. One equilateral triangle shaped park is having 2500V3 m2 area, in this park one man is standing at point A, So for motion A to C to B to D to C the value of displacement and distance will be respectively? (A) 100 m, 10 m (B) 4 m, 200 m (C) 50/3m , 250V3 m (D) 50v3m, 200 + 503 m 100 m u o01 c B +500 +50m.
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Answer:
please refer to the attached
Explanation:
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In given figure the area of an equilateral triangle ABC is 17320.5cm
2
. With each vertex of the triangle as a center, a circle is drawn with a radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π=3.14 and
3
=1.73205)
465240
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Solution
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Given AB=BC=AC
Area of Equilateral △ ABC = 17320.5cm
2
∴
4
3
×AB
2
=17320.5
∴ AB =200cm
Also, AB=2AD
∴ AD=100 cm =radius
Area of sector DAE + Area of sector DBF + Area of sector FCE
We know that area of sector =
360
θ
×π×r
2
=3×
360
60
×3.14×100×100
=15700cm
2
∴ Area of the shaded region = Area of equilateral triangle − Area of all sectors
=17320.5−15700
=1620.5cm
2