Radius as Hypotenuse
Posted on
05
Apr 2022

Radius As Hypotenuse – Problems & Solutions

Welcome back to our fourth article on GMAT circles. Last time we considered inscribed angles and learned that where there is a 90-degree inscribed angle, there is a hypotenuse that is also a diameter of the circle. This time we will explore a class of problems where the radius, rather than the diameter, pulls double duty as a hypotenuse. Let’s dive right in with the following official problem.

1. Radius as Hypotenuse  – GMAT Official Problem

Semicircular archway over a flat street problemThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?

A. √2
B. 2
C. 3
D. 4√2
E. 6

Problem SolutionThis problem is a straightforward application of the Pythagorean theorem. Since we are told that the radius of the semicircle is 6 feet, we can draw a 6-foot radius from center O to the point where height h meets the semicircle. Voila – a right triangle.

h = √(62 – 22)
h = √(36 – 4)
h = √32

This is where you should stop and mark answer choice D since we are taking the square root of a number that is not a perfect square. When we simplify this radical, something will get left inside. Therefore answers B, C, and E are out (Answer A is out because √2 =/= √32), and the correct choice is D.

2. Radius as Hypotenuse Problem 1 

Let’s try something a little different:

In the xy-plane, point (r,s) lies on a circle with center at the origin. What is the value of + s²?

1. The circle has radius 2.
2. The point (2,-2) lies on the circle.

A. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are not sufficient. 

This is the first problem we’ve seen where a circle is placed on the xy-plane. In such problems, it is usually helpful to remember the basic circle principle that every point on the circle (meaning on its edge or perimeter) is equidistant from its center.

Solution

If you’re unfamiliar with these problems, statement 1 may trip you up. Is the radius of the circle sufficient to determine + ? Yes, it is. If you are concerned about the unknown positivity/negativity of the coordinates r and s, recall that the square of any number (except 0) is positive. This means that for any positive/negative combination of r and s, the sum + will have the same value. 

But what you really need here is to see that the expression + matches the famous + from the Pythagorean theorem, and in fact, it functions the exact same way.

Radius as Hypotenuse ProblemIn this setup, the radius is the hypotenuse of the right triangle with legs r and s. Therefore, applying the Pythagorean theorem, the value + represents the square of the radius. So if we know the value of the radius (2), we know the value r² + s², and statement 1 is sufficient.

Statement 2 offers that the point (√2, -√2) lies on the circle. This statement should be “easier” to evaluate than statement 1. Seeing the radicals in the coordinates ought to help you make the connection to the Pythagorean theorem if you didn’t already while evaluating statement 1. But using the principle that every point on a circle is equidistant from its center, we know that this given point (√2, -√2) is the same distance from the center as the point (r, s) in the question. Therefore if we sum the squares of √2 and -√2, the result (4) will also represent the value r² + s² we were asked about.

3. Problem 2

Let’s try one more:

In cross section, a tunnel that carries one lane of one-way traffic is a semicircle with radius 4.2 m. Is the tunnel large enough to accommodate the truck that is approaching the entrance to the tunnel?

1. The maximum width of the truck is 2.4 m.
2. The maximum height of the truck is 4 m.

A. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are not sufficient. 

This one is a little more complex. Sometimes on GMAT quant problems, it is helpful to ask why certain details were specified. In this case, we are told that the tunnel “carries one lane of one-way traffic.” This is important because if it were not the case, the truck would have to drive on one side or the other, and there’s no way it would be able to get through the tunnel. Since there is only one lane going through the tunnel, the truck can “center up” to give itself the best chance of fitting through.

Solution

This is one of those less-common DS problems where each statement on its own is clearly insufficient. If all we know is that the truck is 2.4m wide at its widest point (statement 1), it may still be too tall to fit through the tunnel. If all we know is that the truck is 4m tall at its tallest point, we don’t know whether the truck is narrow enough to make it through the tunnel while being this tall.

Problem 2 - Solution But if we combine statements 1 and 2, we can use the Pythagorean theorem to calculate the max distance of a point on the “centered up” truck from the point at the “center” of the semicircle.

Now here’s the key step: don’t calculate! Running the Pythagorean theorem with our values here would be a waste of time. As long as the value p [from the graphic] is less than 4.2 (the radius of the tunnel), the truck will fit. But for DS, we don’t have to know whether the truck will fit. All we have to know is whether the value p can be calculated, and in this case, it can be. Statements 1 and 2 together are sufficient, and the correct answer choice is C.

 

This concludes our fourth article on the GMAT’s treatment of circles. Next time we will look at circles in two different 3-dimensional shapes: cylinders and spheres.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Inscribed Angles & Inscribed Polygons In The GMAT
Posted on
29
Mar 2022

Inscribed Angles & Inscribed Polygons In The GMAT

Welcome back to our third article on GMAT circles. In the second article, we explored central angles, sectors, and arcs. This time we will introduce another kind of angle: the inscribed angle.

An inscribed angle is an angle drawn by using line segments to connect one point on a circle to two other points on the same circle, as in the graphic below:

Inscribed AngleLike a central angle, an inscribed angle creates a “wedge” shape, like a triangle where one side is rounded. The rounded side is an arc of the circle. For a central angle, the measure of the angle corresponds to the measure of the associated arc in a 1:1 relationship. For an inscribed angle, the measure of the angle corresponds to the measure of the associated arc in a 1:2 relationship. A 30 degree inscribed angle creates a 60-degree arc on the other side of the circle. A 60-degree inscribed angle creates a 12- degree arc on the other side of the circle. And, importantly, a 90-degree inscribed angle creates a 180-degree arc (half a circle or a semicircle) on the other side of the circle.

