<span>(3, 4.5) and (3, 3)
The midsegment of a triangle is a line connecting the midpoints of two sides of the triangle. So a triangle has 3 midsegments. Since you want the midsegment that's parallel to LN, we need to select the midpoints of LM and MN. The midpoint of a line segment is simply the average of the coordinates of each end point of the line segment. So:
Midpoint LM:
((0+6)/2, (5+4)/2) = (6/2, 9/2) = (3, 4.5)
Midpoint MN:
((6+0)/2, (4+2)/2) = (6/2, 6/2) = (3, 3)
So the desired end points are (3, 4.5) and (3, 3)</span>
Answer:
Step-by-step explanation:
k(k+1)(k+2) - 3k(k+1)
k and (k+1) are common to both terms.
k(k+1)(k+2) - 3k(k+1) = k(k+1)[(k+2) - 3] = k(k+1)(k-1)
Area of triangle = 1/2 x base x height
Let the base be "b"
90 = 1/2 x 10 x b
90 = 5b
b = 18 ft
Base = 18 ft
True
∠2 and ∠5 are both vertical at the vertex of the 2 straight lines
