# ACT Math Practice. Geometry and Trigonometry Placement Tests Primary content areas included in the Geometry Placement Test include: » Triangles (perimeter,

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ACT Math Practice

Geometry and Trigonometry Placement Tests Primary content areas included in the Geometry Placement Test include: » Triangles (perimeter, area, Pythagorean theorem, etc.) » Circles (perimeter, area, arcs, etc.) » Angles (supplementary, complementary, adjacent, vertical, etc.) » Rectangles (perimeter, area, etc.) » Three-dimensional concepts » Hybrid (composite) shapes » Right-triangle trigonometry » Special angles (multiples of 30 and 45 degrees)

While this figure is not necessarily drawn to scale, you can almost guess correctly because angles A and C are clearly not the same size; nor are B and C. However, angles A and B are actually equal because they are vertical angles, and angles A and D are equal because they are corresponding angles (same location in the set of four angles). This makes B and D equal as well.

To solve this problem, remember that the angles of a triangle add up to 180 degrees. Also since segments AB and AC are equal, the angles across from them are equal. Therefore, 180-40 = 140/2 = 70. Finally, angles that form a straight line (supplementary angles) equal 180. Thus, 180-70 = 110. 70

Count all of the north-south and east-west and get 200 feet. Then, a diagonal is always longer than a “straight” piece so it has to be longer than 10. If you look at the triangle formed here, it is isosceles with two sides being 10. Use Pythagorean Theorem (10 2 +10 2 = c 2 ) to find out the other side is 10√2 or around 17. 200+17 = 217.

Since the area of a rectangle is length times width, the original garden is 144 square feet in area. The new shape is a square whose area is equal to 144. The formula for the area of a square is s 2. Since s 2 = 144, square root both sides to get s =12.

This problem can be solved in one of two ways. First you can use Pythagorean Theorem, a 2 + b 2 =c 2, and get a 2 + 9 = 36. a 2 = 27. Simplify 27 by breaking it down to 3*3*3. So a = 3√3. The other method to solve is to recognize that these numbers fall into a 30-60-90 degree triangle pattern, where the smallest side is x, the side opposite the 60 degree angle is x times √3 and the hypotenuse (side opposite the 90 degree angle) is 2x.

To solve this problem remember that the length of a sector is a portion of the circumference of a circle, and the circumference is 2∏r. Plus you need to know that a circle is 360 degrees. So here we have a an arc formed by 30 degrees out of 360. Thus, 6 = (30/360)*2∏r. Simplify to 6 = 1/6 ∏r and then divide by 1/6∏to get D.

First you need to find the length of segment AC using Pythagorean Theorem or families of right triangles. A 2 + 144 = 169. A 2 = 25 so the length of AC = 5. Next you need to know your trig definitions for sine, cosine and tangent. Sine is opposite/hypotenuse, Cosine is adjacent/hypotenuse and Tangent is opposite/adjacent. Since we are looking for tangent, we use the side opposite angle A which is 12 over the adjacent side which is 5. So 12/5. Hint: Many of you may have heard the phrase Oscar had a handful of apples to help you remember the trig relationships. 5

To solve this problem, you need to know that the area of a circle is ∏r 2. The area of the small circle is ∏(5) 2, which is 25∏. Because the circles are internally tangent, and B is the center of circle B, the radius of the large circle is 10, making its area 100∏. If we cut circle A out of circle B, then what is left is 100∏- 25∏ which is 75∏.

In order to find the area of a trapezoid, you need the length of both bases and the height. You have b 1 which is 10, and you have the height which is 4 where you have the right angle. With the triangle on the left, use Pythagorean Theorem or families of triangles to find the bottom of the triangle to be 3. Since the length of BC is equal to the length of segment AD, then you can draw a right triangle on the right, and it has the same measurements as the triangle on the left. This means, b 2 is 16. The formula is ½ h(b 1 + b 2 ). ½ *4*(10+16) = 52. 3 3 10

Using the properties of similar triangles: 6 / x = 18/ (x+15). Cross multiply to get 6x +90 = 18x. Solve for x. 12x = 90, x = 7.5

To find the area of triangles, you have to know base and height, and the height has to form a 90 degree angle. The easiest way to find the area of triangle DEG is to find the area of the whole triangle and subtract off the area of triangle EGF. The area of a triangle is ½ bh, so the area of triangle DEG is ½*19*10 = 95. The area of triangle EGF is ½ * 7*10 = 35. Subtract: 95-35 = 60.

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