Transcript File - Coach Matt James

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.
70
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.
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 (102+102 = c2) 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 s2.
Since s2 = 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, a2 + b2 =c2, and get
a2 + 9 = 36. a2 = 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.
5
First you need to find the length of segment AC using
Pythagorean Theorem or families of right triangles. A2 +
144 = 169. A2 = 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.
To solve this problem, you need to know
that the area of a circle is ∏r2. 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∏.
3
3
10
In order to find the area of a trapezoid, you need the length of
both bases and the height. You have b1 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, b2 is 16.
The formula is ½ h(b1 + b2). ½ *4*(10+16) = 52.
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.