An object is located 51 millimeters from a diverging lens. the object has a height of 13 millimeters and the image height is 3.5 millimeters. how far in front of the lens is the image located?
a. 51 millimeters
b. 13.7 millimeters
c. 189 millimeters
d. 1.12 millimeters

B. 13.7 mm

Explanation:

by similarity of rectangular triangles, we have:

(1) Δ ABC and Δ AB'C'

∴ AB/AB' = AC/AC'

∴ hypotenuse (AB)² = AC² + BC²Pythagorean theorem

⇒ AB² = (51 mm)² + (13 mm)²

⇒ AB² = 2770 mm²

⇒ AB = 52.63 mm

(2) Δ AB'C' :

∴ hypotenuse (AB')² = AC'² + B'C'²

∴ B'C' = 3.5 mm

from (1):

⇒ AB' = (AB)(AC') / AC = (52.63)(AC') / (51 mm)

⇒ AB' = 1.032AC'(3)

(3) in (2):

⇒ (1.032AC')² = AC'² + (3.5)²

⇒1.065AC'² = AC'² + 12.25

⇒ 0.065AC'² = 12.25

⇒ AC'² = 188.462

⇒ AC' = √188.462

⇒ AC' = 13.72 mm

i'm assuming that the correct answer is indeed the first one/ "a"..