The ratios of the sides of the 30-60-90 triangle are:
1 : √3 : 2
So the ratio of the perimeter to the hypotenuse is
3+√3 : 2
Use cross-product to find:
perimeter / 7 = (3+√3) / 2
1 : √3 : 2
So the ratio of the perimeter to the hypotenuse is
3+√3 : 2
Use cross-product to find:
perimeter / 7 = (3+√3) / 2
Fact: therefore the ratio of the perimeter to the hypotenuse is the same for all 30-60-90 triangles.
We have arbitrarily created a 30-60-90 triangle with hypotenuse = 2.
So to get a triangle with a hypotenuse of 7, we need to multiply everything by 7/2=3.5.
The sides of the given triangle is therefore (1,√3,2)×3.5, or
(3.5, 3.5√3, 7)
Add up the three numbers (lengths of sides) to get the perimeter.
7/2+21/2sqrt3
21+7sqrt3
7+21sqrt3
21/2+7/2sqrt3
So I don't understand
Remember that 3.5 is the same as 7/2, and so on.
The answer should therefore be
(21+7sqrt(3))/2
=21/2 + (7/2)sqrt(3)
1. D, 15° and 75°
2. B, 40
3. B, 10
4. B, 52 ft.
5. B, DGH ~ DFE; SAS ~
6. A, AA Postulate
7. A, 24
8. C, 6√2 miles; 6√11 miles
9. A, x = 10
10. B, no
11. C, obtuse
12. C, 17√2 ft
13. D, 21/2 + 7/2√3
14. D, 86.19°
15. C, sin A = 21/29, cos A = 20/29
16. B, 2.1 mi
17. A, about 29 miles at 25° south of west
18. C
19. B, P'(-8, -1), Q'(-6, 8), R'(4, 3)
20. D, 288°
21. D, 2
22. B, X
23. B, enlargement; 2
24. B, glide reflection; translate 8 units to the right then reflect across the line y = 4
25. C, 812 in²
26. B, 25,7 ft
27. A, 70cm²
28. 40.8 ft²
29. A, 585 in²
30. A, 8,000
31. B, 40Ï€ in.
32. C, 45Ï€ m
33. D, 4.2025π m²
34. B, 51.8 in.²
35. A, 30
36. A, 472 m²; 486 m²
37. B, 308π in.²
38. C, 57 ft²
39. A, 1,802 m²
40. B, 1:4
41. B, 1,472.6 in.³
42. A, 143.2 ft²
43. C, 1,344.8 m²
44. D, 7:18
45. C, 68
46. B, 60
47. C, 47°
48. A, 44°
49. A, 34°
50. A, (x + 3)² + (y - 2)² = 9
51. C, (x - 2)² + (y + 5)² = 241
52. C, 6:5
53. C, 2/91