+0

# 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

+10
2
1985
9
+1696

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.

May 19, 2015

#6
+26319
+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
+124524
+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
+33053
+15

I get the same as Chris:

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May 19, 2015
#3
+124524
0

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

{Could we  ??? }

May 19, 2015
#4
+33053
0

Of course not - that would be ridiculous!

.

May 19, 2015
#5
+124524
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
+26319
+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
#7
+117762
+10

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

May 19, 2015
#8
+33053
+5

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

.

May 19, 2015
#9
+117762
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