A triangle is bounded by the lines y=0, y=2x and y=-x/2+k, where k is positive. Find the value of k knowing that the area of the triangle is 80

Answer 1

Area, A = base•height/2

The intersection between y=2x and -x/2+k is (2k/5, 4k/5) which makes the height 4k/5.

The intersection between y = -x/2 +k and y = 0 is (2k, 0) which makes the base 2k

80 = 2k(4k/5)/2

80 = 4k²/5

k²=(5/4)(80)=100

k = 10

Answer 2

y=2x and y=-x/2 + k are perpendicular lines

they have negative inverse slopes: -1/2 is the negative inverse of 2

they form a right triangle

sides of the right triangle are 2k, 2k/sqr5 and 4k/sqr5

Area = (2k/sqr5)(4k/sqr5)/2 = 80

(4/5)k^2 = 80

k^2 = (5/4)80 = 200

k = 10

it helps to graph the triangle. It has vertices (0,0), (2k,0) and (2k/5, 4k/5)

(0,0), (20,0) and (4,8)

perpendicular

sides are sqr(16+64) = sqr80

and sqr(16^2+8^2) = sqr320

area = (sqr80)(sqr320)/2 = sqr(25600)/2 = 160/2 = 80