Part IV Review

1: C. Same-side interior angles sum to 180°: 180 − 115 = 65°. Choice A reports the given angle itself instead of computing its same-side interior angle.

2: G. 180 − 48 − 77 = 55°. Choice F comes from an arithmetic slip while subtracting both angles from 180°.

3: A. Use the Law of Cosines because two sides and their included angle are known: c² = 7² + 10² − 2(7)(10)cos(60°) = 149 − 70 = 79. Thus c = √79 ≈ 8.89. Choice B treats the triangle as if the included angle were 90° and uses √(7²+10²).

4: J. 5 + 9 = 14, which is not greater than 15, so the triangle inequality fails. Choice F assumes any three lengths form a triangle without checking the triangle inequality.

5: A. Scale factor = 21 ÷ 6 = 3.5, so EF = 8 × 3.5 = 28. Choice B comes from using an incorrect scale factor instead of 21÷6.

6: F. A 90° counterclockwise rotation sends (x, y) to (−y, x), so (−3, 4) becomes (−4, −3). Translating 2 units right gives (−2, −3). Choice G performs the translation before completing the stated transformation sequence.

7: C. Perimeter scales by the same factor as any side: 16×2.25=36. Choice A comes from using an incorrect scale factor instead of the given 2.25.

8: J. Area scales by the square of the linear ratio: (5/4)² = 25/16, so 32 × 25/16 = 50. Choice F applies the linear ratio directly instead of squaring it for the area scale factor.

9: B. √(7²+24²)=√625=25. Choice A comes from an arithmetic slip while applying the Pythagorean theorem.

10: G. Intersecting chords satisfy the product relationship: 3(8) = 4x. Thus 24 = 4x and x = 6. Choice H reports one of the given chord-segment lengths instead of solving the product equation.

11: A. By the Pythagorean theorem, the opposite side is √(17²−15²) = √64 = 8. Choice B reports the given adjacent side itself instead of solving for the opposite side.

12: H. A contrapositive negates and swaps the hypothesis and conclusion. ‘If square, then congruent diagonals’ becomes ‘If the diagonals are not congruent, then the quadrilateral is not a square.’ Choice F is the converse.

13: B. Completing the square: (x+3)²+(y−4)²=11+9+16=36, so the radius is √36=6. Choice A reports r² instead of taking the square root to find r.

14: F. 120°=2π/3 radians; arc length=9×2π/3=6π. Choice H comes from using an incorrect fraction of the circle's circumference for this central angle.

15: D. After dividing by 36, the squared terms have opposite signs: x²/4 − y²/9 = 1. Opposite signs on the two squared-variable terms identify a hyperbola. Choice A mistakes the equation for an ellipse, overlooking that the two terms have opposite signs, which requires a hyperbola instead.

16: H. A tangent line is always perpendicular to the radius at the point of tangency. Choice F reports an unrelated standard angle instead of applying the tangent-radius perpendicularity property.

17: D. Rectangle area: 12×7=84; semicircle area (r=3.5): 0.5×π×3.5²≈19.24; total≈103.24. Choice A reports the rectangle area alone without adding the semicircle.

18: F. Cylinder volume = πr²h = π(25)(12) = 300π. Choice G omits a factor of 2 somewhere in the cylinder volume formula, giving half the correct value.

19: D. Volume scales by the cube of the linear ratio: (3/2)³ = 27/8, so 16 × 27/8 = 54. Choice A applies the linear ratio directly instead of cubing it for the volume scale factor.

20: G. (9+15)/2×6=72. Choice F comes from an arithmetic slip in the trapezoid area formula.

21: A. √[(−5−3)²+(4−(−2))²]=√[64+36]=√100=10. Choice B comes from an arithmetic slip while applying the distance formula.

22: H. [(2+8)/2, (−7+3)/2]=(5, −2). Choice F subtracts the two x-coordinates and divides by 2 instead of adding them.

23: B. The negative reciprocal of 3/5 is −5/3. Choice A drops the negative sign on the reciprocal.

24: F. The horizontal side from (2,3) to (8,3) has length 6, serving as the base; the height, the vertical distance to (5,9), is 6. Area=(1/2)(6)(6)=18. Choice G comes from an arithmetic slip while computing the base or height of the triangle.

1: A. A 90° counterclockwise rotation follows the rule (x,y)→(−y,x): (3,−2)→(2,3). Choice D applies the 90° clockwise rule instead of the counterclockwise rule.

2: H. Area scales by the square of the linear scale factor: 4²=16. New area=7×16=112. Choice F applies the scale factor directly instead of squaring it.

3: B. By the Law of Cosines: c²=8²+11²−2(8)(11)cos(60°)=64+121−88=97. Choice D omits the −2ab·cos C term entirely, computing 8²+11²=185 as if this were a right triangle.

4: G. By the Law of Sines, a/sin A=b/sin B, so b=10 sin(75°)/sin(45°). Choice F inverts the ratio, dividing by the wrong angle's sine.

5: B. An inscribed angle is half the intercepted arc: 124÷2=62°. Choice C reports the arc itself, forgetting to halve it.

6: G. By the power-of-a-point relationship, tangent²=(external part)(whole length): 10²=4L, so L=100/4=25. Choice F forgets to square the tangent length, solving 10=4L instead of 10²=4L.

7: B. Two pairs of equal angles establish similarity by AA, but similarity alone says nothing about the sides' actual lengths, so congruence is not guaranteed. Choice A assumes equal angles alone are enough for congruence, which requires a side condition as well.

8: G. The two intersection points of equal-radius arcs centered at A and B are each equidistant from A and B, so the line through them is the perpendicular bisector of AB. Choice F uses only one arc, which cannot locate two points to define a line.