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13 Sep 2018, 04:57
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An equilateral triangle of area \(25\sqrt{3}\) square centimeters is made out of a piece of wire. What is the area of the largest circle that can be made by reshaping the same piece of wire?
A. \(\frac{25\pi}{\sqrt{3}}\)
B. \(\frac{225}{\pi}\)
C. 225
D. \(225\sqrt{3}\)
E. \(225\pi\)
A. \(\frac{25\pi}{\sqrt{3}}\)
B. \(\frac{225}{\pi}\)
C. 225
D. \(225\sqrt{3}\)
E. \(225\pi\)
13 Sep 2018, 07:19
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Imo B.
The area of the equilateral triangle can be found out using formula (√3/ 4 ) x (side of equilateral triangle)².
So here (√3/ 4 )*a^2=25√3.
a= 10. The perimeter of triangle is 30. This is the length of the wire.
So that would be the circumference of the circle so 30=2π ⋅ r
So r=15/π .
Area is π*r^2=225/π
The area of the equilateral triangle can be found out using formula (√3/ 4 ) x (side of equilateral triangle)².
So here (√3/ 4 )*a^2=25√3.
a= 10. The perimeter of triangle is 30. This is the length of the wire.
So that would be the circumference of the circle so 30=2π ⋅ r
So r=15/π .
Area is π*r^2=225/π
13 Sep 2018, 07:32
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Bunuel
An equilateral triangle of area \(25\sqrt{3}\) square centimeters is made out of a piece of wire. What is the area of the largest circle that can be made by reshaping the same piece of wire?
A. \(\frac{25\pi}{\sqrt{3}}\)
B. \(\frac{225}{\pi}\)
C. 225
D. \(225\sqrt{3}\)
E. \(225\pi\)
\(\frac{a^2√ 3}{4} = 25\sqrt{3}\)A. \(\frac{25\pi}{\sqrt{3}}\)
B. \(\frac{225}{\pi}\)
C. 225
D. \(225\sqrt{3}\)
E. \(225\pi\)
So, \(a^2 = 100\)
Or, \(a = 10\)
Perimeter of the equilateral triangle \(= 30\)
Now, \(2πr = 30\)
Or, \(πr = 15\)
Or, \(r = 15/π\)
So, Area of the Circle must be \(πr^2 = π*\frac{15}{π}*\frac{15}{π}\) = \(\frac{225}{\pi}\), Answer must be (B)
