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Two equilateral triangles of side 12 cm are placed one on top of another, such that a 6 pointed star is formed. If the six vertices lie on a circle, what is the area of the circle not enclosed by the star?(Give answers to the nearest sqcm.)
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- area of 1 equilateral triangle= [sqrt(3)/4]*a^2=62.28
area of three small equilateral triangle (sides=4)= 20.76
total area covered by triangle= 62.28+20.76=83.04
radius of crcle = 4*sqrt(3)
area of radius=3.14* 4*sqrt(3)*4*sqrt(3)=150.72
so, area of the circle not enclosed by the star= 150.72-83.04= 67.68(aprox 68)
ansr is 68... - 11 years agoHelpfull: Yes(24) No(1)
- 68(approx)
- 11 years agoHelpfull: Yes(4) No(0)
- all the 3 centre will lie at the very same point..
that is at the centroid of the circle..
=(2/3)*12=8
so the area of the circle= pie*r*r
=3.14*8*8
=200.96
=200 m^2 answer - 11 years agoHelpfull: Yes(1) No(3)
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- 11 years agoHelpfull: Yes(0) No(0)
- Area of larger triangle =sqrt(3)/4* side*side,
so, area=36*1.732=62.352
and area of 3 smaller triangles=1.732*3=5.196
so resultant area= 57.156 sq.cm - 11 years agoHelpfull: Yes(0) No(4)
- two equilateral should have same centroid to be formed as a star. so distance of centroid from head of a triangle = 2/3(sqrt((12^2)-(6^2)))=6.67=>radius of circle from any peak of triangle.
Now area of one equilateral triangle=sqrt(3)/4 *(12^2)=62.28
Now area of left three smaller triangle of STAR = 3*sqrt(3)/4*(4^2)=20.76{from geometry 3x=12 => x=4{where x is side smaller equilateral triangle of star }}
Total area of star= 1 equilateral tri + 3 small tri's of star=62. 28+20.76=83.04
Area of circle=3.14*(6.67)^2 = 139.69
Area of circle not enclosed in star=(139.69-83.04=56.65)57 approx.
Enjoy!! - 11 years agoHelpfull: Yes(0) No(2)
- area of 1 triangle=root3/4 a^2=root3/4*12*12 + area of 3 smal triangle 3*root3/4*12*12/3/3=12root3+36root3=48 root3=48*1.7=83
- 11 years agoHelpfull: Yes(0) No(0)
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