1. ## Really hard questions...

1. XY and YZ are diameters of two tangent circles and XY = 10 cm. Find YZ if the area of hte shaded region equals the area of hte unshaded region.

The answer is: 24.14, but how and why?

2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

Notes: These drawings are not EXACT as the real one because I used paint, however they really close.

The answer is 4 cm and 8 cm, but how and why...?

These questions are really hard. Nobody in my class were able to answer them.

2. ## Suggested soln

ok.. here we go...
let's say the dist YZ is x cm, then the diameter of the larger circle is (10+x) cm. And the area is pi [(10+x)/2]^2. Similarly, the area of the smaller circle can be found ie pi(x/2)^2. We can then find an expression for the shaded area and equate that to the area of the smaller circle. You should get a wuadratic equation which can be reduced to
x^2 - 20x -100 = 0.
Solving this would give you the answer as required.

As for the second question, I think you should work on it alonmg similar lines as per question 1.

Hope it helps. Let me know if u need further clarifications.

musMD, Singapore

3. Originally Posted by AlphaRock
[B]...

2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

...
to #2:

I've modified your sketch a little bit.

1. Let x denote the length of the diameter of the upper circle then (12-x) is the length of the diameter of the lower circle.

2. The shaded region is the difference of the large circle and the two small circles:

$A=\pi \cdot \left(\frac{12}2 \right)^2 - \pi \cdot \left(\frac{x}2 \right)^2 - \pi \cdot \left(\frac{12-x}2 \right)^2 = \frac49 \cdot \pi \cdot \left(\frac{12}2 \right)^2$

3. Expand the brackets, collect like terms. You'll get a quadratic equation:

$-\frac12x^2+6x-16=0$

Solve for x.