(05.03 MC) In the figure below, ∠ABC ≅ ∠DEC and ∠GFE ≅ ∠DCE. Point C is the point of intersection between segment AG and segment BD, while point E is the point of intersection between segment AG and segment DF. The figure shows a polygon comprised of three triangles, ABC, DCE, and GFE. Prove ΔABC ∼ ΔGEF. (10 points)

sites:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

what is the information provided?

step-by-step explanation:

the answer is:

mean: 6

mad: 2.14