Rhombus

Solve for x and y using the rules of quadrilateral

x=6, y=9

x=9, y=6

x=2, y=4

x=6, y=10

Explanation

By using the rules of quadrilaterals we know that oppisite sides are congruent on a rhombus. Therefore, we set up an equation to solve for x. Then we will use that number and substitute it in for x and solve for y.

If line 2 and line 3 eventually intersect when extended to the left which of the following could be true?

$\I.\ a = e \II.\ a = f \III.\ e+c = 180$

I only

Cannot be determined

I and II

I and III

I, II, and III

Explanation

Read the question carefully and notice that the image is deceptive: these lines are not parallel. So we cannot apply any of our rules about parallel lines. So we cannot infer II or III, those are only trueif the lines are parallel. If we sketch line 2 and line 3 meeting we will form a triangle and it is possible to make a = e. One such solution is to make a and e 60 degrees.

What is the maximum number of distinct regions that can be created with 4 intersecting circles on a plane?

Explanation

Try sketching it out.

Q3b

Start with one circle and then keep adding circles like a venn diagram and start counting. A region is any portion of the figure that can be defined and has a boundary with another portion. Don't forget that the exterior (labeled 14) is a region that does not have exterior boundaries.

Note: Figure may not be drawn to scale

In rectangle has length and width and respectively. Point lies on line segment $\overline{BD}$ and point lies on line segment $\overline{EF}$ . Triangle has area , in terms of and what is the possible range of values for ?

$0\leq z\leq \frac{xy}{2}$

$0\leq z\leq (xy)$

$(\frac{xy}{2}) \leq z\leq (xy)^2$

$xy\leq z\leq (\frac{x}{2} +\frac{y}{2})$

cannot be determined

Explanation

Notice that the figure may not be to scale, and points and could lie anywhere on line segments $\overline{BD}$ and $\overline{EF}$ respectively.

Next, recall the formula for the area of a triangle:

$area = \frac{1}{2}bh$

To find the minimum area we need the smallest possible values for and .

To make smaller we can shift points and all the way to point . This will make triangle have a height of :

$area = \frac{1}{2}b(0)$

is the minimum possible value for the area.

To find the maximum value we need the largest possible values for and . If we shift point all the way to point then the base of the triangle is and the height is , which we can plug into the formula for the area of a triangle:

$area = \frac{1}{2}xy$

which is the maximum possible area of triangle

Chords $\overline{AB}$ and $\overline{CD}$ intersect at point . $\overline{BX}$ is twice as long as $\overline{AX}$ ; and .

Give the length of $\overline{AX}$ .

$3 \frac{2}{3}$

$5\frac{1}{2}$

$4\frac{1}{2}$

Explanation

If we let , then .

The figure referenced is below (not drawn to scale):

Chords

If two chords intersect inside the circle, then the cut each other so that for each chord, the product of the lengths of the two parts is the same; in other words,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$t \cdot 2t = 2 \cdot 9$

$2t^{2} =18$

$2t^{2} \div 2 =18 \div 2$

$t^{2} = 9$

Taking the positive square root of both sides:

$t = \sqrt{9} = 3$ ,

the correct length of $\overline{AX}$ .

Thingy_5

Refer to the above diagram. Which of the following is not a valid name for $\angle DBF$ ?

$\angle B$

$\angle FBE$

$\angle CBF$

$\angle FBD$

All of the other choices give valid names for the angle.

Explanation

$\angle B$ is the correct choice. A single letter - the vertex - can be used for an angle if and only if that angle is the only one with that vertex. This is not the case here. The three-letter names in the other choices all follow the convention of the middle letter being vertex and each of the other two letters being points on a different side of the angle.

Circle

Use the facts of circles to solve for x and y.

x=11, y= 39.5

x=39.5, y=11

x=10, y=30

x=13, y=10

Explanation

In this question we use the rule that oppisite angles are congruent and a line is 180 degrees. Knowing these two facts we can first solve for x then solve for y.

Then:

Similar triangles

Figures not drawn to scale

The triangles above are similar. Given the measurements above, what is the length of side c?

$\small \small 4\frac{1}{6}$ inches

$\small \small 4$ inches

$\small \small 4\frac{1}{2}$ inches

$\small \small 4\frac{1}{3}$ inches

$\small \small 5$ inches

Explanation

You can find the length of c by first finding the length of the hypotenuse of the larger similar triangle and then setting up a ratio to find the hypotenuse of the smaller similar triangle.

$\small a{^{2}}+b{^{2}}=c{^{2}}$

$\small 6{^{2}}+8{^{2}}=c{^{2}}$

$\small 36+64=c{^{2}}$

$\small 100=c{^{2}}$

$\small c=10$

You also could have found 10 by recognizing this triangle is a form of a 3-4-5 triangle.

The hypotenuse of the bigger triangle is 10 inches.

Now that we know the length of the hypotenuse for the larger triangle, we can set up a ratio equation to find the hypotenuse of the smaller triangle.

$\small \frac{2.5}{6}=\frac{c}{10}$ cross multiply

$\small 6c=25$

$\small \frac{6c}{6}=\frac{25}{6}$

$\small c=4\frac{1}{6}$

Triangle

Use the rules of triangles to solve for x and y.

x=60, y=30

x=30, y=60

x=30, y=30

x=45, y=45

Explanation

Using the rules of triangles and lines we know that the degree of a straight line is 180. Knowing this we can find x by creating and solving the following equation: