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Can someone help with this...

 

 Feb 13, 2019

Best Answer 

 #1
avatar+4419 
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\(\text{Draw a straight line through the centers of the top two circles ending at the rectangle's edges}\\ \text{It should be clear that the length of this line is }4r = 8mm \\ \text{and that it is the length of the horizontal side of the rectangle}\)

 

\(\text{Now draw a line from the center of the top left circle to the center of the bottom circle}\\ \text{It's length is }2r=4mm\\ \text{The vertical distance between the centers is then }4 \cos(30^\circ) = 2 \sqrt{3}mm\\ \text{You can show this by noting the triangle connecting the centers is equilateral and using a bit of trig}\)

 

\(\text{The distance from the top of the rectangle to the top left circle center is }r\\ \text{as is the distance from the bottom to the bottom circle center}\\ \text{Thus the vertical side of the rectangle has length }\\ 2r + 2\sqrt{3} mm =4+2\sqrt{3} mm\)

 

\(\text{The area is just the product of these two numbers}\\ A = (8mm)(4+2\sqrt{3}mm) = 32+16\sqrt{3} ~mm^2\)

.
 Feb 13, 2019
 #1
avatar+4419 
+1
Best Answer

\(\text{Draw a straight line through the centers of the top two circles ending at the rectangle's edges}\\ \text{It should be clear that the length of this line is }4r = 8mm \\ \text{and that it is the length of the horizontal side of the rectangle}\)

 

\(\text{Now draw a line from the center of the top left circle to the center of the bottom circle}\\ \text{It's length is }2r=4mm\\ \text{The vertical distance between the centers is then }4 \cos(30^\circ) = 2 \sqrt{3}mm\\ \text{You can show this by noting the triangle connecting the centers is equilateral and using a bit of trig}\)

 

\(\text{The distance from the top of the rectangle to the top left circle center is }r\\ \text{as is the distance from the bottom to the bottom circle center}\\ \text{Thus the vertical side of the rectangle has length }\\ 2r + 2\sqrt{3} mm =4+2\sqrt{3} mm\)

 

\(\text{The area is just the product of these two numbers}\\ A = (8mm)(4+2\sqrt{3}mm) = 32+16\sqrt{3} ~mm^2\)

Rom Feb 13, 2019

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