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 6.1 Lines and Angles 6.2 Triangles 6.3 Polygons 6.4 Circles

 6.5 Key Formulas 6.6 Review Questions 6.7 Explanations
Explanations

1.      C

Because lines l and m are parallel and line AB is a transversal, the angle whose measure is labeled as 120º is supplementary to .

Now we have a 30-60-90 triangle whose longer leg, AC, is also the distance between lines l and m. Using the :: side ratios for 30-60-90 triangles you can use the hypotenuse length to calculate the lengths of the other two legs. The short leg has a : ratio to the hypotenuse, so its length is 5/2. The long leg has a : ration to the short leg, so its length is 52.

2.      E

Let’s analyze each statement separately.

1. This statement implies that DF = EF. We know that BF = AF because it is given that line CF is the perpendicular bisector of AB, and by definition, F is the midpoint of AB. It is also given that the area of triangle CDB is equal to the area of triangle CEA. These two triangles share the same height, and since the area of a triangle is found by the formula 12 b h, it follows that if their areas are equal, their bases are equal too. If BD = AE, then by subtracting DE from each segment, we have BE = AD and thus EF = DF. So statement I is true.
2. This statement is simply not backed by any evidence. All we know is that BE = AD, EF = DF, and BF = AF. As long as points E and D are equidistant from F, all these conditions hold, so there is no guarantee that they are the midpoints of BF and AF, respectively. E and D could be anywhere along BF and AF, respectively, as long as they are equidistant from F. Thus, this statement is not necessarily true.
3. Triangles CDB and CEA are equal in area; this is given. By subtracting the area of triangle CED from each of these triangles, we see that triangles CEB and CDA must have the same area. This statement is true.

Only statements I and III must be true.

3.      C

The area of a triangle with base x and height h is given by the formula 12xh. The area of a square with sides of length x is x2. Since you know the two shapes have equal areas, you can set the two expressions equal to each other and solve for h:

The correct answer is h = 2x.

4.      D

If ABD is an equilateral triangle, then AD = AB = BD = 4, and all the sides of the rhombus have a length of 4 (by definition of a rhombus, all sides are congruent). Also, by definition of a rhombus, opposite angles are congruent, so . Draw an altitude from a to DC to create a 30-60-90 triangle, and from the length ratio of x : x : 2x among the sides, you can calculate the length of this altitude to be 2. The area of a rhombus is bh, so the area of this rhombus is 4 2 = 8.

5.      D

The length of the arc depends on the circumference of the circle and the measure of the central angle that intercepts that arc. The formula is:

where n is the measure of the central angle that intercepts the arc and r is the radius.

Angle c is the inscribed angle or one-half as large as the central angle that intercepts the circle at the same points. So the measure of this angle is 2cº.

Now simply plug the values into the formula: the length of arc AB is:

2π(8) = =
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
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