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What is a figure with perimeter of 14 and area of 8

User Alex Pi
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1 Answer

5 votes

Answer:

Explanation:

Let's start by using some formulas to relate the perimeter and area of a figure.

For a square, the perimeter is 4 times the length of one side (P = 4s) and the area is the length of one side squared (A = s^2).

If we let s be the length of one side of the square, then we can write two equations:

P = 4s

A = s^2

We're given that the perimeter is 14, so we can substitute that in for P:

14 = 4s

Solving for s, we get:

s = 3.5

Now we can find the area by substituting s = 3.5 into the formula for area:

A = s^2 = 3.5^2 = 12.25

However, this area doesn't match the given area of 8. So, we need to try a different shape.

Let's try a rectangle. For a rectangle, the perimeter is twice the length plus twice the width (P = 2l + 2w) and the area is the length times the width (A = lw).

We're given that the perimeter is 14, so we can write:

14 = 2l + 2w

Simplifying:

7 = l + w

We want the area to be 8, so we can write:

8 = lw

Substituting 7 - w for l (from the previous equation), we get:

8 = w(7 - w)

Expanding:

8 = 7w - w^2

Rearranging and factoring:

w^2 - 7w + 8 = 0

This factors to:

(w - 1)(w - 8) = 0

So the possible values for w are 1 and 8. If w is 1, then l is 6, which means the perimeter is 14 but the area is only 6. If w is 8, then l is also 6, which means the perimeter is 14 and the area is 8.

Therefore, a rectangle with dimensions 6 by 1 (or 1 by 6) has a perimeter of 14 and an area of 8.

User Smeegol
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