Mixed practice: Geometry and Trigonometry

A linear pair totals 180°, so (3x + 7) + (5x − 3) = 180. Thus 8x + 4 = 180 and x = 22. The two angles are 73° and 107°; the smaller is 73°. Reporting 22 would confuse the variable with the requested angle. Revisit Example 1.03.

The surface-area ratio, larger to smaller, is 144/64 = 9/4. For similar solids the length factor is $\sqrt{\text{9/4}}$ = 3/2, so the volume factor is (3/2)³ = 27/8. Multiply the smaller volume by this factor: 40(27/8) = 135. The direct area factor would be inappropriate for volume. Revisit Example 6.14.

Use a representative right triangle with adjacent leg 8 and hypotenuse 17. The opposite leg is $\sqrt{{\text{17}}^{\text{2}}\text{−}{\text{8}}^{\text{2}}}$ = $\sqrt{\text{225}}$ = 15. Therefore sin θ = 15/17. Equivalently, sin² θ = 1 − (8/17)² = 225/289; the positive root is required because θ is acute. Revisit Example 3.08.

Move the constant and complete the square:

x² + 4x + y² − 12y = −15 (x + 2)² + (y − 6)² = −15 + 4 + 36 = 25.

The radius is $\sqrt{\text{25}}$ = 5, not 25. The center is (−2, 6), but it is not the target. Revisit Example 5.08.

The smaller-to-larger length ratio is 2/3, so the area ratio is (2/3)² = 4/9. The smaller area is 81(4/9) = 36. Multiplying by 2/3 would give 54, the distractor created by treating an area like a length. Revisit Example 2.12.

The central angle occupies 108/360 = 3/10 of a full turn. The circumference is 20π, so the arc length is (3/10)(20π) = 6π. The question asks for the coefficient k, so enter 6, not a decimal approximation of 6π. Revisit Example 4.02.

The first two interior angles supplement the given exterior angles: 180 − 124 = 56° and 180 − 138 = 42°. The third interior angle is 180 − 56 − 42 = 82°. Adding exterior angles directly into the interior-angle total would mix different angle types. Revisit Example 1.15.

Write the legs as 2k and 3k. Their area is (1/2)(2k)(3k) = 3k² = 27, so k² = 9. By the Pythagorean theorem,

c² = (2k)² + (3k)² = 13k² = 117.

You do not need the hypotenuse itself, 3$\sqrt{\text{13}}$. Revisit Example 3.14.

The center is the midpoint of the diameter: ((1 + 7)/2, (−4 + 4)/2) = (4, 0). The radius squared is the squared distance from this center to either endpoint: (7 − 4)² + (4 − 0)² = 9 + 16 = 25. The diameter squared would be 100, which explains choice A. Revisit Example 5.07.

The perpendicular height is h = $\sqrt{{\text{13}}^{\text{2}}\text{−}{\text{5}}^{\text{2}}}$ = 12. The cone volume is

V = $\frac{\text{1}}{\text{3}}$π(5)²(12) = 100π.

Thus k = 100. The slant height is needed only to find the perpendicular height; it is not the volume formula’s height. Revisit Example 6.09.

The parallel segment produces △ADE ∼ △ABC. Since AB = 4 + 6 = 10, the small-to-large length ratio is 4/10 = 2/5. Therefore AC = 7.5(5/2) = 18.75. The requested remaining piece is EC = 18.75 − 7.5 = 11.25 = 45/4. The part-to-part equation 4/6 = 7.5/EC also works. Revisit Example 2.09.

An inscribed angle is half its intercepted arc, so the specified arc measures 2(58°) = 116°. The other arc between the same endpoints measures 244°, but it contains the vertex and is not intercepted by the angle. Revisit Example 4.08.

Same-side interior angles formed by a transversal across parallel lines are supplementary. Thus 4x + 12 + 6x − 2 = 180, so 10x + 10 = 180 and x = 17. The actual angles are 80° and 100°, which explains two distractors. Revisit Example 1.06.

The 30°–60°–90° ratio is s : s$\sqrt{\text{3}}$: 2s. A hypotenuse of 10 makes s = 5, so the legs are 5 and 5$\sqrt{\text{3}}$. Their product is 25$\sqrt{\text{3}}$, and the triangle’s area is half of that: 25$\sqrt{\text{3}}$/2. Revisit Examples 3.05 and 3.06.

Tangency to the y-axis means the radius equals the center’s horizontal distance from that axis: r = | − 3| = 3. Therefore A = π(3)² = 9π. The coordinate 4 would determine the distance to the x-axis, not the y-axis. Revisit Example 5.11.

The curved surface contributes 2πrh = 2π(4)(9) = 72π. The single bottom contributes πr² = 16π. The open top contributes no surface, so the total is 88π and k = 88. A closed cylinder would instead have two circular bases. Revisit Example 6.12.

The first triangle has third angle 180 − 38 − 62 = 80°. The second has third angle 180 − 38 − 80 = 62°. Their angles match, so AA similarity is guaranteed. No corresponding side length is fixed, so their sizes and areas need not be equal. They could happen to be congruent, but the information does not require it. Revisit Examples 2.06 and 2.14.

For a sector, A = rs/2. Solve for radius:

r = $\frac{\text{2A}}{\text{s}}$ = $\frac{\text{2(18π)}}{\text{6π}}$ = 6.

The implied central angle is s/r = π radians, so the sector is a semicircle. Its area is indeed half of 36π. Revisit Example 4.15.

Each exterior turn is 180 − 156 = 24°. The exterior turns of a convex polygon total 360°. Because the polygon is regular, every turn is the same, so n = 360/24 = 15. Using 360/156 would confuse interior angles with exterior turns. Revisit Example 1.14.

The complementary-angle identity gives cos(90° − θ) = sin θ = 3/5. There is no need to find θ or to calculate cos θ = 4/5, which is not the requested value. Revisit Example 3.10.

Substitute y = 0 into the circle equation:

(x − 1)² + 4 = 16 x = 1 ± $\sqrt{\text{12}}$ = 1 ± 2$\sqrt{\text{3}}$.

The two points have the same y-coordinate, so their distance is the difference of their x-coordinates: (1 + 2$\sqrt{\text{3}}$) − (1 − 2$\sqrt{\text{3}}$) = 4$\sqrt{\text{3}}$. Revisit Example 5.12.

The outer area is 20(14) = 280. A border of width 2 occupies both ends of each dimension, so the inner rectangle is (20 − 4) by (14 − 4), or 16 by 10. Its area is 160. The frame itself has area 280 − 160 = 120. Subtracting only 2 from each dimension would remove the border on just one side. Revisit Examples 6.04 and 6.05.

The triangle inequality gives |10 − 7| < t < 10 + 7, or 3 < t < 17. The possible integers are 4 through 16, inclusive, so there are 16 − 4 + 1 = 13. The endpoints 3 and 17 would produce degenerate triangles and are excluded. Revisit Example 2.04.

Angle 5π/6 is 150°, in quadrant II, with a 30° reference angle. Thus sin θ = 1/2 and cos θ = −$\sqrt{\text{3}}$/2. Substituting gives 2(1/2) − $\sqrt{\text{3}}$/2 = 1 − $\sqrt{\text{3}}$/2. The negative cosine sign is essential. Revisit Examples 4.13 and 4.14.