         or ...

2. Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and . The area of shaded region can be easily found by using formula of area of sector. Area of shaded region is given by  :    Area of sector  OAFC   -   Area of sector OBED

1. Find the area of the shaded region in Fig, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. We know that RPQ  is 90o as ROQ is the diameter. RQ can be found using Pythagoras theorem.                                               or                                           or                                           Now, the area of the shaded region is given by : = Area of semicircle   -   Area of PQR Area of semicircle is :-                                               or         ...
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