For today’s post, I have two questions for you – both on polygons inscribed in a circle. You must go through the previous post based on this topic before trying these questions.

**Question 1**: Four points that form a polygon lie on the circumference of the circle. What is the area of the polygon ABCD?

Statement I: The radius of the circle is 3 cm.

Statement II: ABCD is square.

**Solution**:

Notice that you have been given that angles B and D are right angles. Does that imply that the polygon is a square? No. You haven’t been given that the polygon is a regular polygon. The diagonal AC is the diameter since arc ADC subtends a right angle ABC. Hence arc ADC and arc ABC are semi-circles. But the sides of the polygon (AB, BC, CD, DA) may not be equal. Look at the diagram given below:

Statement I: The radius of the circle is 3 cm.

This statement alone is not sufficient. Look at the two figures given above. The area in the two cases will be different depending on the length of the sides. Just knowing the diagonal AC is not enough. Hence this statement alone is not sufficient.

Statement II: ABCD is square.

This tells us that the first figure is valid i.e. the polygon is actually a square. But this statement alone doesn’t give us the measure of any side/diagonal. Hence this statement alone is not sufficient.

Using both statements together, we know that ABCD is a square with a diagonal of length 6 cm. This means that the side of the square is 6/?2 cm giving us an area of (6/?2)^2 = 18 cm^2.

**Answer (C)**

Let’s look at a more complicated question now.

**Question 2**: A regular polygon is inscribed in a circle. How many sides does the polygon have?

Statement I: The length of the diagonal of the polygon is equal to the length of the diameter of the circle.

Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

**Solution**:

In this question, we know that the polygon is a regular polygon i.e. all sides are equal in length. As the number of sides keeps increasing, the area of the circle enclosed in the regular polygon keeps increasing till the number of sides is infinite (i.e. we get a circle) and it overlaps with the original circle. The diagram given below will make this clearer.

Let’s look at each statement:

Statement I: The length of one of the diagonals of the polygon is equal to the length of the diameter of the circle.

Do we get the number of sides of the polygon using this statement? No. The diagram below tells you why.

Regular polygons with even number of sides will be symmetrical around their middle diagonal and hence the diagonal will be the diameter. Hence the polygon could have 4/6/8/10 etc sides. Hence this statement alone is not sufficient.

Statement II: The ratio of area of the polygon to the area of the circle is less than 2:3.

Let’s find the fraction of area enclosed by a square.

In the previous post we saw that

Side of the square = ?2 * Radius of the circle

Area of the square = Side^2 = 2*Radius^2

Area of the circle = ?*Radius^2 = 3.14*Radius^2

Ratio of area of the square to area of the circle is 2/3.14 i.e. slightly less than 2/3.

So a square encloses less than 2/3 of the area of the circle. This means a triangle will enclose even less area. Hence, we see that already the number of sides of the regular polygon could be 3 or 4. Hence this statement alone is not sufficient.

Using both statements together, we see that the polygon has 4/6/8 etc sides but the area enclosed should be less than 2/3 of the area of the circle. Hence the regular polygon must have 4 sides. Since the area of a square is a little less than 2/3^{rd} the area of the circle, we can say with fair amount of certainty that the area of a regular hexagon will be more than 2/3^{rd} the area of the circle. But just to be sure, you can do this:

Side of the regular hexagon = Radius of the circle

Area of a regular hexagon = 6*Area of each of the 6 equilateral triangles = 6*(?3/4)*Radius^2 = 2.6*Radius^2

2.6/3.14 is certainly more than 2/3 so the regular polygon cannot be a hexagon. The regular polygon must have 4 sides only.

**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!*