+0  
 
+10
475
9
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

civonamzuk  May 19, 2015

Best Answer 

 #6
avatar+18715 
+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

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

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

heureka  May 19, 2015
Sort: 

9+0 Answers

 #1
avatar+78719 
+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.......

 

CPhill  May 19, 2015
 #2
avatar+26329 
+15

I get the same as Chris:

 Shaded area

.

Alan  May 19, 2015
 #3
avatar+78719 
0

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

 

{Could we  ??? }

 

CPhill  May 19, 2015
 #4
avatar+26329 
0

Of course not - that would be ridiculous!

.

Alan  May 19, 2015
 #5
avatar+78719 
0

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

 

 

CPhill  May 19, 2015
 #6
avatar+18715 
+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+91038 
+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 

Melody  May 19, 2015
 #8
avatar+26329 
+5

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

.

Alan  May 19, 2015
 #9
avatar+91038 
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  :)

Melody  May 19, 2015

3 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details