38. What is the area of a square that has a radius of 5√2 ft (from the center of the 20 square to any vertex)? A. 50 ft² B. 100 ft2 C 100/2² 0.150 ²
38. What is the area of a square that has a radius of 5√2 ft (from the center of the 20 square to any vertex)? A. 50 ft² B. 100 ft2 C 100/2² 0.150 ²
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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38
![### Geometry Practice Problem
#### Question 38:
What is the area of a square that has a radius of \( 5\sqrt{2} \) ft (from the center of the square to any vertex)?
#### Answer Choices:
A. \( 50 \text{ ft}^2 \)
B. \( 100 \text{ ft}^2 \)
C. \( 100\sqrt{2} \text{ ft}^2 \)
D. \( 150 \text{ ft}^2 \)
### Solution:
To solve the problem, we need to recall that the radius of the square (from the center to any vertex) is equal to half the length of the diagonal of the square.
Let's denote the side length of the square as \( s \). The diagonal of the square can be calculated using the Pythagorean theorem in a 45-45-90 triangle formed by two sides of the square and the diagonal.
The formula to calculate the diagonal \( d \) of a square is:
\[ d = s\sqrt{2} \]
Given the radius \( r \) is \( 5\sqrt{2} \) ft, and the radius is half the diagonal:
\[ r = \frac{d}{2} \]
\[ 5\sqrt{2} = \frac{s\sqrt{2}}{2} \]
Solving for \( s \):
\[ 10\sqrt{2} = s\sqrt{2} \]
\[ s = 10 \]
Now, calculate the area \( A \) of the square:
\[ A = s^2 \]
\[ A = 10^2 \]
\[ A = 100 \text{ ft}^2 \]
Therefore, the correct answer is:
**B. \( 100 \text{ ft}^2 \)**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d544791-a77c-4b5e-888b-2e8a4014205c%2F341d0a65-2860-490f-a166-71753a113307%2Ffd9sz4r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Practice Problem
#### Question 38:
What is the area of a square that has a radius of \( 5\sqrt{2} \) ft (from the center of the square to any vertex)?
#### Answer Choices:
A. \( 50 \text{ ft}^2 \)
B. \( 100 \text{ ft}^2 \)
C. \( 100\sqrt{2} \text{ ft}^2 \)
D. \( 150 \text{ ft}^2 \)
### Solution:
To solve the problem, we need to recall that the radius of the square (from the center to any vertex) is equal to half the length of the diagonal of the square.
Let's denote the side length of the square as \( s \). The diagonal of the square can be calculated using the Pythagorean theorem in a 45-45-90 triangle formed by two sides of the square and the diagonal.
The formula to calculate the diagonal \( d \) of a square is:
\[ d = s\sqrt{2} \]
Given the radius \( r \) is \( 5\sqrt{2} \) ft, and the radius is half the diagonal:
\[ r = \frac{d}{2} \]
\[ 5\sqrt{2} = \frac{s\sqrt{2}}{2} \]
Solving for \( s \):
\[ 10\sqrt{2} = s\sqrt{2} \]
\[ s = 10 \]
Now, calculate the area \( A \) of the square:
\[ A = s^2 \]
\[ A = 10^2 \]
\[ A = 100 \text{ ft}^2 \]
Therefore, the correct answer is:
**B. \( 100 \text{ ft}^2 \)**
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