Eight circles of radius 1 have centers on a larger common circle, and adjacent circles are tangent. Find the area of the common circle.

THE AREA OF COMMON CIRCLE IS 2 ( 2 + √ 2 ).

Step-by-step explanation:

As given in the information let us consider eight small circles each having radius 1 and are adjacent to each other

These adjacent circles are tangent to each other.

These 8 small circles form together a big circle which is having radius 'R' to small circles.

Let us consider the angle between big circle and small circle is .

= for one circle

As we are having eight circles angle for eight circles is

= /8

Radius R = 1 / sin

cos 2 = 1 - 2sin² => sin 2 = (1- cos2)/2 and with cos   /8 = 1/2 * √ 2

we get 1/sin   /8  = 4/(2- √ 2)

Therefore area of common circle is

=> 2(2+√ 2).

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!