AMC 10A 2025 (Problem 15)In the figure below, ABEFABEFABEF is a rectangle, AD‾⊥DE‾\overline{AD}\perp\overline{DE}AD⊥DE, AF=7AF=7AF=7, AB=1AB=1AB=1, and AD=5AD=5AD=5. What is the area of △ABC\triangle ABC△ABC?(A) 38\text{(A)}\;\frac38(A)83(B) 49\text{(B)}\;\frac49(B)94(C) 1813\text{(C)}\;\frac18\sqrt{13}(C)8113(D) 715\text{(D)}\;\frac7{15}(D)157(E) 1815\text{(E)}\;\frac18\sqrt{15}(E)8115Related TopicsCoreToolkit 93 — Similar TrianglesMinorToolkit 21 — Quadratic FormulaHints (5)Hint 1(AA)⇒△ABC∼△EDC(AA)\quad\Rightarrow\quad \triangle ABC\sim\triangle EDC(AA)⇒△ABC∼△EDCx7−y=y5−x\frac{x}{7-y}=\frac{y}{5-x}7−yx=5−xyHint 2x2=y2+1x^2=y^2+1x2=y2+1Hint 3y2+17−y=y5−y2+1\frac{\sqrt{y^2+1}}{7-y}=\frac{y}{5-\sqrt{y^2+1}}7−yy2+1=5−y2+1yHint 45y2+1−y2−1=7y−y25\sqrt{y^2+1}-y^2-1=7y-y^25y2+1−y2−1=7y−y25y2+1=7y+15\sqrt{y^2+1}=7y+15y2+1=7y+125(y2+1)=49y2+14y+125(y^2+1)=49y^2+14y+125(y2+1)=49y2+14y+10=24y2+14y−240=24y^2+14y-240=24y2+14y−240=12y2+7y−120=12y^2+7y-120=12y2+7y−12y=−7±49+57624=−7±2524y=\frac{-7\pm\sqrt{49+576}}{24}=\frac{-7\pm25}{24}y=24−7±49+576=24−7±25Since y>0y>0y>0:y=−7+2524=34y=\frac{-7+25}{24}=\frac34y=24−7+25=43Hint 5[ABC]=y⋅12=38[ABC]=\frac{y\cdot1}{2}=\frac38[ABC]=2y⋅1=83Final Answer(A) 38\frac3883
