讲解解题思路---Extraction Content: Question Number: 1. ID: 598661 Question Stem: 如图, 在菱形 ABCD 中, AB=5, tanB=3/4, 过点 A 作 AE⊥BC 于点 E, 现将△ABE 沿直线 AE 翻折至△AFE 的位置, AF 与 CD 交于点 G, 则△CFG 的面积为 ( ) Chart/Diagram Description: Type: Geometric figure. Main Elements: - Rhombus ABCD with vertices labeled A, B, C, D. Sides are not explicitly labeled with length, but indicated in the text. - Line segment AB is shown as a dashed line. Line segment AD is shown as a dashed line. - Line segment BC is shown as a horizontal line containing points B, E, C, and F. Points B, E, C, F are collinear and appear in that order from left to right. Parts of the line segment BC (from B to E) and its extension (from E to F) are shown. - Line segment AE is drawn from A to E, with a perpendicular symbol at E indicating AE ⊥ BC. - Triangle ABE is depicted. - Triangle AFE is depicted, where triangle ABE is folded along AE to obtain triangle AFE. Line segments AF and FE are shown. - Line segment CD is shown, parallel to AB. - Line segment AF intersects line segment CD at point G. - Point G is shown within the rhombus. - Triangle CFG is formed by connecting points C, F, and G. Relative Position and Direction: A is positioned above the line containing B, C, E, F. The line BCF is depicted horizontally. AE is depicted vertically relative to BCF, indicating perpendicularity. AB and CD are parallel diagonals. AD is parallel to BC. Points B, E, C, F are collinear in that order. AF intersects CD at G. Options: A. 9/2 B. 27/16 C. 36/5 D. 108/25