16

According to the information given, this must be a 45-45-90 isosceles right triangle, and the relationship between the sides can be written as x√2.That means that BC = x√2, or 8 = x√2. Solving for x, you get x = \(\frac{8}{\sqrt{2}}\), so each of the legs of the triangle is equal to \(\frac{8}{\sqrt{2}}\).The formula for the area of a triangle is ¹⁄₂(base)×(height), so the area of this triangle is \(\frac{1}{2}\left ( \frac{8}{\sqrt{2}} \right )\left ( \frac{8}{\sqrt{2}} \right )=\frac{64}{4}=16\).

A, D, and F

Triangles ABC and CDE are similar triangles:The angles where the triangles meet are equal, as are the angles marked jº, so the remaining angles must be equal as well. Since similar triangles have proportional sides, any answer choice in the ratio of 4:6:8 will work.Choice (A) is 4:6:8 cut in half, so choice (A) works; remember the figure isn’t drawn to scale, so don’t worry about making CDE smaller than ABC.Choices (D) and (F) are 4:6:8 multiplied by 2 and 4, respectively, so both work as well. None of the remaining choices work:Choices (C) and (G), in fact, violate the Third Side Rule and aren’t even triangles.

C and D

Draw and label the figure, and then set up your scratch paper to plug in the answers. For a given short side, use the 5:3 ratio to find the long side, and use either the Pythagorean theorem or multiples of the familiar 3-4-5 triangles to determine the middle side; since LMN is a right triangle, the two shorter sides can be used as base and height to find the area. Start with choice (C). If the short side is 9, the middle side is 12 and the area becomes 54; this choice is correct, but just barely, and if you try smaller values you will fall out of the area’s range.Eliminate choices (B) and (A). In choice (D), the short and middle sides are 12 and, 16 and the area is 96; this choice is correct. In choice (E), the short and middle sides are 15 and 20, and the area is 150.This is not in the area’s range of 50 to 150, so eliminate it as well as choice (F), which would produce an even larger area.The correct answers are choices (C) and (D).

C, D,E, and F

The Third Side Rule tells you that the third side must be more than the difference of the two other sides and less than their sum. Therefore, the third side must be greater than 5 and less than 19.The two known sides already add up to 19. If you add this to the range for the third side, the perimeter of the triangle is then between 24 and 38 centimeters.Choices (C), (D), (E), and (F) correct.

To find the area of this triangle, you must drop a line segment to make the height. If you call the midpoint of BC point X, you know that BX is equal to 3. If you see that this makes a right triangle with a side of 3 and a hypotenuse of 5, you can use the 3-4-5 triangle rule to get 4 for the height. Otherwise, use the Pythagorean theorem. After you find the height of 4, use the formula for area of a triangle.The base, 6, multiplied by the height, 4, gives you 24. Divide by 2 to get 12.

This question is testing similar triangles. Do you recognize the 3-4-5 triangles? Triangle BCD is a 3-4-5 triangle, and triangle ACE is too, but it’s a similar triangle—a 6-8-10 triangle.That means that CE is 10, which leaves 5 left over for DE.

10

To solve this question, label everything. First label angle ACD as a right angle. Next, label \(\overline{AC}\) = 4. If \(\overline{AC}\) = 125% of \(\overline{AC}\), then \(\overline{BD}=\frac{125}{100}\times 4=\frac{5}{4}\times 4=5\).Since \(\overline{AC}\) and are \(\overline{BD}\) perpendicular,\(\overline{BD}\) can be the base and \(\overline{BD}\) can be the height. Since the formula for the area of a triangle is Area = ¹⁄₂bb the area here equals ¹⁄₂ × 5 × 4 = 10 the correct answer is 10.

15

To solve this question, picture a triangle:

Since the boat travels a total of 3 + 6 = 9 kilometers east, and a total of 12 kilometers south, we can use the Pythagorean theorem to find the total distance. Since a² + b² = c², then 9² + 12² = c², and 81 + 144 = 225 = c². Taking the square root of both sides gives that c = 15, the correct answer.

The ratio of the given leg to the hypotenuse is √3 to 2 in the largest right triangle, so it is a 30-60-90 triangle, and the length of the other leg must be a.The smaller two triangles also contain 90 degree angles, and all three triangles share the left vertex angle, making all three triangles similar with proportional sides. So, the horizontal leg of the smallest triangle is crt3, and the horizontal leg of the medium sized triangle is brt3.To find the area of the shaded region, find the area of the large triangle, subtract the area of the medium sized one, and add back the area of the smallest one. Plug in values: a = 4, b = 2, c = 1.The area of the large triangle becomes \(\frac{1}{2}(4\sqrt{3})(4)=8\sqrt{3}\) The area of the medium triangle becomes \(\frac{1}{2}(2\sqrt{3})(2)=2\sqrt{3}\) The area of the smallest triangle becomes \(\frac{1}{2}(\sqrt{3})(1)=\frac{\sqrt{3}}{2}\) The shaded area is \(8\sqrt{3}-2\sqrt{3}+\frac{\sqrt{3}}{2}=6\sqrt{3}+\frac{\sqrt{3}}{2}=6\frac{1}{2}\sqrt{3}=\frac{13\sqrt{3}}{2}\). When you plug in the three values into each answer, only choice (C) hits your target, making it the correct answer.

Draw your figure as a right triangle atop a rectangle.The hypotenuse represents the path of the image on the wall, and the rectangle’s dimensions represent the height of the stand and its distance from the wall. It should look like this:

The triangle on top is a 30-60-90 triangle with a short side of 16 − 4 = 12; so the side across from the other leg (the distance from the projector to the wall) is so the answer is 12√3 choice (B).