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

Radius (r) = 1 

 

Image result for circle inscribed in a triangle

civonamzuk  May 15, 2015

Best Answer 

 #2
avatar+18712 
+15

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

Image result for circle inscribed in a triangle

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

heureka  May 15, 2015
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6+0 Answers

 #1
avatar+78551 
+5

..............................................................

CPhill  May 15, 2015
 #2
avatar+18712 
+15
Best Answer

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

Image result for circle inscribed in a triangle

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

heureka  May 15, 2015
 #3
avatar+1068 
+5

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

Radius (r) = 1 

Image result for circle inscribed in a triangle

 

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
avatar+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.

.

Alan  May 15, 2015
 #5
avatar+1068 
+5

Heureka's answer is correct!

civonamzuk  May 15, 2015
 #6
avatar+26322 
+8

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

.

Alan  May 15, 2015

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