The sides of a quadrilateral are 3, 4, 5, and 6. Find the length of the shortest side of a similar quadrilateral whose area is 9 times as great.

Remember that similar polygons are ones that have the same ratio for all the sides. So in order to have an area 9 times as great the sides must be 3 times as long (area is linear x linear, or 3 x 3 = 9).

(Check your answer--draw a picture if it helps!)

To find the length of the shortest side of a similar quadrilateral whose area is 9 times as great, we can use the property that if the sides of two similar figures are in a ratio of "a:b", then their areas are in a ratio of "a^2 : b^2".

Let's first find the area of the given quadrilateral. We can use Heron's formula, which states that the area of a quadrilateral with side lengths a, b, c, and d, is given by:

Area = √(s(s-a)(s-b)(s-c)(s-d))

where s is the semi-perimeter, given by:

s = (a + b + c + d)/2

In this case, the sides of the quadrilateral are 3, 4, 5, and 6. So the semi-perimeter is:

s = (3 + 4 + 5 + 6)/2 = 9

Plugging the values into Heron's formula:

Area = √(9(9-3)(9-4)(9-5)(9-6))
= √(9(6)(5)(4)(3))
= √(9 * 720)
= √(6480)
≈ 80.62

Since we want to find a similar quadrilateral whose area is 9 times as great, the area of the similar quadrilateral would be 9 * 80.62 = 725.58.

Now, let's find the sides of the similar quadrilateral. Since the areas of similar figures are in a ratio of the squares of their corresponding side lengths, the ratio of the areas is:

(a^2)/3^2 = 9

Simplifying the equation:

(a^2)/9 = 9
a^2 = 9 * 9
a^2 = 81
a = √81
a = 9

Therefore, the length of the shortest side of the similar quadrilateral is 9.