Fig. 12.26 depicts a racing track whose left and right ends are

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 Multiple Choice QuestionsShort Answer Type

21. Find the area of the shaded region in fig. 12.20 if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°.


Fig. 12.20
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22.

Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.


Fig. 12.21.

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23. Find the area of the shaded region in Fig. 12.22, 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.


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24.

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. 12.23. Find the area of the remaining portion of the square. 



Fig. 12.23

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 Multiple Choice QuestionsLong Answer Type

25.

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Figure. Find the area of the design


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 Multiple Choice QuestionsShort Answer Type

26. In Fig. 12.25, 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.


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27.

Fig. 12.26 depicts a racing track whose left and right ends are semicircular.


Fig. 12.26

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.
(ii) the area of the track.


We have,
The distance between the two inner parallel line segments = 60 m.
Length of each parallel lines = 106 m.
Width of track = 10 m
(i) The distance around the track along its inner edge = sum of length of inner parallel lines + 2 (semi perimeter of inner circular ends)

= (106 + 106) + 2 open parentheses straight pi space straight x space 30 close parentheses
equals space 212 plus 2 space straight x space 22 over 7 space straight x space 30 space equals space 212 space plus space 188.57 space equals space 405.57 space cm
(ii) OD = 30 m and AD = 10 m.
Oa = OD + DA = 30 + 10 = 40 m
Area of the track = Area of rectangle ABCD + Area of rectangle EFGH + 2 (Area of semicircle with radius 40 m - Area of semi-circle with radius 30)

= 106 x 10 + 106 x 10 + 2 open square brackets 1 half straight pi left parenthesis 40 right parenthesis squared minus 1 half straight pi left parenthesis 30 right parenthesis squared close square brackets
equals space 1060 space plus space 1060 space plus space straight pi space left square bracket left parenthesis 40 right parenthesis squared minus left parenthesis 30 right parenthesis squared right square bracket
equals space 2120 space plus space 22 over 7 space left parenthesis 40 plus 30 right parenthesis left parenthesis 40 minus 30 right parenthesis
equals space 2120 space plus space 22 over 7 space straight x space 70 space straight x space 10
equals space 2120 space plus space 2200 space equals space 4320 space straight m squared.
Hence, the area of track is 4320 m2 

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28.

In Fig. 12.27, 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.

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 Multiple Choice QuestionsLong Answer Type

29.

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. 12.28). Find the area of the shaded region (Use straight pi = 3.14 and square root of 3 equals1.73205)



Fig. 12.28

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 Multiple Choice QuestionsShort Answer Type

30.

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.



Fig. 12.29

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