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Geometry part 3

By jake becker posted August 23, 2010

One of the key things to remember with circles is that once you know one piece of information, you know everything about the circle itself. Additional angles and lengths inside are not always so simple, but it is possible to convert circumferences to areas, to radii and diameters without intermediate steps.

Arc Lengths and Sector Areas

Arc Lengths (portions of perimeters) and Sectors (pie slices) seem more complicated than they really are. Both relate directly to the internal angle at the circle’s center, represented by θ in this diagram. Here are the equations:

Sector Area = A = θ/360 * πr²

Arc Length = L = θ/360 * 2πr

θ in these formulas refer to the angle measure in degrees, not radians. If you’re using your calculator, make sure that your calculator is in the right mode.

Literally, all we are doing is finding the fraction of the circle and applying it to either the area or circumference formula, respectively. The pie is always a certain fraction of the circle, meaning the area of the pie and the arc length of the pie is also the same fraction of the circle’s area and circumference respectively.

In the above diagram, a square is inscribed in a circle, which is inscribed in another square. If the area of the larger square is 64 and the area of the green region is 4, which is larger, the green region or the yellow region?

The information we have is:

A(large square) = 64
s(large square) = 8
diameter = 8
r = 4

First, recognize that the figure is symmetrical. So while we may not explicitly be given an angle to find the sector area (not drawn) in which the yellow region resides, we do know its measure. The diagonals of a square intersect at a right angle, so we can deduce that the sector including the yellow region is 1/4 of the area of the circle. Since we know the radius…

A(large sector) = 1/4 * 16π = 4π

To find the yellow region itself, we must subtract the imaginary (not drawn) triangle from 4π. (Note that this imaginary triangle will be twice the green triangle.)

A(imaginary triangle) = 1/2 * r * r = 1/2 * 4 * 4 = 8

A(Yellow Region) = 4π – 8

The area of the green region can be found in two ways. Either we can see that it’s simply one-half of the 8 we just found, OR we can find both sides of the green triangle with the common 45-45-90 1:1:√2 formula. With a hypotenuse of 4, we derive 2√2 for each side, which yields an area of 4.

One More Problem

A circular loop of wire is attached to the two straight wires of a dipole antenna at points X and Y. Point Z is the base of the antenna where the two straight wires meet. The perimeter of the circular loop of wire is 3 feet. The center of the circular loop of wire is 2 feet from point Z. How far is point X from the center of the circular loop of wire?

The key point to remember here is that all radii are equal. That is, the distance from the center of the circle (C, not drawn) to both X and Y is the same.

Before we mentioned any if you have any one piece of information about the circle, then you have it all. This is perfect example. If we know the circumference (perimeter) of the wire, we know the radius, regardless of the presence of Z in the diagram. The information about Z is simply thrown in there to confuse you. The distance to X is actually the radius which is circumference/pi = 3/pi.

Here are just a couple examples, but keep working hard. Givens in geometry provide a series of information, not just what’s stated.

Good luck!

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Geometry Series part 2

By jake becker posted August 20, 2010

To start off, let’s quickly review the essentials. These are formulas/concepts you must know:

a² + b² = c², but only when a right triangle. If you don’t know it’s a right triangle, Pythagorean theorem does not apply!
Common special right triangles include 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples.)
45-45-90 triangles are ALWAYS in the ratio 1:1:√2
30-60-90 triangle are ALWAYS in the ratio 1:√3:2
Angles and opposite sides are in the same relative size order, but are NOT proportional.

Let’s continue with a standard diagram in which we have an equilateral triangle inscribed in a circle, which is inscribed in a square.

The center point of all three figures (triangle, circle, square) are all the same, but this is ONLY true if the triangle is equilateral (all the sides are the same length). Therefore, if given ANY piece of information about the circle, square or triangle, we can derive the rest. We draw a perpendicular line from the center to the side of the triangle.

Note that the hypotenuses of the smaller triangles are equal to the radius of the circle. We also know that the smaller triangles are each 30-60-90 because you are taking the 120-degree internal angle from the circle’s center and cutting it in two. Here are your basic conversions:

r = ½d = ½s, where s is the side of the square.
The sides of the 30-60-90 triangles become ½r : (r√3)/2 : r respectively
The side of the equilateral triangle becomes 2*(r√3)/2 = r√3

If given the area of the square, we should be able to derive essentially any other information.

Area of an Equilateral Triangle

The area of an equilateral triangle equals (s²√3)/4. Memorize this. It will save you the time of drawing a 30-60-90 triangle, solving for the base, finding the height, multiplying and dividing by 2. That was long to write, imagine how long it takes to do!

