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# Homework due tomorrow

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Can someone help with this... Feb 13, 2019

#1
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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
+1

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

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

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

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

Rom Feb 13, 2019