A square on the coordinate plane has as its vertices the points with coordinates , , , and . Give the equation of the circle inscribed inside this square.

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{4 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{2 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = 81$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = \frac{81}{4 }$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = 81$

Explanation

The equation of the circle on the coordinate plane with radius and center is

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

The figure referenced is below:

Incircle 1

The center of the inscribed circle is the center of the square, which is where its diagonals intersect; this point is the common midpoint of the diagonals. The coordinates of the midpoint of the diagonal with endpoints at and can be found by setting $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the following midpoint formulas:

$x_{M} = \frac{x_{1}+ x_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

$y_{M} = \frac{y_{1}+ y_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

This point, $\left (\frac{9}{2}, \frac{9}{2} \right )$ , is the center of the circle. The radius can easily be seen to be half the length of one side; each side is 9 units long, so the radius is half this, or $\frac{9}{2}$ .

Setting $h = k = r = \frac{9}{2}$ in the circle equation:

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \left (\frac{9}{2} \right ) ^{2}$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{4 }$

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

Which of the following lines is perpendicular to the line ?

$\frac{y}{3}-x=3$

Explanation

Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.

The slope of this line is . The negative reciprocal will be $+\frac{1}{3}$ , which will be the slope of the perpendicular line.

Now we need to find the answer choice with this slope by converting to slope-intercept form.

$y=\frac{4}{12}x-\frac{6}{12}$

$y=\frac{1}{3}x-\frac{1}{2}$

This equation has a slope of $+\frac{1}{3}$ , and must be our answer.

A square with a side length of 4 inches is inscribed in a circle, as shown below. What is the area of the unshaded region inside of the circle, in square inches?

Act_math_01

8π - 16

4π-4

8π-4

2π-4

8π-8

Explanation

Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

Screen_shot_2013-03-18_at_10.21.29_pm

In the figure above, is a square and is three times the length of . What is the area of ?

Explanation

Assigning the length of ED the value of x, the value of AE will be 3_x_. That makes the entire side AD equal to 4_x_. Since the figure is a square, all four sides will be equal to 4_x_. Also, since the figure is a square, then angle A of triangle ABE is a right angle. That gives triangle ABE sides of 3_x_, 4_x_ and 10. Using the Pythagorean theorem:

(3_x_)2 + (4_x_)2 = 102

9_x_2 + 16_x_2 = 100

25_x_2 = 100

_x_2 = 4

x = 2

With x = 2, each side of the square is 4_x_, or 8. The area of a square is length times width. In this case, that's 8 * 8, which is 64.

There is a line defined by the equation below:

There is a second line that passes through the point and is parallel to the line given above. What is the equation of this second line?

$y=-\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

Explanation

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = _–_3x + 12

y = –(3/4)x + 3

slope = _–_3/4

We know that the second line will also have a slope of _–_3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = _–_3/4(1) + b

2 = _–_3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

Triangle

Note: Figure NOT drawn to scale.

Refer to the figure above, which shows a square inscribed inside a large triangle. What percent of the entire triangle has been shaded blue?

$44\frac{4}{9} %$

$33 \frac{1}{3} %$

Insufficient information is given to answer the question.

Explanation

The shaded portion of the entire triangle is similar to the entire large triangle by the Angle-Angle postulate, so sides are in proportion. The short leg of the blue triangle has length 20; that of the large triangle, 30. Therefore, the similarity ratio is . The ratio of the areas is the square of this, or $3^{2}:2^{2}$ , or .