Solve the problem, if possible. Round your answer to the nearest tenth, when appropriate.
-The area of a square is $81 \mathrm{~cm}^{2}$. If the same amount is added to one dimension and removed from the other, the resulting rectangle has an area $9 \mathrm{~cm}^{2}$ less than the area of the square. How much is added and subtracted?
A) $4 \mathrm{~cm}$
B) $9 \mathrm{~cm}$
C) $3 \mathrm{~cm}$
D) Cannot be determined without additional information

Question

Quizplus · Accepted Answer

The side of the square is 9 cm (since $9^2 = 81$). Let $x$ be the amount added and subtracted. The new dimensions are $9 + x$ and $9 - x$. The area of the rectangle is $(9 + x)(9 - x) = 81 - x^2$. Given that this area is 9 cm² less than the square's area, we have $81 - x^2 = 72$. Solving for $x$, we get $x^2 = 9$, so $x = 3$.

Solve the Problem, If Possible 81 cm281 \mathrm{~cm}^{2}81 cm2 If the Same Amount Is Added to One Dimension

Solve the Problem, If Possible $81 \mathrm{~cm}^{2}$ If the Same Amount Is Added to One Dimension