Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?
(A) 72
(B) 576
(C) 4032
(D) 4608
(E) 6336
Help with triangle question
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When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.islandgurl918 wrote:Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?
(A) 72
(B) 576
(C) 4032
(D) 4608
(E) 6336
We need to determine how many ways we can combine L, M and N to form a triangle. For each point, we need to choose an x value and a y value.
Point L:
x value: -3≤x≤4, giving us 8 choices.
y value: 3≤y≤11, giving us 9 choices.
Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 8 shirts and 9 ties, and we need to determine how many outfits can be made:
(number of choices for x)*(number of choices for y)= 8*9 = 72 choices for L.
Point N:
x value: In order to construct a right triangle, N must have the same x coordinate as L (so that N is directly above L and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for N's x value.
y value: If L and N have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 9 choices for y when we chose L, so we have 9-1=8 choices for N's y value.
(number of choices for x)*(number of choices for y)=1*8=8 choices for N.
Point M:
y value: For LM to be parallel to the x axis, L and M must share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for L's y value.
x value: If L and M have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 8 choices for x when we chose L, so we have 8-1=7 choices for M's x value.
(number of choices for x)*(number of choices for y)=1*7=7 choices for M.
So we have 72 choices for L, 8 choices for N, and 7 choices for M. We need to determine how many ways we can combine L, N and M to make a triangle. It's as though we have 72 shirts, 8 ties, and 7 pairs of pants, and we need to determine the number of outfits that can be made:
(number of choices for L)*(number of choices for N)*(number of choices for M) = 72*8*7 = 4032.
The correct answer is C.
Hope this helps!
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This is same as asking that in how many ways a right angled triangle can be constructed such that one side is parallel to x axis and the x and y coordinates satisfy the inequality -3<=x<=4 and 3<=y<=11. Also these values of x and y have to be integers.
Let Sx = {-3, -2, -1, 0, 1, 2, 3, 4}.
Let Sy = {3, 4, 5, 6, 7, 8, 9, 10, 11}.
If we select 2 points from Sx say x1, x2 and 2 points from Sy say y1 and y2, we can get two lines: (1) joining (x1,y1) and (x2,y2)
(2) joining (x1, y2) and (x2, y1).
Take line (1) . You can make two right angled triangles on it with one side parallel to x axis.
(a) M(x1, y1), L(x2, y1), N(x2, y2).
(b) N(x1, y1), L(x1, y2), M(x2, y2).
Take line (2). You can make two right angled triangles on it with one side parallel to x axis.
(a) M(x1, y2), L(x2, y2), N(x2, y1)
(b) N(x1, y2), L(x1, y1), M(x2, y1).
So if we select 2 points from Sx and 2 points from Sy, we can construct 4 triangles with it satisfying the above given conditions.
Number of ways of selecting 2 points from Sx and 2 from Sy is 8C2 * 9C2 .
So the number of possible triangles with given conditions that can be constructed is 8C2 * 9C2 * 4 = 4032.
Let Sx = {-3, -2, -1, 0, 1, 2, 3, 4}.
Let Sy = {3, 4, 5, 6, 7, 8, 9, 10, 11}.
If we select 2 points from Sx say x1, x2 and 2 points from Sy say y1 and y2, we can get two lines: (1) joining (x1,y1) and (x2,y2)
(2) joining (x1, y2) and (x2, y1).
Take line (1) . You can make two right angled triangles on it with one side parallel to x axis.
(a) M(x1, y1), L(x2, y1), N(x2, y2).
(b) N(x1, y1), L(x1, y2), M(x2, y2).
Take line (2). You can make two right angled triangles on it with one side parallel to x axis.
(a) M(x1, y2), L(x2, y2), N(x2, y1)
(b) N(x1, y2), L(x1, y1), M(x2, y1).
So if we select 2 points from Sx and 2 points from Sy, we can construct 4 triangles with it satisfying the above given conditions.
Number of ways of selecting 2 points from Sx and 2 from Sy is 8C2 * 9C2 .
So the number of possible triangles with given conditions that can be constructed is 8C2 * 9C2 * 4 = 4032.
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x value: -3≤x≤4, giving us 7 spaces. Between two points on the number line
x value: -3≤x≤4, giving us 8 choices. In total points on the number line.
y value: 3≤y≤11, giving us 8 spaces. Between two points on the number line
y value: 3≤y≤11, giving us 9 choices. In total points on the number line.
Then 7 * 8 * 8 * 9 = 4032
You can apply this with any question on this sort answer will be correct always......... ( especially in right angels. )
[/b]
x value: -3≤x≤4, giving us 8 choices. In total points on the number line.
y value: 3≤y≤11, giving us 8 spaces. Between two points on the number line
y value: 3≤y≤11, giving us 9 choices. In total points on the number line.
Then 7 * 8 * 8 * 9 = 4032
You can apply this with any question on this sort answer will be correct always......... ( especially in right angels. )
[/b]
Saurabh Goyal
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You are right , I didn't notice at all I have seen one same question in OG and there its equal that's why may be,vijchid wrote:The original question had the following constraint
-3 < x < 4 and 3 < y < 11.
How come all of you decided to change the constraint -3<=x<=4 and 3<=y<=11?
but ya you are right it was a big mistake,
but now the problem is even more
answer is 5 * 6 * 6 * 7 = 1260
and we don't have that option.......... :roll:
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