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# The dotted diagonal AC (Fig. 42) has the length of twice the radius. Find the area of the emblem only.

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The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

civonamzuk  May 15, 2015

#2
+18712
+15

The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

$$\\A_{circle}=\pi r^2 \\ A_{square}=(r \sqrt{2})^2=2r^2\\ A_{goblet}=\frac{ A_{circle} - A_{square} }{4} = \frac{\pi r^2 - 2r^2}{4}\\\\ A = 4\cdot \left[ A_{quadrant}-2\cdot A_{goblet}- A_{triangle} \right]\\\\ A = 4\cdot \left[ \frac{\pi r^2 }{4} -2\cdot \left( \frac{\pi r^2 - 2r^2}{4} \right) - \frac{ \left( r\sqrt{2}-r \right)^2 }{2} \right]\\\\ A = \pi r^2 -2\cdot \left(\pi r^2 - 2r^2\right) - 2\cdot \left( r\sqrt{2}-r \right)^2\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 2\cdot \left( r\sqrt{2}-r \right)^2\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 2\cdot \left(2r^2-2\sqrt{2}r^2 +r^2 \right)\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 4r^2 +4\sqrt{2}r^2-2r^2\\\\ A = -\pi r^2 +4\sqrt{2}r^2-2r^2\\\\ A=r^2\cdot (4\sqrt{2}-2-\pi)$$

heureka  May 15, 2015
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#1
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..............................................................

CPhill  May 15, 2015
#2
+18712
+15

The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

$$\\A_{circle}=\pi r^2 \\ A_{square}=(r \sqrt{2})^2=2r^2\\ A_{goblet}=\frac{ A_{circle} - A_{square} }{4} = \frac{\pi r^2 - 2r^2}{4}\\\\ A = 4\cdot \left[ A_{quadrant}-2\cdot A_{goblet}- A_{triangle} \right]\\\\ A = 4\cdot \left[ \frac{\pi r^2 }{4} -2\cdot \left( \frac{\pi r^2 - 2r^2}{4} \right) - \frac{ \left( r\sqrt{2}-r \right)^2 }{2} \right]\\\\ A = \pi r^2 -2\cdot \left(\pi r^2 - 2r^2\right) - 2\cdot \left( r\sqrt{2}-r \right)^2\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 2\cdot \left( r\sqrt{2}-r \right)^2\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 2\cdot \left(2r^2-2\sqrt{2}r^2 +r^2 \right)\\\\ A = \pi r^2 -2\pi r^2 + 4r^2 - 4r^2 +4\sqrt{2}r^2-2r^2\\\\ A = -\pi r^2 +4\sqrt{2}r^2-2r^2\\\\ A=r^2\cdot (4\sqrt{2}-2-\pi)$$

heureka  May 15, 2015
#3
+1068
+5

The dotted diagonal AC (Fig. 42)  has the length of twice the radius. Find the area of the emblem only.

AC = 2r       r = 1

Area of the square is:  (sqrt(2))2  = 2.000u2

Triangles:  (lkC + oAf) = (sqrt(2) -1)2 = 0.171572875253809862u2

Half circle area is:  r2pi/2 = 1.5707963267948966u2

(2.000u2 - 1.570796326794896u2 - 0.17157287525380986u2)*2= 0.515261595902587u2

civonamzuk  May 15, 2015
#4
+26322
+5

You don't specify the value of the radius in the question civonamzuk, but if r = 1 then heureka's answer evaluates to the same as the one you give.

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Alan  May 15, 2015
#5
+1068
+5

civonamzuk  May 15, 2015
#6
+26322
+8

Really?  That's a new one on me!  (Forgive the pun!)

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Alan  May 15, 2015

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