Question

the area of a square frame is 55 square inches. find the length of one side of the frame. explain.
a. to the nearest whole inch
b. to the nearest tenth of an inch
explain.
<55<, so <√55<. since 55 is closer to than to, √55 is closer to (type integers or decimals.)
b. the length of one side of the frame to the nearest tenth of an inch is in. explain.

Explanation:

Step1: Recall the area formula for a square

The area formula of a square is $A = s^{2}$, where $A$ is the area and $s$ is the side - length. Given $A = 55$ square inches, we need to solve for $s$, so $s=\sqrt{55}$.

Step2: Find the nearest whole - number approximation

We know that $7^{2}=49$ and $8^{2}=64$. Since $49<55<64$, then $7 < \sqrt{55}<8$. The difference between $55$ and $49$ is $55 - 49=6$, and the difference between $64$ and $55$ is $64 - 55 = 9$. Since $55$ is closer to $49$ than to $64$, $\sqrt{55}\approx7$ to the nearest whole inch.

Step3: Find the nearest tenth approximation

Using a calculator, $\sqrt{55}\approx7.416$. Rounding to the nearest tenth, we look at the hundredths place. Since the digit in the hundredths place is $1$ (less than $5$), we round down. So $\sqrt{55}\approx7.4$ to the nearest tenth of an inch.

Answer:

a. 7 inches. Explanation: $49<55<64$, so $7 < \sqrt{55}<8$. Since $55$ is closer to $49$ than to $64$, $\sqrt{55}$ is closer to $7$.
b. 7.4 inches. Explanation: $\sqrt{55}\approx7.416$, and rounding to the nearest tenth gives 7.4 as the digit in the hundredths place is 1 which is less than 5.