Analyzing Figures

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SAT Subject Test in Math I › Analyzing Figures

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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

CORRECT

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?

CORRECT

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 ?

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

CORRECT

$0\leq z\leq (xy)$

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

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

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

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

CORRECT

$\small \small 4$ 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