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. 1100
C. 9900
D. 10000
E. 12100

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.

We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.

Point P:
x value: -4≤x≤5, giving us 10 choices.

y value: 6≤y≤16, giving us 11 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 10 shirts and 11 ties, and we need to determine how many outfits can be made:

(number of choices for x)*(number of choices for y)=10*11=110 choices for P.

Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.

y value: If P and Q 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 11 choices for y when we chose P, so we have 11-1=10 choices for Q's y value.

(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.

Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.

x value: If P and R 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 10 choices for x when we chose P, so we have 10-1=9 choices for R's x value.

(number of choices for x)*(number of choices for y)=9*1=9 choices for R.

So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:

(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.

The correct answer is C.

Thanks Mitch:)

Singh

Here we are talking about a right triangle whose two sides are parallel to x and y axis.

One side is parallel to y axis =>> Two vertices will have same x coordinate.
One side is parallel to X axis =>> Two vertices will have same y coordinate.

Number of x coordinates = 10*9 = 90
Number of y coordinates = 11*10 = 110

Total number of coordinates possible = 110*90 = 9900 i.e C

PS: I took almost 4 minutes to solve this problem

We need to determine the number of possible ways to construct line PR and PQ.

(1)Number of ways to construct line PR is to select pairs of x-coordinate from -4 <=x<=5 = 10 * 9 = 90

(2)Number of ways to construct line PQ is to select pairs of y-coordinate from 6 <=x<=16 = 11 * 10 = 110

Total ways to construct traingle = (1) * (2) = 90 * 110 = 9900



(Since PR is || to x-axis and PQ is || to y axis, We do not have to consider any combinations where the traingle is tilted at any angle from the axis)