Find the area of a triangle whose sides are along the lines x=-5 y=0 and 3x+5y=20
Answers
Answer:
area = 245 / 6
≈ 40.83 (approx)
Step-by-step explanation:
The triangle is a right angle triangle, as the side along y=0 (the x-axis) is horizontal and the side along x = -5 is vertical.
The vertex with the right angle is at A = (-5, 0) where these two lines meet.
Let B be the other vertex on the line y = 0, so B = (x,0) for some x. As B is on the line 3x+5y=20, we have
3x + 5×0 = 20 => 3x = 20 => x = 20/3.
So B = ( 20/3, 0 ).
The base of our triangle is then
base = 20/3 - (-5) = 20/3 + 5 = 20/3 + 15/3 = 35/3.
Let C be the other vertex on the line x = -5, so C = (-5, y) for some y. As C is on the line 3x+5y=20, we have
3×-5+5y=20 => -15+5y=20 => 5y = 35 => y = 7.
So C = ( -5, 7 ) and the height of our triangle is
height = 7 - 0 = 7.
Finally,...
Area = 1/2 × base × height = 1/2 × 35/3 × 7 = 245 / 6 ≈ 40.83