emahmud wrote:
I don't understand the
OG explanation of the following problem. Is there anyone to help me out?
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
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
Answer:
RELATED VIDEO