Hello. I couldn’t find the math thread and I really need some help CC.
I do not understand these three questions nor do I understand the explanations in the book. I will try to post the question (not sure about image) and answer.
Tell me if I need to repost it somewhere else please.
- “The 8-sided figure below is divided into 12 congruent isosceles right triangles. The total area of the 12 triangles is 96 square centimeters. What is the perimeter, in centimeters, of the figure?”
- It is of three horizontal squares divided into triangles in the middle. The third square has another square above it to connect three more squares.
I dont see how I can post a pic.
- "Line segments GH, JK, and LM are parallel and intersect line segments FL and FM as shown in the figure below. The ratio of the perimeter of triangle FJK to the perimeter of triangle FLM is 3:5, and the ratio of FH to FM is 1:5? What is the ratio of GJ to FG?
*It is all one giant triangle. It has two lines in the middle
- “A group of die-hard baseball fans has purchased a house that gives them a direct view of home plate, although their view of the rest of the field is largely impeded by the outfield wall. The house is 30 meters tall, and their angle of vision from the top of the building to home plate has a tangent of 7/6. What is the horizonal distance, in meters, from home plate to closest wall of the fans’ house?”
Answers)
14— K 56
44— J 2:1
54 — F 35
These come from the 2013 version of Act prep. The number of the problem is next to it.
Thanks so much CC!
@TheYellowBiscuit
- The area of each isosceles triangle is 96/12 = 8 sq cm. Then we must have (1/2)s^2 = 8 cm^2 where s is a leg length of one of those triangles, so s = 4 cm.
I don’t understand your description of the diagram, but you should be able to figure out the rest from there.
- Triangles FGH, FJK, and FLM are similar by angle-angle-angle.
The first part of the statement tells us that FJ:FL = 3:5 since the perimeters of their corresponding triangles are in the ratio 3:5. Without loss of generality, let FJ = 3 and FL = 5 (so that JL = 2).
The second part of that statement tells us FH:FM = 1:5, so FG:FL is also in the ratio 1:5. Using FL = 5, we have FG = 1. Then GJ = FJ - FG = 3 - 1 = 2.
Finally the desired ratio GJ:FG is 2:1.
- This question seems really poorly worded (what is "angle of vision?") unless there is some sort of accompanying diagram. It isn't obvious to me if the answer they want is 307/6 = 35 or 306/7 = 25 5/7.
@MITer94
Thank you so much! I now understand #14 and will take another look at the AAA (and others) rules again.
For #54 there is a picture but I (dont know/cant) post a picture. On the diagram the angle of vision is from the top left corner of the building to the field.
Sorry my descriptions are confusing.
Thanks for the help.
Most trig problems of this type on the ACT involve SOHCAHTOA. In this case, the wall of the house is the adjacent side (they are at the top of the house looking toward home plate). That means that the distance from the wall to home plate is the opposite side of the right triangle. Tangent = Opp/Adj = 7/6. Since the house is 30 meters high, the proportion becomes x/30 = 7/6. Solving for x gives 210/6 = 35. The whole bit about the outfield wall is just a distraction. Not a great problem even with the diagram.
@TheYellowBiscuit in any case, the question basically asks if you know the definition of the tangent of an angle - not a very interesting problem anyway. A few ACT problems involve trig theorems such as the law of sines or cosines.
Also, I don’t know anyone who’s bought a house that is 30 m tall.
@MITer94
@TestRekt
Thanks for the help.
Yeah I know what tangent is, but the jump to doing a proportion is . . . I dont know. It doesn’t make sense to me.
@TheYellowBiscuit well, the trig functions can all be defined as ratios (proportions) of sides of a right triangle.
Same thing goes with equating slope of a line in the xy-plane and the tangent of the corresponding angle.