Five Golden Circles

The total area is $70$ . Now let the area of one circle be $C$ . Then notice that the total area is $5C - 4 \times 5$ as we have five circles minus four intersections.

Equating and solving gives $C = \boxed{18}$ as required.

Suppose $A$ is the area of a circle. Since all circles are congruent, the area of the light gold and gold must be $35\times 2=70$ . The intersection area is $5$ , then we get

$(A-5)+5+(A-10)+5+(A-10)+5+(A-10)+5+(A-5)=70$

$5A-20=70$

$5A=90$

$A=\frac{90}{5}=18$

The area of the figure =2x35=70 then the area of the five circles= 70+4x5=90 then the area of one circle =90/5=18

$A + (A - 5) + (A - 5) + (A - 5) + (A - 5) = 2 \times 35 \Rightarrow A = \boxed{18}$

Figure total area is $70$ that is equal to $5C-4I$ = $5C-4(5)$

Solve

$5C-4(5)=70$

$5C=90$

$C=18$

The golden area = the blue area = 35

Golden area + Blue area = 2x35 = 70

Area of 4 intersecting area = 4x5 = 20

Total area of 5 individual circles = 70+20 =90

Area of one circle = 90/5 = 18

To find the total area of the diagram, l multiplied the given area by $2$ and got $70$ . Since the circles overlap four times, $4 \times 5$ is the amount of overlap. Thus, without accounting for overlap the area is $90$ which gives us an area of $18$ for each circle since $\frac {90} {5} = 18$ .

((35 * 2)+(5*4))/5 = 18

I knew the total area was 70. Then I subtracted 20 for the overlapping parts. Then I imagined similar "overlapping" sections on the far left and far right circles so I subtracted another 10. This meant the "middle" section of each circle had to be 40/5 or 8. Then I added the 8 for the center of one circle plus 10 for the 2 overlapping pieces in one circle to get 18.

We are given the area under Line A which is from the bottom of the first circle to the top of the last and with this given information we can discover that the total area is double the golden area: G=Golden Area T=Total area, G=35 and 2G=T, so 70=T and since there is 4 equal overlaps in each circle and the area of each overlap is 5 so if we expand the figure to 5 tangent circles we had 4 over laps so the new area F=70+5(4)=90 and since we have 5 circles the area of a circle is Q=90/5=18

So the area of one of these circles is 18

Let the radius of each circle be 'r'. Then the common tangent below any two near circles is: √2 r so that the sides of the golden triangle are 4√2 r and 2r which gives us an expression for the golden area as :4 √2 r² + π r²/2 - 4[√2 r² -π r²/2 + 5/2] = 35 and A = π (r²). We get: r = 3√(2/π) and A = 18

Area of the circle = Ac Area of the overlapping thing =Ao We know that Ao is 5 so and and the golden area is 35 and if you look as all pieces together you see that this has an area of 2,5Ac -2Ao => 2,5Ac -2Ao=35 2,5Ac-10= 35 /+10 2,5Ac =45 /:2,5 Ac=18

I hope you can't understand that what Im trying to say. (I'm not an native speaker ;()

The golden area is 35 which is half of total area so total area is 70 no create same overlapping circles one before the first circle and second after the last circle now in total we will have 6 overlapping areas (convex lens shaped area) and 5 concave lens shaped area total area is 70 from which we know that 30 is taken by convex lens so 5 times the areaa of concave lens = 40 so area of one concave lens is 8 a circle is made up of one concave lens and 2 convex lens so 8+5+5=13