If the area of the square = 64 and we needed to find the area of the triangle, we just use the conversions above:

d = 8
r = 4
side of triangle = 4√3

Area of triangle = [(4√3)²√3]/4 = 16*3*√3 / 4 = 4*3*√3 = 12√3

Angle Relationships

Another important rule is that the interior angle created from of two radii extending to the outside of the circle is exactly twice the measure of any angle on the circle extending to those same points. In the image above, 2b = a.

Let’s take a look at this practice question:

In the figure above, a circle is inscribed in a square. If the perimeter of the square is 32 and x=35, what is the area of the shaded region?

We can determine r by knowing that the length of the square’s side. If s = 32/4 = 8, then d = 8 and r = 4. But we still need to know the interior angle. Since x=35, we know that the interior angle of the shaded region is 2(35) = 70. So A(shaded) = (2x/360) * πr² = 70/360 * 16π = 28π/9

There are infinite variations of these concepts. Be flexible in your reasoning, and practice makes perfect!

Good luck!

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Geometry Series Part 1: Circles Inscribed in Squares

By jake becker posted August 19, 2010

In this series, we will cover many types of geometric scenarios encountered on the SAT and ACT. A basic knowledge of simple formulas (area, perimeter, etc) is essential, but there are numerous shortcuts to geometry questions that will save you time. Today, we’ll explore circles inscribed squares.

Some things to remember

The center of the square is the same point as the center of the circle
Draw lines! Depending on what the question asks for, draw in lines that create simple shapes. (Squares can be turned into triangles, for example).
Shared angles will normally not be explicitly stated, unless necessary.
Trust the pictures, but not too much. Inferences must be drawn from fact. Just because it looks like 90-degrees doesn’t mean it is! (Many of these common inferences will be detailed in this series)
Lengths cannot be negative.

For circles:

d=2r and all lines from the center to the exterior equal r.
C = 2πr = πd
A = πr²
Tangent lines create right angles with the radius that meets that tangent.
If you know r, you know everything about the circle!
Use π = 22/7 with caution. Remember 22/7 > π.
The diagonal equals (length of a side * √2), since it creates 45-degree angles.
The intersection of the diagonals creates a right angle.
When a circle is inscribed inside a square, the side equals the diameter. (Inscribed means that the circle fits perfectly inside the square with its edges touching the side of the square like in the diagram)

Test your SAT math skills with this SAT multiple choice practice question.

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All About Circles

By amanda chen posted January 15, 2010

There are four main things you need to know about circles to tackle any SAT Math question.

The definition of Diameter and Radius. For a pictorial illustration, see the diagram below.
The formula for Circumference given by (diameter)(p)
The formula for Area given by (p)(radius²)
Knowing what fraction of the circle the sector is. Note that a sector is what the ‘slice’ in the circle, an example sector is the yellow wedge show in the green circle. The white part of the circle is also known as a sector, even though it doesn’t look like your typical ‘slice’.

Let’s try and apply the concepts to see if you understand.

Suppose I told you that the diameter of a circle is 12 cm. What would be its circumference? And what would be its area? Simply apply the formulas to get:

circumference – (diameter)(π)-(12)(π)-12π

area-(π)(radius)²-(π)(6)²-36π

Don’t forget that radius is half of the diameter, so if the diameter is 12 cm, the radius is 6 cm.

The question might add another step by telling you that there is another circle with diameter 6 cm and ask how many times bigger, in terms of area, is the circle with diameter 12 cm than this circle? Just because one has diameter 12 cm and the other has diameter 6 cm does not mean that the bigger circle is twice as big.

Use the formulas to figure out that the area of the new circle is 9π cm² while the area of the big circle is 36π cm², as we found out earlier. This means that the bigger circle is 4 times as large as the smaller circle.

The last concept involves applying your knowledge of circumference and area to sectors. Sectors are basically a fraction of the entire circle. If I told you that the yellow sector in the circle above had an arc of 45° that means that it is 1/9 of the entire circle, because a circle has 360°. The implications of this are two fold:

The arc length of the yellow wedge is also 1/9^th of the circumference

The area of the yellow wedge is also 1/9^th of the area of the circle

Thus, if I told you that the radius of that circle was 10 cm. Then its circumference would be 10π cm and its area would be 25π cm². Correspondingly, the arc length of the sector would be 10π/9 cm and the area of the sector would be 25π/9 cm².

If the question wants you to find the perimeter of the yellow wedge, all you have to do is add the radius twice to the arc length you found to get (10 + 10π) cm.

Remember these four concepts in mind and you’ll have the tools to solve any circle problem! For practice, you could try solving these practice problems.

A circle has area 50π cm². A second circle has half the area of the first circle. What is the diameter of the second circle? (Ans: 10 cm)
A circle has a shaded sector. The sector is 1/6 of the whole circle. What is the area of the circle if the sector has area 6π cm²? What is the diameter of the circle? (Ans: 36π cm²; 12 cm)

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