Question

Regular pentagons A and B are similar. The apothem of pentagon A is equal to the radius of pentagon B. Compare their areas.

This question is really confusing. Any help would be great!

Answer 1

When two polygons are similar, their corresponding sides are proportional, and their corresponding angles are congruent. The areas of pentagons A and B will be in the same ratio as the squares of their side lengths.

Step-by-step explanation:

When two polygons are similar, their corresponding sides are proportional, and their corresponding angles are congruent.

In this case, regular pentagons A and B are similar. The apothem of pentagon A is equal to the radius of pentagon B.

Since the apothem of pentagon A is equal to the radius of pentagon B, their corresponding sides are proportional. This means that the ratio of their areas will be equal to the ratio of the squares of their corresponding side lengths.

Given that the apothem and radius are equal, the ratio of their areas will be equal to the square of the ratio of their side lengths:

Area of pentagon A / Area of pentagon B = (Side length of pentagon A / Side length of pentagon B)2

So, the areas of pentagons A and B will be in the same ratio as the squares of their side lengths.

Answer 2

What the question means is
Regular pentagon A is a larger regular pentagon B, in such a way that
the apothem of A coincides with the "radius" of pentagon B, which is actually the distance from the centre to a vertex of B.

The figure attached shows the situation, where regular pentagon A is the lighter figure, and pentagon B is the darker one inside A.

The ratio of the areas depends on the ratio of the apothems.
The ratio of an apothem to the "radius" is the ratio sin(108/2)=sin(54), since the interior angle of a pentagon is 108 degrees.

The ratio of areas of B to A is the square of the ratio of the apothems, namely sin^2(54)=0.809017^2=0.654508