Latest TCS Aptitude Question SOLUTION: 1.There is a circle with 2 triangles inscribed in it. in opposite direction making a star. The triangle is equilateral of side 12. u have to tell the area of the remaining portion

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(#M40021743) TCS Logical Reasoning question Keep an EYE

1.There is a circle with 2 triangles inscribed in it. in opposite direction making a star. The triangle is equilateral of side 12. u have to tell the area of the remaining portion of the circle. ans

Asked In TCS bhakti (11 years ago)

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Ankita Ghosh  (11 years ago)

from formula-
(sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude
so altitude is=6sqrt(3)
now the small triangle below that altitude having the altitude is
(sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude
sp altitude=2sqrt(3)
so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
so radius is = 4sqrt(3)
remaining area= 22.7 * r * r – 12* (sqrt(3)/4) * a * a where a=4
so ans is=67.58

Is this solution Helpfull? Yes (39) | No (8)  

Anton Menezes  (11 years ago)

(Draw a figure with two equilateral triangles in opposite directions in a circle, such that internally it contains 12 smaller equilateral triangles of side 4 cm each)

The required area = Area of circle – Area occupied by the equilateral triangles ------- eq (1)
As seen in the figure, the overlapping equilateral triangles of side 12cm can be divided into 12 smaller equilateral triangles of side 4cm each.
Also, from the figure, Radius of the circle = 2 x height of the smaller equilateral triangle.
Height of equilateral triangle = (√3/2)a
In the smaller equilateral triangle, the side of the triangle = 4 cm.
Height of the smaller equilateral triangle = (√3/2) x 4 = 2√3 cm.
Hence, radius of the circle = 2 x 2√3 = 4√3 cm.
Area of circle = π r2 = π x (4√3)2 = 48 π cm2. ------- eq (2)

Area occupied by the equilateral triangles = Area of 12 smaller equilateral triangles.
Area of equilateral triangle = (√3/4)a2 .
Here, a = 4cm; Area of smaller equilateral triangle = (√3/4) x 42 = 4√3 cm2.
Area of 12 small equilateral triangles = 12 x 4√3 = 48√3 cm2. ------- eq (3)
Substituting eq (2) and (3) in (1):
Hence, the required area = 48 π - 48√3 = 48(π-√3) cm2.

Is this solution Helpfull? Yes (11) | No (1)  

sambasiva rao  (11 years ago)

Area of a triangle is= (1/2)*b*h eq----->1
b(Base)=12
h(height)=?
we know that area of equilateral is =sqrt(3)*(b^2)/4 eq---->2
there fore area= (1.73*144)/4
= 62.28

equilate eq(1) and eq(2) we get
62.8 = 6*h

h = 10.38

we know that h= radius of circle
r = 10.38
so area of circle is = pie* r^2
= (22/7)*(10.38*10.38)
= 338.63

remaining area is = area of cirle - (2*area of triangle)
= 338.63 - 2*62.8
= 213
answer is----- 213

Is this solution Helpfull? Yes (8) | No (16)  

Pravin  (11 years ago)

area of equilateral triangle=b*h.

here height of triangle can be find by pythagoras theorem.
h=sqrt(144-36)=6*sqrt(3)

therefore area of triangle=6*sqrt(3)*12=124.704
therefore area of two eq triangles=124.7047*2=249.408

area of circle=pi*r(sq).
here radius=height of eq triangle.
so area of circle=339.428

so the remaining area in circle=area of circle-area of two eq triangle
=> 339.428 - 249.408 = 90.02057

Is this solution Helpfull? Yes (1) | No (12)  

smith kumar saha  (11 years ago)

320.52 is the answer

Is this solution Helpfull? Yes | No (8)  

sambasiva rao  (11 years ago)

Area of a equilateral triangle is= sqrt(3)*squre(base)/4
area of triange is = sqrt(3)*12*12/4 = 62.28

semi perimeter s = (a+b+c)/2
s= (12+12+12)/2 = 18
we know that if a equilateral triangle inscribed in a then radius of that circle is = semi perimeter of triangle/sqrt(3)
there fore radius(r) = 18/sqrt(3) = 6* sqrt(3) = 10.38
so area of circle is = pie* r^2
= (22/7)*(10.38*10.38)
= 338.63

2 equilateral triangles bisects means they form another 3 small equilateral triangles with base 12/3= 4

so area of 3 small triangles = 3*(sqrt(3)*4*4/4) = 20.76

remaining area is = area of cirle - (area of triangle) - 3*(area of small triangle)
= 338.63 - 62.8 - 20.76
= 255.07
answer is----- 225.07

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Vijay  (11 years ago)

Ans: 32*pi

we need to find out sum of the areas of 6 small archs with radius 4 and angle 120, in that cirlce.

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ramya  (9 years ago)

anton can you please xplain how you got 12 equilateral small triangles? bcoz i get only 6 small triangles

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