<p>I can't seem to figure these took math questions out:
(I took them for preparing for the act, 2009-2010:)</p>
<p>Unforuntately for one of the questions, the diagram refuses to copy and paste, sorry. :/</p>
<ul>
<li>The diagram for the circle can be found on page 33.
URL: <a href="http://www.act.org/aap/pdf/preparing.pdf%5B/url%5D">http://www.act.org/aap/pdf/preparing.pdf</a></li>
</ul>
<p>The transmitter site of radio station WGGW and the
transmitter site of another radio station, WGWB, are
on the same highway 100 miles apart. The radio signal
from the transmitter site of WGWB can be received
only within a radius of 60 miles in all directions from
the WGWB transmitter site. For how many miles along
the highway can the radio signals of both stations be
received?
(Note: Assume the highway is straight.)
F. 8
G. 12
H. 40
J. 44
K. 48</p>
<p>Answer: G.</p>
<p>The diagram for the triangle is also found on page 33.</p>
<p>In the figure below, the vertices of ABC have (x,y)
coordinates (4,5), (5,3), and (1,3), respectively. What
is the area of ABC ?
A. 4
B. 4 square root 2
C. 4square root 3
D. 8
E. 8square root 2</p>
<p>Answer: A</p>
<p>Thank you in advance!</p>
<p>56.
GG tower has 52 mile radius, GW has 60 mile radius. Use 50 miles as halfway mark.
(60-50) = 10. GW goes 10 miles over 50 mile mark.
(52-50)= 2, GG goes 2 miles over 50 mile mark.
(10 + 2)= 12 Add.</p>
<ol>
<li>After plotting points on graph and making triangle, you should see that (1,3) and (5,3) make hyp.
Distance in x value between (1,3) and (5,3) is 4.
This is 45-45-90, so to find one side, just divide hyp by square root of 2.
Now each of the other two sides should equal 2 square root of two.
Now just find area of triangle (.5)(2 square root of two)(2 square root of two)
= (.5)(4)(2) = 4</li>
</ol>
<p>Here is another strategy for 56 (Draw a line marked 100 if you like):</p>
<p>Subtract 52 from 100: 100-52=48
This is the area that only the 60 mile station covers</p>
<p>Subtract 48 from 60: 60-48=12</p>
<p>This is the area that they both cover.</p>
<p>You could also use the same strategy by subtracting 60 from 100, then 40 from 52.</p>
<p>Sorry if i’m a bit off topic here but wasn’t this question in last year’s booklet?</p>
<p>Yeah, op states at the beginning that it is in the 2009-10 preparing for the ACT packet</p>
I realize this post is super old, but on the off chance that anyone is using the 2009-2010 ACT study guide, I felt it necessary to respond.
For 59, asking about the area of the triangle, an above poster said it’s a 45-45-90 triangle. That is not accurate. If this was a 45-45-90 triangle, the two legs would be congruent. They are not.
In order to solve the problem, you must divide the triangle into two smaller triangles.
If you draw a line vertically from A to the hypotenuse, you create a triangle that has a base measurement of 1. You then use the distance formula to determine the distance between A and B, which is square root of 5. Then you use Pythagorean Theorem to determine the length of the other side, which is 2. From there, you calculate the area of the small triangle as 1.
The other triangle you’ve created has a base of 2 (that we just discovered, as it’s the leg of the other triangle), and a side leg length of 3. We then use the area formula to determine the area of that triangle is 3.
Adding the two areas gives you the final answer of 4.
Pickles solution, though correct, is WAY too complicated. Yes divide the triangle into to smaller ones- A to hyp… This shows a base of of 1(x values: 5-4) and height of 2 (y values:5-3). Area= 1/2B*H= 1.
Next triangle: Auto height of 3(x values:5-1- 1(other base)) and base=2 from previous triangle. Area=1/2B*H=3
3+1=4. Done son