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In which area model is a² − 16 directly visible?

Answer and explanation

Correct answer: A rectangle with sides a + 4 and a − 4

The correct answer is C: a rectangle with sides a + 4 and a − 4. Use the difference-of-squares identity (p + q)(p − q) = p² − q². Here p = a and q = 4, so (a + 4)(a − 4) = a² − 4² = a² − 16. The area model therefore has two side lengths, one increased by 4 and the other decreased by 4. Option A represents (a + 4)² = a² + 8a + 16, which contains positive middle terms and is not the given expression. Option B represents (a − 4)² = a² − 8a + 16, which also has a middle term and a positive constant. Option D has area 16a, not a² − 16. Notice that the difference-of-squares expression has no middle term because the +4a and −4a contributions cancel. Memory cue: a² − b² always factors as (a+b)(a−b), so think of a rectangle with sum and difference as its sides.

Related tags

Difference Of SquaresArea ModelAlgebraic IdentitiesVisual Models Of IdentitiesExploring Algebraic IdentitiesMathematicsClass 9 Mcq

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