1. Problem: If line M and N are both perpendicular to line S, then M and N are said to be ___. - When two lines are both perpendicular to the same line, they are parallel to each other. Answer: parallel 2. Problem: If AB⟂CD at E, which of the following is true about ∠AED and ∠BED? - I. They are linear pair - II. They are supplementary - III. They are right angles Since AB is perpendicular to CD at E, angles ∠AED and ∠BED are right angles (90°). - Right angles form a linear pair and are supplementary (sum to 180°). Answer: I, II, and III 3. Problem: Which of the following is ALWAYS true? - Vertical pairs of angles are supplementary? No, they are congruent. - Vertical pairs of angles are complementary? No. - Linear pairs of angles are congruent? No. - Linear pairs of angles are supplementary? Yes, they sum to 180°. Answer: Linear pairs of angles are supplementary. 4. Problem: Two parallel lines are cut by a transversal, forming angle H and angle K. If the two angles are exterior angles on the same side of the transversal, what is the measure of angle H if the measure of angle K is 50°? - Exterior angles on the same side of the transversal are supplementary. - So, $\angle H + \angle K = 180^\circ$ - $\angle H = 180^\circ - 50^\circ = 130^\circ$ Answer: 130° 5. Problem: X is the midpoint of segment WZ and Y is the midpoint of segment XZ. If YZ = 5, then how long is segment WZ? - Since X is midpoint of WZ, $WX = XZ = \frac{WZ}{2}$. - Y is midpoint of XZ, so $XY = YZ = \frac{XZ}{2}$. - Given $YZ = 5$, so $XZ = 10$. - Since $XZ = 10$, $WZ = 2 \times XZ = 20$. Answer: 20 6. Problem: If the sum of the supplement and the complement of an angle is 130 degrees, what is the angle? - Let the angle be $x$. - Supplement: $180^\circ - x$ - Complement: $90^\circ - x$ - Sum: $(180 - x) + (90 - x) = 130$ - Simplify: $270 - 2x = 130$ - $2x = 140$ - $x = 70^\circ$ Answer: 70° 7. Problem: Which of the following is the sum of the angles of a hexagon? - Sum of interior angles of polygon with $n$ sides: $180(n-2)$ - For hexagon, $n=6$ - Sum: $180(6-2) = 180 \times 4 = 720^\circ$ Answer: 720° 8. Problem: What is the measurement (in radians) of an interior angle of a regular octagon? - Sum of interior angles: $180(n-2) = 180(8-2) = 1080^\circ$ - Each interior angle: $\frac{1080}{8} = 135^\circ$ - Convert to radians: $135^\circ \times \frac{\pi}{180} = \frac{3\pi}{4} \approx 2.36$ radians Answer: 2.36 rad 9. Problem: Octagon ABCDEFGH is regular. If EF and GH are extended until they intersect, what is the measure of the angle formed? - Exterior angle of regular octagon: $\frac{360^\circ}{8} = 45^\circ$ - Lines EF and GH are extended; the angle formed is supplementary to the exterior angle. - Angle formed: $180^\circ - 45^\circ = 135^\circ$ - But options do not include 135°, closest is 53°, 60°, 38°, 90°. - Reconsider: EF and GH are non-adjacent sides; the angle formed is the angle between two sides separated by one side. - The angle between two sides separated by one side in a regular polygon is $\frac{360^\circ}{n} \times 2 = 90^\circ$. Answer: 90° 10. Problem: If the sum of the measures of the five exterior angles of a hexagon is 300 degrees, what is the measure of the 6th exterior angle? - Sum of exterior angles of any polygon is $360^\circ$. - Sum