Geometry: Lengths, Angles, Perimeters
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1. The area of circle O is 20π. The sum of the areas of square I and square II is:

(A) 60
(B) 80
(C) 90
(D) 120
(E) cannot be determined

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With any circle problem, the first thing to do is get the radius. If the area is 20π the radius is √20 and the diameter is 2√20. If the small square side is a and the big square side is b, the pythagorean theorem (SATan's favorite) says a2+b2 = (2√20)2 = 80. You don't need each individual area, just the sum of the areas of the squares and that's 80. The answer is (B).

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Note: Figure not drawn to scale.

2. In the figure above, AE = DE and BE = CE. The area of triangle AED is 5 times the area of triangle BEC. The length of AD is 100. What is the length of AB?

(A) 10
(B) 20
(C) 30
(D) 40
(E) 50

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Since the small triangle (BEC) is one-fifth the area of the big triangle and has the same height, the base must be one-fifth as long. Since AD is 100, BC must be 20. That leaves 40 for AB and 40 for CD. So the answer is (D).

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Note: Figure not drawn to scale.

3. In the figure above, AD = 10, DE = 12, EA = 9, BC = 3, EA is parallel to FB, and DE is parallel to CF. What is the length of FB?

(A) 2.5
(B) 2.7
(C) 2.9
(D) 3.1
(E) 3.3

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Since the triangles are similar (by angle-angle) 3/10 = FB/9 which means FB = 27/10 or 2.7. The answer is (B).

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4. Regular octagon ABCDEFGH has sides of length s. What is the length of AD in terms of s?

(A) s · √2
(B) s · (√2 + 1)
(C) s · √3
(D) s · (√3 + 1)
(E) s · √3/2

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The angles of an octagon are 135 (you should know all the shapes up to octagons). If you draw AD and draw a perpendicular from B to AD, you end up with at 45-45-90 triangle (this is the SAT so that pretty much had to happen). The hypotenuse is s so the sides are s/√2. That means AD = s/√2 + s + s/√2 = 2s/√2 + s = s√2 + s which is the same as choice (B).

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Note: Figure not drawn to scale.

5. In the diagram above, AC is a diameter of the circle, AB = 2, BC = 8, and DE is perpendicular to AC. What is the length of DE?

(A) 4
(B) 6
(C) 8
(D) 10
(E) cannot be determined

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It's a circle problem so you'll need the radius. Since the diameter (AB + BC) is 10, the radius is 5. Call the center P and draw PE which is 5. Note that BP is 3. Aha! It's SATan's favorite 3-4-5 triangle. So BE is 4 and DE is 8. The answer is (C).

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Note: Figure not drawn to scale.

6. In the diagram above, lines l, m, and n are parallel to one another, x, y, and z are the measures of angles, and AB = BC. Which of the following is true?

I. x = y
II. AB = AC
III. y = z

(A) none
(B) I, only
(C) II, only
(D) I and III, only
(E) I, II, and III

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The obtuse angle at C is 120 because the lines are parallel. That makes the BCA equal to 60. Since the triangle is isoceles, BAC and BCA are equal. So all the angles in the triangle are 60 and it is equilateral. Because of alternate interiors x and y are both 60. So I, II, and III are all true and the answer is (E).

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Note: Figure not drawn to scale.

7. In the figure above, AC is parallel to EF, BC = CD, angle BCD = 100°, and angle BDE = 70°. What is the measure in degrees of angle FED?

(A) 120
(B) 130
(C) 140
(D) 150
(E) 160

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Since angle BCD is 100, the two base angle must each be 40 which means that angle CDE = 70+40 = 110. You don't need angles A and F. All you need to know is that they add up to 180 because of the two parallel lines (they are same side interior angles). It's a five-sided figure so A + F + C + CDE + FED = 540. Now just plug in 180 + 100 + 110 + FED = 540 and get FED = 150. The answer is (D).

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Note: Figure not drawn to scale.

8. In the figure above, AB = 1, AC = √3, and BC = 2. What is the measure in degrees of angle x?

(A) 25
(B) 30
(C) 35
(D) 40
(E) cannot be determined

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They are telling you that ABC is a 30-60-90 triangle. The small angle at C is 30 and the bigger angle adjacent to it is 75 (75 + 30 + 75 = 180). That means x = 25 (25+80+75=180). The answer is (A).

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9. The regular pentagon above has center O and equal sides and equal angles. What is the measure in degrees of angle 1?

(A) 12
(B) 15
(C) 16
(D) 18
(E) 20

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The central angles of a regular pentagon are all 72 (360/5). Angle 1 is the base angle of an isoceles triangle whose vertex angle is 144 (72+72). That means the base angles add up to 36 (144+36=180) and must each be 18. The answer is (D).

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Note: Figure not drawn to scale.

10. If z=w–x and v=(w–x)/2 and all lines are either vertical or horizontal, then the perimeter of the figure is:

(A) 4w+3x
(B) 2(w+x+y+z+v)
(C) w+y+z+2v+4x
(D) 3w+y+2x
(E) none of these

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This one is vicious. The unlabeled vertical line segment on the bottom is w–z which equals x. The perimeter is:

2v+5x+z+y+w = w–x+5x+w–x+y+w = 3w+3x+y.

Note that y isn't related to any of the other variables so there's no way to get rid of it. The answer definitely has exactly one y in it. So it isn't A or B. Choice C is close but you'd have to have 5x to make it right. Choice D is close but you need it to be 3x. So the answer is (E).

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