Geometry problem with 2 circles and a triangle

Question

I tried to solve this problem:

If $D$ is the diameter of the larger circle and $d$ is the diameter of the small circle, then $d+D$ equals

(A) $a+b$ (B) $2(a+b)$ (C) $\sqrt{(a+b)}$ (D) $2\sqrt{ab}$ (E) $\sqrt{2(a^2+b^2)}$

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But I did not know how to do it so I looked at the answers and I saw $E$ looked convincing because it is the only one that has square powers and $D$ (from the diagram) is $\sqrt{a^2+b^2}$ so I chose $E$. When I looked at the solutions, it said that $A$ was the answer and as I was reading the solutions. It said that I must drop perpendiculars from the center of the small circle to the sides of each triangle : That is what I assumed I must do. Then it said that the distance from the right angle to the points of contact is $\frac{1}{2}d$. Then it said that the distances from the other to angles to the points of contact are $a - \frac{1}{2}d + b-\frac{1}{2}d$. I don't understand this part, what does it mean? Help would be appreciated.

Answer 1

the sum of the distances from the other t***W***o angles to the points of contact

= red solid line (or the red dotted line) + green solid line (or the green dotted line)

= $(a - \frac{1}{2}d) + [b-\frac{1}{2}d]$

= $a - \frac{1}{2}d + b-\frac{1}{2}d$

Answer 2

The bottom left corner of the triangle has a quadrilateral with every angle $90^{\circ}$ - a rectangle therefore. And you have two sides equal to $\frac d2$, so it is a square with side $\frac d2$. Mark all the sides of the square with $\frac d2$, and you should be able to see what is going on.

(Note that the triangle is right-angled because it is built on the diameter $D$- I note you haven't marked that known angle)