of 5 exterior angles: 300° - 6th exterior angle: $360^\circ - 300^\circ = 60^\circ$ Answer: 60° 11. Problem: How many sides does a regular polygon have if an exterior angle measures 45°? - Number of sides: $n = \frac{360^\circ}{\text{exterior angle}} = \frac{360}{45} = 8$ Answer: 8 12. Problem: In an equilateral triangle, what is the difference between the sum of the exterior angles and the sum of the interior angles? - Sum of exterior angles: $360^\circ$ - Sum of interior angles: $180^\circ \times (3-2) = 180^\circ$ - Difference: $360^\circ - 180^\circ = 180^\circ$ Answer: 180 degrees 13. Problem: A polygon has 170 diagonals. How many sides does it have? - Number of diagonals formula: $\frac{n(n-3)}{2} = 170$ - Multiply both sides by 2: $n(n-3) = 340$ - $n^2 - 3n - 340 = 0$ - Solve quadratic: $n = \frac{3 \pm \sqrt{9 + 1360}}{2} = \frac{3 \pm \sqrt{1369}}{2} = \frac{3 \pm 37}{2}$ - Positive root: $\frac{3 + 37}{2} = 20$ Answer: 20 14. Problem: What is the range of the possible lengths of the third side of a triangle whose sides measure 8 and 13? - Triangle inequality: $|8 - 13| < x < 8 + 13$ - $5 < x < 21$ Answer: 5 < x < 21 15. Problem: In triangle XYZ, side XY is 13 cm long and side XZ is 21 cm long. How many possible lengths can side YZ have if its measure is a whole number? - Possible lengths: $|13 - 21| < YZ < 13 + 21$ - $8 < YZ < 34$ - Whole numbers from 9 to 33 inclusive: $33 - 9 + 1 = 25$ Answer: 25 16. Problem: What kind of triangle is formed by sides of lengths 12, 10, and 15? - Check using Pythagorean theorem: - Largest side: 15 - $15^2 = 225$ - Sum of squares of other sides: $12^2 + 10^2 = 144 + 100 = 244$ - Since $225 < 244$, triangle is acute. Answer: Acute 17. Problem: Which of the following side lengths belong to an acute triangle? - Check each set using Pythagorean theorem: - 5,7,9: $9^2=81$, $5^2+7^2=25+49=74$, $81>74$ obtuse - 6,9,12: $12^2=144$, $6^2+9^2=36+81=117$, $144>117$ obtuse - 6,8,10: $10^2=100$, $6^2+8^2=36+64=100$, right triangle - 7,9,11: $11^2=121$, $7^2+9^2=49+81=130$, $121<130$ acute Answer: 7 cm, 9 cm, 11 cm 18. Problem: In triangle PQR, if the measure of the exterior angle of angle Q is 128 degrees and angle R is 37 degrees, what is the measurement of angle P? - Exterior angle at Q: $128^\circ$ - Interior angle Q: $180^\circ - 128^\circ = 52^\circ$ - Sum of angles in triangle: $P + Q + R = 180^\circ$ - $P + 52 + 37 = 180$ - $P = 180 - 89 = 91^\circ$ Answer: 91 degrees 19. Problem: Two angles of a triangle measure 60° and 80°. The bisectors of these angles meet at B. What is the measure of the obtuse angle formed at B? - The angle between bisectors of angles $A$ and $C$ is $90^\circ + \frac{B}{2}$. - Given angles: 60°, 80°, so third angle $B = 180 - 60 - 80 = 40^\circ$ - Angle at B formed by bisectors: $90^\circ + \frac{40^\circ}{2} = 90^\circ + 20^\circ = 110^\circ$ Answer: 110° 20. Problem: The altitude of an equilateral triangle is 12 cm. How long is each side? - Altitude formula: $h = \frac{\sqrt{3}}{2} s$ - $12 = \frac{\sqrt{3}}{2} s$ - $s = \frac{12 \times 2}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ Answer: 8√3 cm 21. Problem: Which of the following postulates states that, if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent? Answer: SAS Postulate 22. Problem: The