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If perimeter is 16 inches what is area

Find the area 

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4 Answers

A few notes:

As the other tutors said, more information is needed to find the exact area as different shapes have different areas.

However there are nice facts that can be shown about the area given the perimeter.

1. If the shape is apolygon with n sides, the maximum area possible occurs if the polygon is regular.

   For example in your question, if n = 4 (quadrilateral), the maximum area occurs if it is a square with 4 in. on a side yielding 16 sq. in. for the area.

2. If any shape is allowed, even ones with curved sides, then the largest area is that of a circle.

Thus we can get the maximum area achievable by using P = 2πr for the perimeter (in this case circumference) and A = πr2.

We find that r = P/(2π) so A = π(P/(2π))2 = P2/(4π). Any positive area less than this is also possible.

So in this problem the largest area possible is (16 in.)2/(4π) = 64/π sq. in. ≈ 20.37 sq. in. is important to give all the information for any problem. The shape of the polygon is essential, perimeter = p if the figure is a circle, the you'll need the formula of the circumference c = pi r^2(pi r squared) square s = side p = 4s = s + s + s s rectangle p = 2(length + width) = l + l + w + w triangle s + side p = s + s + s so basically adding all the measure of the sides together, the sum is the perimeter. Key is the measure of the sides are all equal, then

Michelle, the shape of the figure is necessary to determine the area. If the figure is a square with a perimeter of 16 inches, then each side of the square is 4 inchest in length. Area is found by multiplying the length x the width. For a square, the length and width are the same.

A = length x width

A = 4 x 4

A = 16 sq. in.

This is difficult to answer without knowing the type of object but assume its a rectangle.


16 inches=area around the rectangle=sum of all four sides; for example if the length was 5 inches and the width was 3 inches, the area would be 5+3+5+3=16

The area would be the length times the width; 5*3 or 15