Let $ABCD$ be a rectangle where $\Delta PAB$ is isosceles. The radius of each of the smaller circles is $3$ cm and the radius of the bigger circle is $4$ cm. Find the length and breadth of the rectangle.
What I Tried: Here is the picture:-
I know that radius of the inscribed circle is $\frac{\Delta}{S}$.
So let $BC = x$ , $CP = y$. I know that $(DP = CP = y)$ as $\Delta BCP \cong \Delta ADP$ from $RHS$ congruency.
So with the circle of radius $3$, I have :-
$$3 = \frac{\Delta}{S}$$
$$\rightarrow 3 = \frac{(\frac{xy}{2})}{(\frac{x+y+\sqrt{x^2+y^2}}{2})}$$
Similarly for the bigger circle of radius $4$, I cannot find the area with the $\frac{1}{2}$ * base * height formula, as the height is not known. So instead I used the Heron's Formula for this, and that's where I need help as it gets complicated. So :-
$$4 = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
Here I have $$s = \frac{2\sqrt{x^2 + y^2} + 2y}{2} = \sqrt{x^2+y^2} + y$$
That gives me $$4 = \sqrt{\frac{y^2 * (\sqrt{x^2 + y^2} - y)}{\sqrt{x^2 + y^2} + y}}$$
Now solving both these equations is going to be over-complicated. Although the answer to the length and breadth of the rectangle is given as $24,9$ respectively (The Geogebra picture also shows it) . How to go solve such a big equation to get such a simple answer? Is there any other way or not?
I also doubt if I have done any mistakes along the way, can anyone help me out?
I have sliced the rectangle in half using symmetry.
The key observation is the pair of similar triangles, $\triangle QPF$ and $\triangle OBL$. Their ratio is the same as the ratio of the radii, which is $3:4$. We can see they are similar by using properties of tangents, and the fact that $OPCB$ is a rectangle.
Now we have $QP:BO = 3:4$. We also have $QC = 3$. This gives the length of this half-rectangle: $12$.
Using your formula $3 = \dfrac \triangle s$ we can find the width.