Which identity is proved by equating the area of a square with side (a+b) and the sum of its inner parts?

Correct answer: a² + 2ab + b² = (a+b)²

The direct answer is C: a² + 2ab + b² = (a+b)². Imagine a large square whose side is a+b. Divide each side into lengths a and b. The large square then contains one square of area a², one square of area b², and two rectangles, each having area ab. Adding the four inner areas gives a²+ab+ab+b² = a²+2ab+b². The area calculated from the outside is (a+b)², so both expressions are equal. A is the difference-of-squares identity and needs a square with a difference of lengths. B describes a single rectangle formed by lengths a and a+b, not the complete four-part square. C exactly matches the visual partition. D is not generally true because (a−b)² equals a²−2ab+b², not a²−b². Memory cue: a square of a sum has two identical middle rectangles, hence 2ab.