External tangents to a circle intersect to form the vertices of a square whose length of a side is equal to the circle's diameter. If the area inside the square but outside the circle is 42 square units, what is the sum of the areas of the square and circle?

Let the diameter of the circle be d, which is also the side length of the square. The area of the square is d² and the area of the circle is π(d/2)² = πd²/4. Using 22/7 for π, the area inside the square but outside the circle is d² - (22/28)d² = d² - (11/14)d² = (3/14)d². Setting this equal to 42 gives d² = 196. The sum of the areas of the square and the circle is 196 + (11/14)(196) = 196 + 154 = 350. The sum of the areas is 350 square units.

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