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Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal to
- a)7/6 cm
- b)5/6 cm
- c)½ cm
- d)5/11 cm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Four equal discs are placed such that each one touches two others. If ...
Let the radius be r
Area of empty space = Area of the square – 4 × Area of quadrant
⇒ Side of square = 2 × radius = 2r
⇒ 150/847 = Side2 – 4 × ¼ × π r2
⇒ 150/847 = (2r)2 – 22/7 r2
⇒ 150/847 = r2 (4 – 22/7)
⇒ 150/847 = 6r2/7
⇒ r2 = 25/121
∴ radius, r = 5/11 cm
Most Upvoted Answer
Four equal discs are placed such that each one touches two others. If ...
Let the radius of each disc be $r$. We can draw a diagram as shown:
[asy] unitsize(1cm); pair A,B,C,D,E,F; A=(0,0); B=(2,0); C=(1,sqrt(3)); D=(1,0); E=(D+C)/2; F=(D+B)/2; draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); label("$r$",(A--D),S); label("$r$",(B--D),S); label("$r$",(C--D),N); label("$r$",(C--E),NW); label("$r$",(B--F),NE); label("$r$",(A--D),S); label("$r$",(B--D),S); label("$r$",(C--D),N); label("$r$",(C--E),NW); label("$r$",(B--F),NE); [/asy]
The area enclosed by the discs is equal to the area of an equilateral triangle with side length $2r$ minus the area of three sectors of the circles with radius $r$ and central angles $120^\circ$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}(2r)^2=\frac{3\sqrt{3}}{2}r^2$, and the area of each sector is $\frac{120}{360}\pi r^2=\frac{1}{3}\pi r^2$. Therefore, the area enclosed by the discs is $\frac{3\sqrt{3}}{2}r^2-3\cdot\frac{1}{3}\pi r^2=\left(\frac{3\sqrt{3}}{2}-\pi\right)r^2$. We are given that this area is $\frac{150}{847}$, so we have the equation $\left(\frac{3\sqrt{3}}{2}-\pi\right)r^2=\frac{150}{847}$. Solving for $r$ gives $r^2=\frac{\frac{150}{847}}{\frac{3\sqrt{3}}{2}-\pi}=\frac{150}{847}\cdot\frac{2}{\frac{6\sqrt{3}}{2}-2\pi}=\frac{300}{847\left(\sqrt{3}-\pi\right)}$. Multiplying both the numerator and denominator of the fraction by the conjugate of the denominator gives $r^2=\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}$. We are asked to find $r$, so we take the square root of both sides. Since $r$ must be positive, we take the positive square root: $r=\sqrt{\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}}=\boxed{\textbf{(D)}\ \sqrt{\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}}}$.
[asy] unitsize(1cm); pair A,B,C,D,E,F; A=(0,0); B=(2,0); C=(1,sqrt(3)); D=(1,0); E=(D+C)/2; F=(D+B)/2; draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); label("$r$",(A--D),S); label("$r$",(B--D),S); label("$r$",(C--D),N); label("$r$",(C--E),NW); label("$r$",(B--F),NE); label("$r$",(A--D),S); label("$r$",(B--D),S); label("$r$",(C--D),N); label("$r$",(C--E),NW); label("$r$",(B--F),NE); [/asy]
The area enclosed by the discs is equal to the area of an equilateral triangle with side length $2r$ minus the area of three sectors of the circles with radius $r$ and central angles $120^\circ$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}(2r)^2=\frac{3\sqrt{3}}{2}r^2$, and the area of each sector is $\frac{120}{360}\pi r^2=\frac{1}{3}\pi r^2$. Therefore, the area enclosed by the discs is $\frac{3\sqrt{3}}{2}r^2-3\cdot\frac{1}{3}\pi r^2=\left(\frac{3\sqrt{3}}{2}-\pi\right)r^2$. We are given that this area is $\frac{150}{847}$, so we have the equation $\left(\frac{3\sqrt{3}}{2}-\pi\right)r^2=\frac{150}{847}$. Solving for $r$ gives $r^2=\frac{\frac{150}{847}}{\frac{3\sqrt{3}}{2}-\pi}=\frac{150}{847}\cdot\frac{2}{\frac{6\sqrt{3}}{2}-2\pi}=\frac{300}{847\left(\sqrt{3}-\pi\right)}$. Multiplying both the numerator and denominator of the fraction by the conjugate of the denominator gives $r^2=\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}$. We are asked to find $r$, so we take the square root of both sides. Since $r$ must be positive, we take the positive square root: $r=\sqrt{\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}}=\boxed{\textbf{(D)}\ \sqrt{\frac{300\left(\sqrt{3}+\pi\right)}{2519-847\pi}}}$.
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Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer?
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Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer?.
Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer?.
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Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is 150/847 square centimeter, then the radius of each disc is equal toa)7/6 cmb)5/6 cmc)½ cmd)5/11 cmCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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