A regular polygon is a polygon with all sides equal in length, and all angles equal in measure.
The total number of degrees in the interior of an -sided polygon is
For example, a six-sided polygon (or hexagon) has (6 – 2)·180 = 4·180 = 720 degrees in its interior. Therefore each angle of a regular hexagon has
= 120 degrees.
For those of us that do not like to memorize formulas, there is a quick visual way to determine the total number of degrees in the interior of an -sided polygon. Simply split the polygon up into triangles and quadrilaterals by drawing nonintersecting line segments between vertices. Then add 180 degrees for each triangle and 360 degrees for each quadrilateral. For example, here is one way to do it for a hexagon.
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Since the hexagon has been split up into 2 triangles and 1 quadrilateral, the hexagon has 2(180) + 360 = 720 degrees. This is the same number we got from the formula.
To avoid potential mistakes, let me give a picture that would be incorrect.
The above figure cannot be used to compute the number of interior angles in the hexagon because segment ̅̅̅̅ is “crossing through”
segment ̅̅̅̅̅.
Now let’s draw a segment from the center of the hexagon to each vertex of the hexagon.
We see that the central angles formed must add up to 360 degrees.
Therefore each central angle is 60 degrees as shown in the figure above.
In general, the number of degrees in a central angle of an -sided polygon is .
It is worth looking at a regular hexagon in a bit more detail.
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Each of the segments just drawn in the previous figure is a radius of the circumscribed circle of this hexagon, and therefore they are all congruent. This means that each triangle is isosceles, and so the measure of each of the other two angles of any of these triangles is
= 60. Therefore each of these triangles is equilateral. This fact is worth committing to memory.
Now try to solve each of the following problems. The answers to these problems, followed by full solutions are at the end of this lesson. Do not look at the answers until you have attempted these problems yourself.
Please remember to mark off any problems you get wrong.
L EVEL 2: G EOMETRY
3. In the figure above, one side of a triangle is extended. Which of the following is true?
(A) (B) (C) (D) (E)
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L EVEL 4: G EOMETRY
4. If a square has a side of length and a diagonal of length , what is the value of ?
(A) 5 (B) 10 (C) 20 (D) √ (E) √
5. In the figure above, the center of the circle is and ̅̅̅̅
is tangent to the circle at . What is the area of the shaded region to the nearest tenth?
6. What is the area of a square whose diagonal has length √ ?
7. In the figure above, if , , and , what is the value of ?
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L EVEL 5: G EOMETRY
8. In the triangle above, and . What is the value of ?
9. In the figure above, what is the equation of line ? (A)
(B) √ (C) √ (D) √ (E) √
10. In the figure above, is a regular hexagon and . What is the perimeter of rectangle to the nearest tenth?
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11. A diagonal of a rectangle forms an angle of measure with each of the two longer sides of the rectangle. If the length of the shorter side of the rectangle is 12, what is the length of the diagonal?
(A) 26 (B) 24 (C) 18 (D) √ (E) √
12. In the figure above, what is in terms of ? (A)
(B) (C) (D) (E)
Answers
1. 46 5. 2.9 9. E
2. E 6. 25 10. 29
3. D 7. 27.5 11. B
4. D 8. 12 12. D
Full Solutions
4.
* We begin by drawing a picture
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Solution using a 45, 45, 90 triangle: Note that since the sides of a square are congruent, each triangle is an isosceles right triangle. This is the same as a 45, 45, 90 right triangle. So we need x + 10 = √ We can now either solve this algebraically or by starting with choice (C) and using our calculator. We will do the more difficult algebraic method and leave the second method to the reader.
x + 10 = √ x + 10 = √ √ √ = 10 – 5√
x(√ = 10 – 5√
x = – √ √ ~ 7.07
Putting 5√ into your calculator gives the same output. So the answer is choice (D).
Solution using the Pythagorean Theorem: We have (x + 5)2 + (x + 5)2 = (x + 10)2
x2 + 10x + 25 + x2 + 10x + 25 = x2 + 20x + 100 2x2 + 20x + 50 = x2 + 20x + 100
x2 = 50
x = √ = √ = √ √ = 5√ , choice (D).
Some clarification: (x + 5)2 = (x + 5)(x + 5) = x2 + 5x + 5x + 25. Therefore (x + 5)2 = x2 + 10x + 25.
5.
* A tangent line to a circle is perpendicular to the radius so that the triangle is a 30, 60, 90 right triangle. Since the side opposite 30 is 6, the side opposite 60 is √ . In a right triangle we can think of the two legs as the base and the height. So the area of the triangle is
A = bh = (6)· √ . = √ .
