The base of a triangle is 17. The other two sides are integers and one of the sides is twice as long as the other. What is the longest possible length of a side of the triangle?
a) 17 b) 32 c) 33 d) 35 e) longest side can be any length.
If the sides of a regular pentagon are extended through the vertices. then the plane will be divided into how many nonoverlapping regions?
a) 6 b) 1 c) 13 d) 15 e) 16
2007-08-12 17:42:01 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(b) 32.
The longest side being twice the other side limits the triangle to 32 and 16. If the shorter of the other two sides were 17, the longest side would be 34 and the figure is not a triangle, since the longest side (34) is equal to the sum of the other two.
(e) 16.
I count:
(1) The pentagon itself
(5) Triangles like a five-pointed star around the pentagon
(5) Triangular areas that touch vertex-to-vertex to the five star points, where the lines that form the previous triangles cross.
(5) Areas in between those secondary triangles.
2007-08-12 17:47:46 · answer #1 · answered by McFate 7 · 0⤊ 3⤋
In order for a figure with side lengths a, b, and c to be a triangle:
a + b > c
b + c > a
a + c > b
If the base of a traingle is 17 and the other two sides are in a ratio of 1:2 (one is twice as long as the other), we can narrow the choices given in order to find an answer:
If the longest side is 35 or 33, then one of the traingle sides will not be of integer length (35 and 33 cannot be divided by 2 evenly). Therefore, we eliminate these two choices.
If the longest side is 34, then the side that is half the length of this is 17. This means there are two sides of length 17 and one of 34.
17 + 17 is not greater than 34.
So, we can eliminate choice e) longest side can be any length.
If the longest side is 32, then the smaller side would be 16 (half of 32). This would make the three sides of the traingle at lengths 16, 17, and 32.
16 + 17 > 32
This is a possible answer because this is true. Because 32 is a greater side length than the last remaining choice, this is our answer.
b) 32
For the second question:
If the verticies are extended on both sides: then you just have to draw a diagram. I counted e) 16 pentagons.
Otherwise, if they are extended on one side, I counted a) 6
Hope this helps!
2007-08-12 18:15:01 · answer #2 · answered by zero_max12 2 · 0⤊ 0⤋
For your first problem, remember the triangle inequality: The sum of any two sides of a triangle will always be greater than the third side.
Can the two sides you're looking for be 10 and 5? 20 and 10? 40 and 20?
10 and 5? 10 + 5 = 15 which is < 17 so not a triangle
40 and 20? 20 + 17 < 40 so not a triangle
20 and 10? It can be a triangle, though not the answers you're looking for.
Do some searching to find the limits.
For the second problem, I'd draw a picture and see how many pieces you can count. . . Maybe start with a triangle and look for patterns.
Good luck!
2007-08-12 17:56:10 · answer #3 · answered by douglas 2 · 1⤊ 0⤋
14 + 12 + 5 = 31 so 2 pupils are wearing the two a) 14+12-2 = 24 b) 14-2 = 12 c) 12-2 = 10 Please notice that alot of people will answer 26 for the 1st question, in spite of the shown fact that that's inaccurate: 29 - 5 = 24 (minus of people who arent wearing purple OR blue 14+12 = 26 there is 26 blue and purple colorations yet purely 24 pupils this suggests 2 human beings could desire to be wearing the two colorations as I basically pronounced there is totally 24 human beings so what proportion purple and blue = 24 purple not blue = purple human beings much less the folk who wearing the two blue not purple = blue human beings much less the folk who wearing the two
2016-12-15 13:27:35 · answer #4 · answered by Anonymous · 0⤊ 0⤋
b)32, it has to be an even number without even working the problem, you know that
don't have a clue to the last question though,
Good luck!
2007-08-12 18:40:37 · answer #5 · answered by savagevisions 2 · 0⤊ 0⤋
b and e
2007-08-14 09:08:11 · answer #6 · answered by Anonymous · 0⤊ 1⤋