29

Draw a rough sketch of the wall, the circles, and the spaces. Notice that there is one space for every circle, plus one more space at the end.The area of each circle is 36π, so r = 6, and the diameter of each circle is 12 inches.Convert the length of the wall into inches: 31 × 12 = 373 inches, plus the extra 6 inches equals 379 inches.You know that the wall in covered in a certain number of circles plus spaces. Let the distance covered by a circle and a space be represented as (12 + x), and the number of circles be represented as y, so now you have y(12 + x) .You also know that there is an extra space, so add an extra x to the end, and this is now the total length of the wall. So, y(12 + x) + 1 = 379 .The question tells us that x must be an integer, and that you need the greatest number of circles, and thus you want x to be as small as possible. Avoid solving the equation and try plugging in 1 for x, since it’s the smallest positive integer. Now the equation becomes y(12 + 1) + 1 = 379, and you can solve for y . So, 13y + 1 = 379 and 13y = 378, leading to y = 29.

Note that OD must be a diameter because it is the longest possible line segment crossing the circle. OA and OD (draw it in) are both radii, and therefore equal in length (3), and both of them are equal to AD.Therefore, triangle OAD is equilateral, and the measure of ∠AOD is 60°.The central angle for sector OABC is 120° (the supplement to 60°), making this sector’s area the area ¹⁄₃ of the circle:(¹⁄₂)3²π = 3π.Because π is slightly greater than 3, 3π is slightly greater than 9, giving you choice (A) for the answer.

40

Draw and label the figures, and then set up your scratch paper to Plug In. If circle O has a radius of 6, it has an area of πr² = 36π; circle M, then, has a radius of 2 and an area of 4π. If sector AOB has an area of 4π out of a total area of 36π, then the sector takes up \(\frac{4\pi }{36\pi }=\frac{1}{9}\) of the entire circle, and ∠AOB represents ¹⁄₉ of 360°.The correct answer is thus ¹⁄₉ × 360 = 40.

Plug in a value for the radius of the circle, say r = 2, making the diameter 4;\(\overline{AB}\) and \(\overline{CD}\) are both diameters of the circle, so Quantity A is 8.The circumference of the circle is 4π ≈ 12, so Quantity B is greater.

The diameter of the larger circle, in inches, is ¹⁄₂, so the radius is .Therefore, the area of the larger circle is \(\pi \left ( \frac{1}{2} \right )^{2}=\frac{\pi }{4}\), and the area of the smaller circle is half this area,^π⁄₈. Setting this amount equal to the area formula allows you to determine the radius of the smaller circle:πr² = π⁄₈; r = ^√2⁄₄.Therefore, the diameter is ^√2⁄₂. Subtract this amount from 1 (the diameter of the larger circle):\(1-\frac{\sqrt{2}}{2}=\frac{2-\sqrt{2}}{2}\)

Try plugging in 5 for x. If circle A has an area of 9π, it has a radius of 3.Circle B then has an area of 9π × 5 = 45π.Circle C has an area of 45π × 5 = 225π, with a radius of 15.Therefore, the ratio of circle A’s radius of 3 to circle C’s radius of 15 is 1:5 or 1:x. Alternatively, note that circle C’s area is the area of circle A times x², making the ratio of the areas 1:x².The ratio of the radii should be the square root of this ratio, because area is πr², giving you the ratio 1:x.Both solution methods prove that the quantities are equal.

A tangent to a circle forms a right angle with a radius drawn to the point of tangency. If CO is the hypotenuse of ∆OBC, then you know that the legs of the right triangle must be shorter than 5. Since OB is the radius of the circle, you know that the radius of the circle must be less than 5, so the circumference must be less than 10π.

3.4

First, draw and label the figure.Each of the triangles formed by the origin and the two vertices has legs of 2√2 and 2√2. Since each one is an isosceles right triangle—in other words, a 45-45-90 triangle—the sides are in the ratio x:x: x√2, and the long side of each is 2√2 × √2= 4.The long side of a triangle is also the side of the square, so the area of the square is 16. Since the side of the square is the same as the diameter of the circle, the diameter is 4, the radius is 2, and the area of the circle is 4π.The area inside the square but outside the circle, then, is 16 – 4π; use an approximation for π to get 16 – (4 × 3.14) = 3.44.Rounded to the nearest tenth, the answer is 3.4.

Plug in 4 for circle B’s diameter; thus circle A’s diameter is 32.The radius of A is 2, and the radius of B is 16; circle B has an area of 4π and circle A has an area of 256π.The ratio is 256π:4π, which reduces to 64:1.

Draw in either diagonal of the square, which also is the diameter of the circle.You have now created two isosceles right triangles, so the length of the diagonal/diameter is 4√2, and the radius is 2√2.The area of the circle is π(2√2)²= 8π.