A square picture frame has an outer perimeter of 28 and an inner perimeter of 20. What is the shortest distance between any two vertices?
A. 5
B. √2
C. 2√2
D. 7
E. 7√2
The answer is
B: √2
To find the shortest distance between any two vertices of the square picture frame, we need to calculate the length of the diagonal of the square.
Let's denote the length of one side of the square as "x". Since the outer perimeter of the frame is 28, we know that the sum of the four sides is 28:
4x = 28
Dividing both sides of the equation by 4, we find:
x = 7
Thus, the length of one side of the square is 7.
We also know that the inner perimeter of the frame is 20. Since the frame is square, the inner length of each side is equal to the outer length minus twice the frame width. Let's denote the frame width as "w":
7 - 2w = 5
Solving for "w", we get:
w = 1/2
So, the frame width is 1/2.
Now, we can calculate the length of the diagonal using the Pythagorean theorem. The length of the diagonal (d) of a square with side length "s" is given by:
d = √(s^2 + s^2)
Plugging in the value of "x" (7):
d = √(7^2 + 7^2)
= √(2 * 7^2)
= √(2 * 49)
= √98
= 7√2
Therefore, the shortest distance between any two vertices of the square picture frame is 7√2. Hence, the answer is option E.
4 s = 20
s = 5