In triangle PQR, PQ = 12 units and PR = 16 units. Point S is on side ...
To find the distance between the midpoints of segments QS and PR, we need to find the coordinates of points Q, R, S, and M.
Let's assume that point P is at the origin (0,0) on the coordinate plane. Since PQ = 12 units, point Q must be at (0,12). Similarly, since PR = 16 units, point R must be at (16,0).
Since PQ = PS, point S must be at the same y-coordinate as Q, which is (0,12).
To find the coordinates of the midpoint M of PR, we can use the midpoint formula:
M = ((x1 + x2)/2, (y1 + y2)/2)
Substituting the coordinates of P and R into the formula, we get:
M = ((0 + 16)/2, (0 + 0)/2)
M = (8, 0)
Now that we have the coordinates of points Q, R, S, and M, we can find the distance between the midpoints of segments QS and PR.
Finding the midpoint of segment QS:
To find the midpoint of segment QS, we need to find the average of the x-coordinates and the average of the y-coordinates of points Q and S.
x-coordinate of midpoint = (0 + 0)/2 = 0
y-coordinate of midpoint = (12 + 12)/2 = 12
Therefore, the midpoint of segment QS is (0,12).
Finding the midpoint of segment PR:
To find the midpoint of segment PR, we need to find the average of the x-coordinates and the average of the y-coordinates of points P and R.
x-coordinate of midpoint = (0 + 16)/2 = 8
y-coordinate of midpoint = (0 + 0)/2 = 0
Therefore, the midpoint of segment PR is (8,0).
Calculating the distance between the midpoints:
Using the distance formula, we can calculate the distance between the midpoints of segments QS and PR:
distance = √((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates of the midpoints into the formula, we get:
distance = √((8 - 0)^2 + (0 - 12)^2)
distance = √(8^2 + (-12)^2)
distance = √(64 + 144)
distance = √208
distance ≈ 14.42
Therefore, the distance between the midpoints of segments QS and PR is approximately 14.42 units.
In triangle PQR, PQ = 12 units and PR = 16 units. Point S is on side ...
Suppose T is the midpoint of QS.
PS = PQ or Δ PQS is an isosceles triangle. Therefore, PT is perpendicular to QS. Consider Δ PTR. Angle T is 90 degrees. Therefore, T lies on the circumference of the circle with PR as diameter or diameter of the circle is 16. MT is the radius of the circle. Therefore, the distance between the midpoints of segments QS and PR is equal to radius of the circle = 8 units. Hence, (C).
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