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Right triangle PQR is to be constructed in the xy-plane so t
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19 Aug 2020, 12:09
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Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties could be constructed?
Re: Right triangle PQR is to be constructed in the xy-plane so t
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19 Aug 2020, 13:38
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Carcass wrote:
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties could be constructed?
(A) 110 (B) 1,100 (C) 9,900 (D) 10,000 (E) 12,100
TOUGH!!
Take the task of building triangles and break it into stages.
Stage 1: Select any point where the right angle will be (point P). The point can be selected from a 10x11 grid. So, there 110 points to choose from. This means that stage 1 can be completed in 110 ways.
Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R. The 2 legs of the right triangle are parallel to the x- and y-axes. The first point we select (in stage 1) dictates the y-coordinate of point R. In how many ways can we select the x-coordinate of point R? Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1). So, there are 9 coordinates to choose from. This means that stage 2 can be completed in 9 ways.
Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q. The 2 legs of the right triangle are parallel to the x- and y-axes. The first point we select (in stage 1) dictates the x-coordinate of point Q. In how many ways can we select the y-coordinate of point Q? Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1). So, there are 10 coordinates to choose from. This means that stage 3 can be completed in 10 ways.
So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = 9900
Re: Right triangle PQR is to be constructed in the xy-plane so t
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20 Aug 2020, 16:17
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-4<=x<=5 6<=y<=16
Let's find out how many possible integer values of x and y we'll have available to choose from
We have 5-(-4)+1=10 x-values to choose from 16-6+1=11 y-values to choose from
Event1: choose an x coordinate for P Event2: choose a y coordinate for P Event3: choose an x coordinate for R Event4: choose a y-coordinate for R Event5: choose an x-coordinate for Q Event6: choose a y-coordinate for Q
Possible outcomes of each Event |Event1|: 10 |Event2|: 11 |Event3|: 9 (b/c we can't choose the same coordinate as P in Event1) |Event4|: 1 (since PR is parallel to the x-axis we have to choose the same y-coordinate as P in Event2) |Event5|: 1 (since the right angle occurs at P, Q must be at the same x-coordinate as P in Event1) |Event6|: 10 (since we can't choose the same y-coordinate as P, we have to chose from the remaining 10)
Notice that the number of outcomes of each event are independent of each other. Regardless of what outcome take place the number of outcomes of each event is the same.
We can use Fundamental Principle of Counting.
Total Outcomes Overall= |Event1|*|Event2|*|Event3|*|Event4|*|Event5|*|Event6| = 10*11*9*1*1*10 = 9900
Re: Right triangle PQR is to be constructed in the xy-plane so t
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23 Aug 2020, 03:17
1
x-axis: P and Q will always have same x coordinate so lets not consider Q in this scenario
P : can have values from -4 to 5. So, P can be placed in any one of the 10 coordinates R : can also have values from -4 to 5 but P has already taken a point and R cannot take the same position because then it wont be a triangle. So R can be placed in one of the 9 coordinates.
So in x-axis alone you can have 10*9 bases(PR) of the triangle.
y-axis : P and R will always have same x coordinate so lets not consider R in this scenario
P : can have values from 6 to 16. So, P can be placed in any one of the 11 coordinates Q : can also have values from 6 to 16 but P has already taken a point and Q cannot take the same position because then it wont be a triangle anymore. So Q can be placed in one of the 10 coordinates.
So in y-axis alone you can have 11*10 bases(PQ) of the triangle.
Thus total triangle you can have is 9*10*11*10 = 9900.
Re: Right triangle PQR is to be constructed in the xy-plane so t
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10 Mar 2024, 02:31
1
To approach this, we have to find ways to select each of the coordinates (x,y) of Points P, Q, and R.
Such as P, Q, and R follow the x and y constraints
Anyways, first let's find out the total possible points we can select for both x, and y
for x, we have 10 possible points for y, we have 11 possible points
Now, Let's check the number of ways Q can take values.
X coordinate of Q = 10C1 ways Y coordinate of Q = 11C1 ways
Q is now dealt with
For P, the x coordinate is now fixed (as PQR is a right angle triangle and QP is perpendicular to PR, which is parallel to the X axis) because QP is parallel to the y-axis
but we still need to decide a y coordinate -
Therefore, Y coordinate of P = 10C1 ways (One point is occupied by Q)
Now for R. We know, its y coordinate is fixed as PR is parallel to X axis
Therefore, for its Y coordinate = Y coordinate of R = 9C1 ways (one point is occupied by P)
Therefore, total combinations = [(X coordinate of Q = 10C1 ways) * (Y coordinate of Q = 11C1 ways)] * [Y coordinate of P = 10C1 ways] * [Y coordinate of R] = 10*11*10*9 = 9,900