SOLUTION: Find the area of the larger circle in {{{ cm^2 }}} if {{{ r=sqrt(2) }}} cm. https://ibb.co/PjYkRmL

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Question 1199330: Find the area of the larger circle in +cm%5E2+ if +r=sqrt%282%29+ cm.
https://ibb.co/PjYkRmL

Found 3 solutions by greenestamps, ikleyn, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


ANSWER: C

To see this, draw segments from the center of the small circle to the center of the large circle, and to the points of tangency of the small circle with the x- and y-axes. Use those to determine that the radius of the large circle is

sqrt%282%29%2Bsqrt%282%29%2Asqrt%282%29=2%2Bsqrt%282%29

Then the area of the circle is (pi)(r^2) = pi%282%2Bsqrt%282%29%29%5E2, answer C.

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

I just solved this problem and answered this question several days ago.

See the link

https://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.1199274.html



Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

 

To find the area of the larger circle, we need to find its radius.

The red and green triangles are similar 45-45-90 isosceles right triangles.
Since the green hypotenuse is given to be sqrt%282%29, it is a standard
1-1-sqrt%282%29 isosceles right triangle, and its legs are 1 each.

The radius of the larger circle is the sum of the red and green hypotenuses.
The green hypotenuse is sqrt%282%29 and is a radius of the small circle, So, the 
vertical red side is also sqrt%282%29, because it is also a radius of the smaller circle. 

We set up a proportion between the sides of the two similar triangles:

matrix%281%2C2%2Cgreen%2Chypotenuse%29%2Fmatrix%281%2C2%2Cred%2Chypotenuse%29%22%22=%22%22matrix%281%2C3%2Cgreen%2Cvertical%2Cside%29%2Fmatrix%281%2C3%2Cred%2Cvertical%2Cside%29

sqrt%282%29%2Fmatrix%281%2C2%2Cred%2Chypotenuse%29%22%22=%22%221%2Fsqrt%282%29

Cross-multiply and get

matrix%281%2C2%2Cred%2Chypotenuse%29%22%22=%22%222

matrix%281%2C4%2Cradius%2Cof%2Clarger%2Ccircle%29%22%22=%22%22matrix%281%2C2%2Cred%2Chypotenuse%29%22%22%2B%22%22matrix%281%2C2%2Cgreen%2Chypotenuse%29

matrix%281%2C4%2Cradius%2Cof%2Clarger%2Ccircle%29%22%22=%22%222%22%22%2B%22%22sqrt%282%29

Using Area%22%22=%22%22pi%2Aradius%5E2,

Area%22%22=%22%22pi%282%2Bsqrt%282%29%29%5E2 cm2

Edwin