If the ratio of the areas of two similar rectangles is 100/81, then what is the ratio of the perimeters?
√(100/81) = ?
To find the ratio of the perimeters of two similar rectangles, we need to understand that the ratio of corresponding sides of similar figures is equal.
Let's assume that the length and width of the first rectangle are 'a' and 'b', respectively. Similarly, let the length and width of the second rectangle be 'ka' and 'kb', where 'k' is the common scale factor.
Given that the ratio of the areas of the two rectangles is 100/81, we can write:
(a * b) / (ka * kb) = 100/81
Simplifying the equation, we have:
(ab) / (k^2 * ab) = 100/81
Canceling out 'ab' from both sides, we get:
1 / k^2 = 100/81
Next, we can cross-multiply to solve for 'k':
81 = 100 * k^2
Divide both sides by 100:
81/100 = k^2
Taking the square root of both sides:
sqrt(81/100) = k
Simplifying:
9/10 = k
So, the scale factor 'k' is equal to 9/10.
Now, let's find the ratio of the perimeters. The perimeter of the first rectangle is given by:
P1 = 2(a + b)
And the perimeter of the second rectangle is:
P2 = 2(ka + kb)
Substituting the values, we have:
P1 / P2 = (2(a + b)) / (2(ka + kb))
Canceling out the common factors of 2, we get:
P1 / P2 = (a + b) / (ka + kb)
Substituting the given values of 'a', 'b', and 'k', we have:
P1 / P2 = (a + b) / (9/10 * a + 9/10 * b)
P1 / P2 = (10a + 10b) / (9a + 9b)
Simplifying further:
P1 / P2 = 10(a + b) / 9(a + b)
Canceling out the common factor of (a + b), we have:
P1 / P2 = 10/9
Therefore, the ratio of the perimeters of two similar rectangles with an area ratio of 100/81 is 10/9.