The four half-cylinders are equivalent to two cylinders of radius 1, whose total volume will therefore be 2(π1²6) = 12π.The answer is choice (C).

9π

First, draw the circle inside a square.Because the square has an area of 36, each side is 6.This means that the diameter of the circle is 6 and the radius is 3. Using the circle area formula, the answer is 9π.

The diameter of the circle is 12, so the radius is 6, and the area is 36π.The total number of parts in the ratio is 3 + 4 + 5 = 12, so each part covers an area of \(\frac{36\pi }{12}\) = 3π.The largest ratio part is 5 times this amount, or 15π.

For Quantity B, the side of the square is the same length as the diameter of the circle.The diameter is twice the radius, so the radius is 6. Plug this into the formula for area: A = πr2 to find that A = 36π. Quantity B is greater.

Start by plugging in a radius for the smaller circles; try r = 2.The circumference of each circle is 2πr = 4π, and the sum of all three circumferences is 12π.Because the diameter of circle O is equal to the sum of the 3 shorter diameters, the diameter of circle O is 4 + 4 + 4 = 12, its radius is 6, and its circumference is 12π, so the quantities are equal.

Plug in a value for p. If p = 8, then the side of the square is 2 and the area is 4. If the circumference of the circle is 8, then the radius is ⁴⁄_π, and the area is \(\frac{16}{\pi }\)—approximately 5. Quantity B is larger. Plug in another value for p and you will find that Quantity B remains larger.

To find the shaded region, subtract the unshaded region (the triangle and semicircle) from the entire triangle.The area of an equilateral triangle of side x is \(\frac{x^{2}\sqrt{3}}{4}=\frac{8^{2}\sqrt{3}}{4}=16\sqrt{3}\).Triangle BCD is also equilateral, and has sides of length 4, so its area is \(\frac{x^{2}\sqrt{3}}{4}=4\sqrt{3}\).The radius of the circle is 2, so the area of the semicircle is \(\left ( \frac{1}{2} \right )\pi r^{2}=\left (\frac{1}{2} \right )\pi 2^{2}=2\pi\). So the answer is 16√3 - 4√3 - 2π = 12√3 - 2π.

Draw a fourth triangle and semicircle, and you can see that the figure shown represents 1¹⁄₂ circles and ³⁄₄ of a square.Because the three diameters are perpendicular and congruent, they represent three sides of a square; the isosceles right triangles shown constitute three of the four triangles in the completed square.The area of a circle with diameter of 8 (and radius of 4) is πr² = 16π.1¹⁄₂ times this area is 24π.Eliminate choices (A), (D), and (E) because they do not contain 24π.The diameter of each semicircle is the length of the side of the square.The area of the entire square would be s² = 8² = 64.
³⁄₄ of this area is 48. Adding the two areas together gives you the expression in choice (C).

Plug in 10 for the height and radius of the cylinder. So Quantity A is πr²h = π9²11 = 891π. Quantity B is π8²12 = 768π

C, D, and F

Consider all the possible different integer pairs for the dimensions of the rectangle.You cannot try the integer pair of 1 and 9 or 5 and 5, because you know that x < y. If the rectangle has sides of 4 and 6, you can solve for the diagonal (equal to the circle’s diameter) with the Pythagorean theorem, which gives you [latex]\sqrt{52}[/latex] or 2[latex]\sqrt{13}[/latex], correct choice (C). If the rectangle has sides of 3 and 7, the diagonal is \(\sqrt{58}\), correct choice (D). If the rectangle has sides of 2 and 8, the diagonal is \(\sqrt{58}\), or 2\(\sqrt{58}\), correct choice (F).