Math Geom Chapter 7 Review Key

Geometry CP Review: Chapter 7 Write in simplest radical form. 1. √48 / √64 = 4√3 / 8 = √3/2 2. 5 / 2√3 * √3 / √3 = 5√3 / 6 3. √(24/2) = √24 / √2 * √2 = √48 / 2 = 4√3 / 2 = 2√3 Find the value of x. Write your answer in simplest radical form. 4. 10² + 16² = x² 100 + 256 = x² 356 = x² x = √356 x = 2√89 5. x² + 36² = 48² x² + 1296 = 2304 x² = 1008 x = √(4 * 252) x = 2√252 x = 12√7 Is the triangle acute, right, or obtuse? 6. 9² + 14² ≶ (7√5)² 81 + 196 ≶ 245 277 > 245 Acute! 7. Find the approximate area of the triangle. [Image of a right triangle with base 6 and angle 40 degrees] tan(40) = x/6 6tan(40) = x x ≈ 5.03 A = (1/2)bh A ≈ (1/2)(6)(5.03) A ≈ 15.1 m² Find the exact value of each variable. Write your answer in simplest radical form. (Special Right Triangles) 8. [Image of a 45-45-90 triangle] x = √32 / √2 x = 4√2 y = 4√2 9. [Image of a 30-60-90 triangle] x = √33 y = 2√11 10. [Image of a 30-60-90 triangle] x = √15 y = 2√5 11. [Image of a 45-45-90 triangle] x = √21 y = 2√21 12. [Image of a 30-60-90 triangle] x = √10 y = 2√10 13. An isosceles triangle has a perimeter of 42 and a base of 12. Find the exact length of the altitude to the base. [Image of an isosceles triangle with base 12 and equal sides 15] 36 + h² = 15² 36 + h² = 225 h² = 189 h = √189 h = 3√21 14. A square has a perimeter of 60ft. Find the exact length of its diagonal. [Image of a square with side length 15] 15√2 15. A rectangle has an area of 210 feet and a width of 30 feet. Find the exact length of its diagonal. [Image of a rectangle with width 30 and length 7] 210 = 30x 7 = x 7² + 30² = d² 49 + 900 = d² d² = 949 d = √949 16. A rectangle has a diagonal that is 10 feet long and a width of 6 feet. Find the exact area of the rectangle. [Image of a rectangle with width 6 and diagonal 10] 6² + l² = 10² 36 + l² = 100 l² = 64 l = 8 A = lw A = 8 * 6 A = 48 ft² Find the exact value of each variable. Write your answer in simplest radical form. 17. [Image of a right triangle with hypotenuse 10 and one leg 5] 5² + b² = 10² 25 + b² = 100 b² = 75 b = 3√5 18. [Image of a 30-60-90 triangle] x = 2√2 y = 2√2√3 y = 2√6 19. What is the exact perimeter of an equilateral triangle with an altitude of 15 inches? [Image of an equilateral triangle with altitude 15] (15/√3) * 2 = 10√3 Each side = 10√3 3 sides = 30√3 in Find the value of x to the nearest tenth. 20. [Image of a right triangle with angle 41 degrees and adjacent side 12.75] tan(41) = x/12.75 x = 12.75 * tan(41) x ≈ 10.9 21. [Image of a right triangle with angle 39.1 degrees and hypotenuse 45.8] sin(39.1) = x/45.8 x = 45.8 * sin(39.1) x ≈ 28.9 22. [Image of a right triangle with angle 40 degrees and adjacent side 0.92] tan(40) = x/0.92 x = 0.92 * tan(40) x ≈ 0.8 Evaluate the value of angle D to the nearest tenth of a degree. 23. [Image of a right triangle with angle D, opposite side 11.2, and adjacent side 8.7] tan(D) = 11.2 / 8.7 D = tan⁻¹(11.2/8.7) ≈ 52.2° 24. [Image of a right triangle with angle D, opposite side 0.89, and adjacent side 0.99] sin(D) = 0.89 / 0.99 D = sin⁻¹(0.89/0.99) ≈ 64.0° 25. [Image of a right triangle with angle D, hypotenuse 31.68, and adjacent side 20.75] cos(D) = 20.75 / 31.68 D = cos⁻¹(20.75/31.68) ≈ 49.1° Find the exact value of x. 26. [Image of a right triangle with hypotenuse 12, and one leg 6] short leg = 6, long leg = x x = 6√3 27. [Image of a right triangle with hypotenuse 10, and one leg 8] 8² + x² = 10² 64 + x² = 100 x² = 36 x = 6 28. [Image of a right triangle with hypotenuse 16, and one leg 8] 8² + x² = 16² 64 + x² = 256 x² = 192 x = 8√3 29. You are hiking up a mountain peak. You begin hiking at a trailhead whose elevation is about 7500 feet. The trail ends near the summit whose elevation is about 2.5 miles. The horizontal   distance between these two points is about 18,200 feet as shown. Estimate the angle of elevation from the trailhead to the summit. Round your answer to the nearest tenth of a degree.   [Image of a right triangle with vertical side 5700, horizontal side 18200, and angle of elevation θ] 2.5 mi * 5280 ft/mi = 13200 ft tan(θ) = 5700 / 18200 θ = tan⁻¹(5700/18200) ≈ 17.4°