New SAT Math Workbook

Additional properties would be needed to make it a

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Unformatted text preview: ary to ∠CEF. ∠ AEC = 80° 3. (E) The exterior angle is equal to the sum of the two remote interior angles. 4 x = 100 x = 25 Angle A = 3x = 75° 4. (D) The other base angle is also x. These two base angles add to 2x. The remaining degrees of the triangle, or 180 – 2x, are in the vertex angle. (E) ∠1 = ∠3 ∠2 + ∠3 = 180° ∠2 = 50° 5. 4. (C) Since ∠BEG and ∠EGD add to 180°, halves of these angles must add to 90°. Triangle EFG contains 180°, leaving 90° for ∠EFG. (C) ∠A = ∠C 4 x − 30 = 2 x + 10 2 x = 40 x = 20 5. ∠ A and ∠ C are each 50°, leaving 80° for ∠ B. ∠1 = ∠3 ∠2 = ∠4 ∠1 + ∠2 = ∠3 + ∠4 But ∠3 + ∠4 = 180°. Therefore, ∠1 + ∠2 = 180° www.petersons.com Geometry 227 Exercise 7 1. 2. (C) A hexagon has 6 sides. Sum = (n – 2) 180 = 4(180) = 720 (D) Opposite sides of a parallelogram are congruent, so AB = CD. x + 4 = 2 x − 16 20 = x AD = BC = x − 6 = 14 Exercise 8 1. (C) Tangent segments drawn to a circle from the same external point are congruent. If CE = 5, then CF = 5, leaving 7 for BF. Therefore BD is also 7. If AE = 2, then AD = 2. BD + DA = BA = 9 2. (D) Angle O is a central angle equal to its arc, 40°. This leaves 140° for the other two angles. Since the triangle is isosceles, because the legs are equal radii, each angle is 70°. (E) The remaining arc is 120°. The inscribed 1 angle x is its intercepted arc. 2 3. (B) AB = CD x + 8 = 4x − 4 12 = 3x 3. AB = 12 x=4 BC = 12 CD = 12 4. (A) If all sides are congruent, it must be a rhombus. Additional properties would be needed to make it a square. 4. 5. (B) A rhombus has 4 sides. Sum = (n – 2) 180 = 2(180) = 360 (C) Rectangles and rhombuses are both types of parallelograms but do not share the same special properties. A square is both a rectangle and a rhombus with added properties. 5. 1 ( 40° + AC ) 2 100° = 40° + AC 60° = AC 50° = (D) An angle outside the circle is the 2 difference of its intercepted arcs. 1 www.petersons.com 228 Chapter 13 Exe...
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