1. Inscribed Angle – GMAT Official Guide Problem

GMAT problems rarely use the term “inscribed angle” or feature an inscribed angle in isolation. Usually, the inscribed angle is part of an inscribed polygon, a polygon drawn inside a circle so that its vertices are points on the circle. Take a look at this official GMAT problem:

Inscribed Angle Official GMAT ProblemIn the figure shown, the triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

 

 

A. 15π
B. 12π
C. 10π
D. 7π
E. 5π

The problem refers not to an angle inscribed in a circle but to a triangle inscribed in a semicircle. Still, knowing the “1:2” factor of relationship between an inscribed angle and its associated arc is the key to solving this problem. Your logic might go something like this:

  1. This is a semicircle or a 180-degree arc.
  2. The angle at point B “opens up” to the straight edge of the semicircle, which is like the diameter of a circle. Another semicircle or 180-degree arc could be drawn across from this angle so that it makes a whole circle with the existing piece.
  3. Since the measure of an inscribed angle is 1/2 the measure of the arc it “creates” on the other side of the circle, the angle at point B is a 90-degree angle, and the triangle is a right triangle. 

At this point, your attention should return to the given information about the lengths of line segments AB and BC, which we now know to be the legs of a right triangle. These legs have lengths 6 and 8, which have a 3:4 relationship. Therefore we are looking at a 3-4-5 triangle, and the length of the hypotenuse is 10.

Finally, you must recall that this hypotenuse is the diameter of the circle. Therefore the diameter of the whole circle is 10. However, marking answer choice C would be a mistake, since we were asked for the length of arc ABC, where arc ABC is a semicircle (half a circle). So your final step is to divide your diameter of 10π by 2, leading you to the correct answer choice: E.

2. Inscribed Square – GMAT Official Guide Problem

Let’s try another problem, this time with an inscribed square:

Inscribed Square GMAT Official Guide ProblemThe figure shows a drop-leaf. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?
A. 1/2
B. 
√2/2
C. 1
D. √2
E. 2

Notice that the problem doesn’t mention “a square inscribed in a circle,” but that is nonetheless what we have here. Many GMAT quant problems create scenarios that correspond to some mathematical phenomenon without using the math language. In this case, the fact that we are dealing with a square inscribed in a circle is relatively easy to see.

As in the previous problem, we are asked for a value of the circle (this time it is the radius instead of an arc length) but given only information about the inscribed shape: a square. As in the previous problem, the key is realizing that with any 90-degree inscribed angle, the line segments forming the angle are legs of a right triangle whose hypotenuse is also a diameter of the circle.

Using the Pythagorean theorem, the hypotenuse of this triangle (or the diagonal of the inscribed square) is √2. As before, forgetting to divide this value by 2 (since we were asked for the radius, not the diameter) will lead you to an incorrect answer choice. Don’t trip at the finish line. The value you need is √2 /2, answer choice B.

Here is a related problem:

Square inscribed in a circle problemIf rectangle ABCD is inscribed in the circle above, what is the area of the circular region?

A. 36.00
B. 42.25
C. 64.00
D. 84.50
E. 169.00

Again, we are asked for a value of the circle (its total area) but given only information about the inscribed rectangle. For our purposes, this rectangle is just as good as the square in the previous problem. With the square, we only needed the length of one side, because we know that all four sides are the same length. With a rectangle, we need both the length and the width in order to calculate the diagonal – the diameter of the circle – via the Pythagorean theorem. If you know your Pythagorean triples (like 3-4-5), you may realize immediately that the diagonal of this rectangle is 13.

D = √(5² + 12²)
D = √(25 + 144)
D = √169
D = 13

Now that we have the circle’s diameter, we can solve for its area. The radius of the circle is 13/2 or 6.5, and since Area = r², the square of 13/2 or 6.5 will be the coefficient of in the correct answer choice. It would be a waste of time to fully multiply out 6.5 * 6.5. We know that it will be of form __.25, and the only answer choice that matches this is B.

3. Data Sufficiency – GMAT Official Guide Problem

Let’s transition to data sufficiency for one final problem. Using the diagonal/diameter relationship in the previous problems, it would be possible to construct a variety of DS problems. But some DS inscription problems rely on another property of inscribed polygons.

Square ABCD is inscribed in circle O. What is the area of square region ABCD?

  1. The area of circular region O is 64π.
  2. The circumference of circle O  is 16π.

Solution

To answer this problem, all you need to know is that there is only one way to inscribe a square in a circle. The vertices of the square must lie on the circle. The perimeters and areas of the square and the circle will scale together. This means that if we know any value for either shape, we can calculate every value for both of them. Therefore each statement on its own is sufficient, and the answer to this problem is D.

As long as any polygon can be established as regular (having sides of equal length and angles of equal measure), there is only one way to inscribe it in a circle. A square is a regular quadrilateral, so this works for squares every time. But this same problem could have used a regular pentagon, a regular hexagon, or any regular polygon you like, and the correct answer would still be D. The regularity of the polygon is sufficient – and necessary – for this to work. If the regularity of the polygon cannot be established, then there are an infinite number of ways to inscribe it in a circle, each with its own unique area and perimeter.

It is also possible to flip the relationship and inscribe a circle inside a polygon. A related term is circumscription. The shape on the inside is inscribed in the shape on the outside. The shape on the outside is circumscribed around the shape on the inside. GMAT problems where the circle is on the inside usually use a square, so that the diameter of the circle is equal to the length of each side of the square. Such problems tend to be of lower difficulty level.

 

This concludes our third article on the GMAT’s treatment of circles. Next time we will look at what happens when the radius – rather than the diameter – pulls double duty as the hypotenuse of a right triangle.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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