The area of an equilateral triangle ABC is 17320.5 cm2 . With ea

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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.

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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

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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


Let the side of the equilateral triangle be 'a' cm, then,


Let the side of the equilateral triangle be 'a' cm, then,Area of tria
Area of triangle

equals space fraction numerator square root of 3 over denominator 4 end fraction left parenthesis side right parenthesis squared
rightwards double arrow space space 49 square root of 3 equals fraction numerator square root of 3 over denominator 4 end fraction straight x space straight a squared
rightwards double arrow space straight a squared space equals space 49 space straight x space 4
rightwards double arrow space straight a space equals space 14 space cm
It is given that radius is equal to half the length of the side

i.e.          r space equals space 14 over 2 equals 7 space c m
It is given that
Area of A r e a space o f space increment space left parenthesis A B C right parenthesis space equals space 49 square root of 3 space c m squared
a r e a space o f space 3 space s e c t o r s space equals space 3 space x space πr squared space straight x space straight theta over 360
space space space space space space space space space space space space space space space space space space space space space space space space equals space 3 space straight x space 22 over 7 straight x space 7 space straight x space 7 space straight x space 60 over 360
space space space space space space space space space space space space space space space space space space space space space space space space space equals space 77 space space cm squared
Hence, required area
= Area of triangle - Area of 3 sectors
equals open parentheses 49 square root of 3 minus 77 close parentheses space cm squared equals space 7.77 space cm squared
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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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