sides of a triangle are 37 dm and 58 dm respectively. If the perimeter of the triangle is 174 dm, how long is the third side? - Third side = $174 - 37 - 58 = 79$ dm Answer: 79 dm 23. Problem: In triangle BCD, BC = 25 m, and CD = 10 m. The perimeter of the triangle maybe: - Third side $x$ must satisfy triangle inequality: - $|25 - 10| < x < 25 + 10$ => $15 < x < 35$ - Perimeter options: 69, 70, 71, 72 - $x = ext{perimeter} - 25 - 10 = ext{perimeter} - 35$ - Check each: - 69 - 35 = 34 (valid) - 70 - 35 = 35 (not less than 35) - 71 - 35 = 36 (invalid) - 72 - 35 = 37 (invalid) Answer: 69 m 24. Problem: A triangular lot bounded by sides 13 m, 11 m, and 9 m. Find the area. - Use Heron's formula: - $s = \frac{13 + 11 + 9}{2} = 16.5$ - Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16.5(16.5-13)(16.5-11)(16.5-9)}$ - $= \sqrt{16.5 \times 3.5 \times 5.5 \times 7.5} \approx \sqrt{2380.3125} \approx 48.8$ Answer: 48.8 sq. m 25. Problem: The sides of a triangle are 5 cm, 6 cm, and 8 cm. Find the length of the altitude to the shortest side. - Shortest side: 5 cm - Area using Heron's formula: - $s = \frac{5+6+8}{2} = 9.5$ - Area $= \sqrt{9.5(9.5-5)(9.5-6)(9.5-8)} = \sqrt{9.5 \times 4.5 \times 3.5 \times 1.5} \approx \sqrt{224.4375} \approx 14.98$ - Altitude $h = \frac{2 \times \text{Area}}{\text{base}} = \frac{2 \times 14.98}{5} = 5.99$ Answer: 5.99 26. Problem: Two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other is 3 units less than its base. Find the altitudes if the areas differ by 21 square units. - Let base = $b$ - Altitudes: $b+3$ and $b-3$ - Areas: $\frac{1}{2} b (b+3)$ and $\frac{1}{2} b (b-3)$ - Difference: $\frac{1}{2} b (b+3) - \frac{1}{2} b (b-3) = 21$ - Simplify: $\frac{1}{2} b (b+3 - b + 3) = 21$ - $\frac{1}{2} b (6) = 21$ - $3b = 21$ - $b = 7$ - Altitudes: $7+3=10$ and $7-3=4$ Answer: 4 & 10 27. Problem: In triangle ABC, D and E are points on AC and BC, respectively, such that DE || AB. If AC = 11, DC = 6 and BC = 15, what is the length of BE? - Since DE || AB, triangles CDE and CAB are similar. - Ratio: $\frac{CD}{AC} = \frac{CE}{AB} = \frac{DE}{CB}$ - $CD = 6$, $AC = 11$, so ratio $= \frac{6}{11}$ - $BE = BC - CE$ - $CE = \frac{6}{11} \times BC = \frac{6}{11} \times 15 = \frac{90}{11} = 8.18$ - $BE = 15 - 8.18 = 6.82$ Answer: 6.82 28. Problem: Which statement about quadrilaterals is true? - All quadrilaterals have four sides. Answer: All quadrilaterals have four sides. 29. Problem: Which of the following statements are true? I. A square is a rectangle. II. A square is a rhombus. III. A parallelogram is a rhombus. IV. A rectangle is a parallelogram. - I: True - II: True - III: False (not all parallelograms are rhombus) - IV: True Answer: I, II, and IV only 30. Problem: Which quadrilateral must have diagonals that are congruent and perpendicular? Answer: square 31. Problem: Given an equilateral △XYZ with midpoints A,B,C respectively, what figure is formed by quadrilateral XABZ? Answer: trapezoid 32. Problem: Which of the following has two diagonals that are perpendicular bisectors of each other? Answer: rhombus 33. Problem: What can be said about these statements? I. Any rectangle has two congruent diagonals. II. Any rhombus has two perpendicular diagonals. - I: True - II: True Answer: Both statements are true.