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We already found that the radius of the circle is √ . Thus the area of the circle is
A = πr2 = π( √ )2 = π·36·3 = 108π.
The sector shown is of the entire circle. So the area of the sector is A = ( )·108π = 9π.
The area of the shaded region is the area of the triangle minus the area of the sector.
A = √ – 9π ~ 2.90258065 Rounding to the nearest tenth gives us 2.9.
Remarks: (1) We know that the sector is of the circle because there are 360 degrees in a circle and = .
(2) We can more formally find the area of the sector by using the following ratio:
Sector Circle Angle 30 360 Area x 108π
3240π = 360x x = = 9π See Lesson 19 for more details.
6.
We begin by drawing a picture
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Solution using a 45, 45, 90 triangle: Since all sides of a square have equal length, an isosceles right triangle is formed. An isosceles right triangle is the same as a 45, 45, 90 triangle. So we can get the length of a side of the triangle just by looking at the formula for a 45, 45, 90 right triangle. Here s = 5. The area of the square is then A = s2 = 52 = 25.
Solution using the Pythagorean Theorem: If we let s be the length of a side of the square, then by the Pythagorean Theorem
s2 + s2 = (5√ )2 2s2 = 50
s2 = 25 s = 5 Thus, the area of the square is A = s2 = 52 = 25.
Remark: We did a bit more work than we had to here. The area of the square is A = s2. We already found that s2 = 25. There was no need to solve this equation for s.
* Using an area formula: The area of a square is A = where d is the length of the diagonal of the square. Therefore in this problem
A = = ( √ ) = = 25.
7.
* We are given that ̅̅̅̅ is perpendicular to ̅̅̅̅. It follows that the measure of angle PQR is 90 degrees. Since x = 35, the measure of angle QPR is 180 – 90 – 35 = 55. Now, since PQ = PS, the angles opposite these sides are congruent. So angle PQS has measure – = 62.5. Therefore y = 90 – 62.5 = 27.5.
Here is a picture of the triangle with some of the angles filled in.
8.
Solution: We redraw the three triangles next to each other so that congruent angles match up.
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We now set up a ratio, cross multiply, and divide: . So 36 = 3x, and therefore x = 12.
9.
* We begin by forming a 30, 60, 90 triangle. If we let x = 1 in the special triangle given to us at the beginning of each math section of the SAT we get the following picture.
Note that we plotted the point by going right 1, then up √ . The slope of the line is m = √ = √ . Since the line passes through the origin, we have b = 0. Thus, the equation of the line in slope-intercept form is
y = mx + b = √ x + 0.
So y = √ x, choice (E).
10.
* Since the hexagon is regular, BC = EF = CD = 6. Now let’s add a bit to the picture.
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Again, note that the hexagon is regular. So each angle of triangle BOC is 60 degrees. Thus, triangle BOC is equilateral. Therefore OF = OC = BC = 6.
Since EF = 6 and FC = 12, triangle CEF is a 30, 60, 90 triangle. It follows that CE = 6√ . Since BCEF is a rectangle, BF = 6√ as well. Therefore the perimeter of rectangle is 6 + 6 + 6√ + 6√ = 12 + 12√ ~ 28.97. To the nearest tenth the answer is 29.
11.
* We begin by drawing a picture.
The side opposite the 30 degree angle has length 12. Thus, the hypotenuse has length 24, choice (B).
12.
* Solution using Strategy 30: First note that
xw + xz + yw + yz = x(w + z) + y(w + z) = (x + y)(w + z).
Now, k = x + y, k = w + z, and so (x + y)(w + z) = k·k = k2. So the answer is choice (D).
Remark: Note that the angle labeled k is an exterior angle of both triangles. We have used Strategy 30 twice here, once for each triangle.
Solution using strategy 4: Let’s choose values for x, y and z, say x = 40, y = 50, and z = 30. Each unlabeled interior angle is 180 – 40 – 50 = 90, and so w = 180 – 90 – 30 = 60. Now,
xw + xz + yw + yz = (40)(60) + (40)(30) + (50)(60) + (50)(30) = 8100 Since the angle labeled with k is supplementary with the unlabeled angle, k = 180 – 90 = 90. So let’s plug k = 90 into each answer choice.
(A) 2025 (B) 45 (C) 90 (D) 8100 (E) 32,400
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Since (A), (B), (C), and (E) came out incorrect, the answer is choice (D).