## Inscribed and circumscribed

One more sophisticated type of geometric diagram involves polygons “inside” circles or circles “inside” polygons. When a polygon is “inside” a circle, every vertex must lie on the circle:

In this diagram, the irregular pentagon ABCDE is **inscribed** in the circle, and the circle is **circumscribed** around the pentagon. We can also say: the circle **circumscribes** the pentagon. The word “inscribed” describes the inside shape, and the word “circumscribed” describes the outside shape. Here’s another diagram with the polygon on the outside.

Notice, now, that each side of this irregular pentagon is **tangent** to the circle. Now, the pentagon is **circumscribed** around the circle, and the circle is **inscribed** in the **pentagon**. In both cases, the outer shape circumscribes, and the inner shape is inscribed.

## Triangles

As is the case repeatedly in discussions of polygons, triangles are a special case in the discussion of inscribed & circumscribed. **Every single possible triangle can both be inscribed in one circle and circumscribe another circle**. That “universal dual membership” is true for no other higher order polygons —– it’s only true for triangles. Here’s a small gallery of triangles, each one both inscribed in one circle and circumscribing another circle.

Notice that, when one angle is particularly obtuse, close to 180°, the size difference between the circumscribe circle and the inscribed circle becomes quite large. Notice, also: in the case of a right triangle, the second image, the hypotenuse of the triangle is the diameter of the circumscribed circle. We will return to this point.

## Quadrilaterals

Many quadrilaterals can be neither inscribed in a circle nor circumscribed by a circle: that is it say, it is impossible to construct a circle that passes through all four vertices, and it is also impossible to construct a circle to which all four sides are tangent.

Some quadrilaterals, like an oblong rectangle, can be inscribed in a circle, but cannot circumscribe a circle. Other quadrilaterals, like a slanted rhombus, circumscribe a circle, but cannot be inscribed in a circle.

An elite few quadrilaterals can both circumscribe one circle and be inscribed in another circle. Of course, the square (below, left), the most elite of all quadrilaterals, has this property. Another example is the “right kite” (below, right), a kite with a pair of opposite right angles:

While this “dual membership” is true for all triangles, it’s limited to some special cases with quadrilaterals.

## Higher Polygons

What was true for quadrilaterals is also true for all higher polygons.

a. Most, the vast majority, can neither circumscribe a circle nor be inscribed a circle.

b. Some can be inscribed in a circle, but cannot circumscribe a circle.

c. Some can circumscribe a circle, but cannot be inscribed in a circle.

d. An elite few can both circumscribe a circle and be inscribed in a circle.

That last category, the elite members, always includes the regular polygon. Just as all triangles have this “dual membership”, so do all regular polygons. Here’s a gallery of regular polygons, with both their inscribed circle and their circumscribing circle.

Obviously, as the number of sides increases, the sizes of the two circles become closer and closer.

GMAT questions about inscribed and circumscribed polygons are rare, and may test both your understanding of terminology and your visualization skills by describing the geometric situation (e.g. “rectangle JKLM is inscribed in a circle”) and ** not** giving you a diagram.

## A special case: a triangle inscribed in a semicircle

This is one special case the GMAT loves. It appears in the OG13 (DS #118) and could easily appear somewhere on the Quant section of your real GMAT.

If all you know is that KL is the diameter of the circle, that’s sufficient to establish that ∠J = 90° and that triangle JKL is a right triangle with KL as the hypotenuse. Alternately, if all you know is that triangle JKL is a right triangle with KL as the hypotenuse, that’s sufficient to establish that arc KJL is a semicircle and KL is a diameter. This is a powerful set of ideas because the deductions run both ways and because it inextricably connects two seemingly disparate ideas.

BTW, this post is the fourth in a series of five articles. Here’s the whole series.

1) Introduction to Circles on the GMAT

2) GMAT Geometry: Circles and Angles

3) Circle and Line diagrams on the GMAT

4) Inscribed and Circumscribed Circles and Polygons on the GMAT

5) Slicing up GMAT Circles: Arclengths, Sectors, and Pi

## Practice question

1) In the diagram above, S is the center of the circle. If QS = 5 and QR = 6, what is PQ?

A. 7

B. 8

C. 9

D. 10

E. 11

## Practice question explanation

1) First of all, QS is a radius, so if QS = 5, that means PS = SR = 5 and the diameter PR = 10. Furthermore, because PR is a diameter, that means triangle PQR is a right triangle, with ∠PQR = 90°. We know two sides of this right triangle: QR = 6, and PR = 10, so we can use the Pythagorean Theorem to find the third side.

(PQ)^2 + (QR)^2 = (PQ)^2

(PQ)^2 + (6)^2 = (10)^2

(PQ)^2 + 36 = 100

(PQ)^2 = 100 – 36 = 64

PQ = sqrt{64} = 8

Answer = **B**

how can we know the raduia of a circle when it has a 60-60-60 triangle inside.

Is it possible to INSCRIBE a circle in ANY quadrilateral, not counting rectangles. If not, why? Please give example.

Thanks very much

Dear Lou,

It’s possible to inscribe a circle in a square, in any rhombus, and even in any kite, but typically a circle will not touch all four sides of most rectangles, most trapezoids, and most irregular quadrilaterals. In particular, if there’s anything “long & skinny” about the quadrilateral, such as a long skinny rectangle, then any circle small enough to fit across the “skinny” length is not going to be big enough to reach the “long” distance.

Mike 🙂

Great post sir.

You have been an inspiration, in my life. i am also , following your blog post related to meditation and yoga . it really does helps me out, to perform better in the exams. tremendously calms my mind.

Regarding the question above, a quicker way to solve this post is, since the line segment qs meets at point s, which is the hypotenuses of the triangle, we can conclude that, QS has to be half the value of PR. Since QS is 5, PR has to be 10. 🙂 then applying the Pythagorean theorem! 🙂 bam wam thank you mam! problem solved under 10 seconds! 🙂

Dear Manish,

That’s precisely what I did in my solution. You see, I have to show a few more steps, because I am demonstrating not only for the students who get the solutions instantly, but also for the students who need to see step-by-step solutions. It’s good that you can see quick solutions. Remember, my friend, that demonstrating the quickest possible solution is not necessarily the same as teaching.

Mike 🙂

Dear Mike,

Oh. i am sorry. maybe i overlooked the explanation part which you mentioned above. maybe,i was bit too excited.

P.S:

Thank you for all the efforts you put in @magoosh. these blog post help me a lot! 🙂

Manish,

My friend, I certainly know what it is to get excited about math. 🙂 In terms of the GMAT, that’s a healthy feeling to cultivate! 🙂 You are quite welcome, my friend! It means a lot to me to know that you have found these blogs helpful.

Mike 🙂