The diagram shows two touching semicircles of radius 1 and parallel diameters AB and CD. What is the square of the distance AD? Let's first draw the height between the parallel top and bottom lines. Using that, we can obtain a lot of right triangles that are useful. Note also that the distance between the two centers of the circles is 2, because the two centers of the circles are separated by 2 radii, and 2 radii together is 2. So now that we've drawn this image, we can make use of many right triangles. Specifically, looking at the picture on the right, we know that the distance AD that we seek is x, and x is determined by the height, 1, and the length, 2 plus root 3. Using the Pythagorean theorem, we know that x squared equals 1 squared plus the quantity of 2 plus root 3 squared. Solving this, we find that x squared is 1 plus 4 plus 3 plus 4 root 3, which we can simplify as 8 plus 4 root 3. Because it asks for the square of the distance AD, we only need to leave it as x squared, and we found that x squared equals 8 plus 4 root 3, and therefore, our correct answer is B.

semicircles

geometry

Pythagorean theorem

distance calculation

right triangles