geometry

[gleaming jungle]

For (a) u can prove that ACP and DBP are similar triangles by using AAA and then it's solved.

Where is the figure of (b) coz it says some red circle?

[gleaming jungle]

I may give a sketch here. Denote the intersection of AB and CD as P. Then u can assume that AP/BP=a. Then the area of the larger red circle/the area of the smaller one=a^2, and the whole circle has area (1+a)^2 size of the smallest red circle. Then using the knowing we can get the conclusion.

[gleaming jungle]

why (1/2)^2? sry i didnt get u

i'm trying to represent area in white as sth times the area in the smaller red circle and then get the multiple by solving the equation by using the knowing

[bold hollow]

My approach to solve part b is to set the radius of the big red circle to b and the small red circle to a, and then compare all of the areas in terms of a and b to create a solvable equation.

So for instance the area of the big red circle would be b^2•π and the area of the small one would be a^2•π. Because these radii are collinear, we can use it to find the area of the biggest circle's area. a, the radius of the small red circle, plus b, the radius of the big red circle, gives us the radius of the biggest circle (I think you can see why). This means that the area of the biggest circle is (a+b)^2•π. For the white area, we can see it is the area of the big circle minus the areas of the red circles: (a+b)^2•π-(a^2•π+b^2•π). This simplifies to 2abπ, but we also are told that the white area is equal to the area of the bigger red circle, b^2•π. With this, we get b^2•π=2abπ (simplifies to b=2a), which we can use to solve the rest.

Feel free to let me know if you need further explanation.

[patent barn]

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