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+10
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avatar+1068 

From the figure shown below, DE is the diameter of circle A and BC is the radius of circle B. If DE = 60 cm and AC = 10 cm, find the area of the shaded region.

 

025-overlapping-circles.gif

 May 19, 2015

Best Answer 

 #6
avatar+21819 
+15

From the figure shown below, DE is the diameter of circle A and BC is the radius of circle B. If DE = 60 cm and AC = 10 cm, find the area of the shaded region

$$h=\overline{AC}=10~\rm{cm}
\qquad r_{A} = \frac{ \overline{DE} }{2}= 30~\rm{cm}
\qquad r_{B} = \overline{CB} \\\\
r_A^2 = h\cdot(2\cdot r_B-h)\qquad \Rightarrow\qquad
r_B = \frac{h+\dfrac{r_A^2}{h} }{2}=\frac {10+90}{2}= 50 ~\rm{cm}\\\\
\small{\text{Angle DBE $=B
\qquad B\ensurement{^{\circ}} = 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} $}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{B\ensurement{^{\circ}} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} } - r_A\cdot(r_B - h)
\right]
$}}$$

$$\small{ A_{Blue}= \pi\cdot r_B^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot r_A^2}{2}+ r_A\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{30}{50} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot 30^2}{2}+ 30\cdot(50 - 10)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2\cdot
0.7951672353008667 - \pi \frac{\cdot 30^2}{2}+ 1200\right]
$}} \\\\
\small{ A_{Blue}= 6031.5121678758663628914 ~\rm{cm^2}
$}}\\\\
A_{Blue} \approx 6031.5122 ~\rm{cm^2}$$

.
 May 19, 2015
 #1
avatar+98044 
+15

I'll try this one......!!!

We need to find the radius of the large circle.BC = BD = BE

 

Now....DAC is a right triangle with DC = √[30^2 + 10^2] = 10√10 cm

 

And by the Law of Sines   ....   sin90 / 10√10   =  sin DCA /30  .....sin-1 (3/√10) = DCA = DCB = about 71.565°......and in triangle  DBC, DB = CB....so angle DBC = 180 - (2)71.565 = about 36.87°

 

And by the Law of Sines, again ........  10√10 / sin 36.87 = DB / sin 71.565

So DB = (10√10) sin71.565 / sin 36.87 =  50 cm

So the area of the big circle is 2500pi cm^2

 

And 1/2 of the area of the small circle =  (1/2)(30)^2 pi  =  450pi cm^2

 

And to find the area bound by DCEA, we have area of sector BDE - area of triangle BDE =  (1/2)(50)^2 (1.287002217586568 ) - (1/2)(60)(40)   =  1608.75277198321 -1200  = 408.75277198321 cm^2

 

So....the shaded area = area of the big circle - (1/2) area of the small circle - area of DCEA =

[ 2500pi - 450pi - 408.75277198321 )  =  about 6031.51 cm^2

 

Don't know if it's correct or not.......

 

 May 19, 2015
 #2
avatar+27529 
+15

I get the same as Chris:

 Shaded area

.

 May 19, 2015
 #3
avatar+98044 
0

Well.......we couldn't both be wrong.....!!!!!

 

{Could we  ??? }

 

 May 19, 2015
 #4
avatar+27529 
0

Of course not - that would be ridiculous!

.

 May 19, 2015
 #5
avatar+98044 
0

LOL!!!....remember......in olden days...just because two people agreed that the Earth was flat.....didn't necessarily make it so........

 

 

 May 19, 2015
 #6
avatar+21819 
+15
Best Answer

From the figure shown below, DE is the diameter of circle A and BC is the radius of circle B. If DE = 60 cm and AC = 10 cm, find the area of the shaded region

$$h=\overline{AC}=10~\rm{cm}
\qquad r_{A} = \frac{ \overline{DE} }{2}= 30~\rm{cm}
\qquad r_{B} = \overline{CB} \\\\
r_A^2 = h\cdot(2\cdot r_B-h)\qquad \Rightarrow\qquad
r_B = \frac{h+\dfrac{r_A^2}{h} }{2}=\frac {10+90}{2}= 50 ~\rm{cm}\\\\
\small{\text{Angle DBE $=B
\qquad B\ensurement{^{\circ}} = 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} $}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{B\ensurement{^{\circ}} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ 2\cdot\arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 360\ensurement{^{\circ}} } - \frac{1}{2}\cdot (2\cdot r_A)\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot r_B^2 - \pi \frac{\cdot r_A^2}{2}
-\left[
\pi \cdot r_B^2 \cdot \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} } - r_A\cdot(r_B - h)
\right]
$}}$$

$$\small{ A_{Blue}= \pi\cdot r_B^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{r_A}{r_B} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot r_A^2}{2}+ r_A\cdot(r_B - h)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2
\left(
1- \frac{ \arcsin{ \left( \dfrac{30}{50} \right)} } { 180\ensurement{^{\circ}} }
\right)
- \pi \frac{\cdot 30^2}{2}+ 30\cdot(50 - 10)
\right]
$}} \\\\
\small{ A_{Blue}= \pi\cdot 50^2\cdot
0.7951672353008667 - \pi \frac{\cdot 30^2}{2}+ 1200\right]
$}} \\\\
\small{ A_{Blue}= 6031.5121678758663628914 ~\rm{cm^2}
$}}\\\\
A_{Blue} \approx 6031.5122 ~\rm{cm^2}$$

heureka May 19, 2015
 #7
avatar+99122 
+10

Are three wise men enough to convince you Chris ?     :)

I have added this to the "Great Answers To Learn From" sticky thread    Thanks Guys 

 May 19, 2015
 #8
avatar+27529 
+5

It could just be that heureka is a member of the Flat-Earth Society like Chris and myself!

.

 May 19, 2015
 #9
avatar+99122 
0

Yes Alan, that is quite possible  -  I am in The Flat Earth Society too!

 

It seems that the youngsters have their 'Music Society' and us oldies (no offence Heureka) have the 'Flat Earth Society'  

 

$$Zero$$     oh!  I mean    $$Ze^gro^{es}$$     could have fun with this  :)

 May 19, 2015

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