In a rectangle ABCD, AC is one of the diagonals. If AC+AB=3AD and AC-AD=4 units, then what is the area of the triangle ?

Explanation

Let AD=x, AB=y, and AC=d. We have d+y=3x and d-x=4 \implies d=x+4. Substituting d, x+4+y=3x \implies y=2x-4. In right \triangle ADC, x^2+y^2=d^2 \implies x^2+(2x-4)^2=(x+4)^2. Solving this quadratic gives x=6. Thus, y=8. The area of the rectangle is 6 \times 8 = 48. (Note: The English text asks for the area of the "triangle", which is a typo for "rectangle" as confirmed by the Hindi version "आयत").

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