Last week we looked at questions on polygons inscribed in a circle. This week, let’s look at questions on circles inscribed in regular polygons. As noted earlier, it’s important to keep in mind that regular polygons are symmetrical figures. You need very little information to solve for anything in a symmetrical figure.

**Question 1**: A circle is inscribed in a regular hexagon. A regular hexagon is inscribed in this circle. Another circle is inscribed in the inner regular hexagon and so on. What is the area of the tenth such circle?

Statement I: The length of the side of the outermost regular hexagon is 6 cm.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

**Solution**: Thankfully, in DS questions, we don’t need to calculate the answer. We just need to establish the sufficiency of the given data. Note that we have found that there is a defined relation between the sides of a regular hexagon and the radius of an inscribed circle and there is also a defined relation between the radius of a circle and the side of an inscribed regular hexagon.

When the circle is inscribed in a regular hexagon,

Radius of the inscribed circle = (√3/2)* Side of the hexagon

When a regular hexagon is inscribed in a circle,

Side of the inscribed regular hexagon = Radius of the circle

So all we need is the side of any one regular hexagon or the radius of any one circle and we will know the length of the sides of all hexagons and the radii of all circles.

Statement I: The length of the side of the outermost regular hexagon is 6 cm.

If length of the side of the outermost regular hexagon is 6 cm, the radius of the inscribed circle is (√3/2)*6 = 3√3 cm

In that case, the side of the regular hexagon inscribed in this circle is also 3√3 cm. Now we can get the radius of the circle inscribed in this second hexagon and go on the same lines till we reach the tenth circle. This statement alone is sufficient.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Note that a hexagon has diagonals of two different lengths. The diagonals that connect vertices with one vertex between them are smaller than the diagonals that connect vertices with two vertices between them. Length of AC will be shorter than length of AD. Given the length of a diagonal, we do not know which diagonal it is. Is AC = 12 or is AD = 12? The length of the side will be different in the two cases. So this statement alone is not sufficient.

**Answer (A)**

Keep in mind that you don’t actually need to solve for an answer is DS; in fact, in some questions you will not be able to solve for the answer under the given time constraints. All you need to do is ensure that given unlimited time, you will get a unique answer.

**Question 2**: Four identical circles are drawn in a square such that each circle touches two sides of the square and two other circles (as shown in the figure below). If the side of the square is of length 20 cm, what is the area of the shaded region?

(A) 400 – 100π

(B) 200 – 50π

(C) 100 – 25π

(D) 8π

(E) 4π

**Solution**: First let’s recall that squares and circles are symmetrical figures. The given figure is symmetrical.

We don’t know any formula that will help us get the area of the curved shaded grey shape in the center. In such cases, very often what you need is to find the area of one region and subtract the area of another out of it. Here, if we subtract the area of the four circles out of the area of the square, the leftover area includes the shaded region but it includes other regions (around the corners etc) too. This is where symmetry helps us.

Notice that we can split the figure into four equal regions to get four smaller squares. Now focus on the diagram give below which shows you one such smaller square. The area around the four corners of the smaller squares is equal i.e. the area of the red region = area of the blue region = area of the yellow region = area of the green region.

Our shaded grey region has four such equal areas so

Area of the shaded grey region = Area of the smaller square – Area of one circle

Area of the shaded grey region = (10)^2 – π(5)^2 = 100 – 25π

**Answer (C)**

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*