Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (0, 5), F (5, 5), and G (1, 1). He has 20 units of fencing. Where could Alex place point H so that he does not have to buy more fencing?

(−3, 1)
(−3, −1)
(−5, 1)
(−5, −1)

(-3,1)

Step-by-step explanation:

First we need to know the distance between E and F, and the distance betweem F and G:

d(E,F) = sqrt((5-0)^2 + (5-5)^2) = 5

d(F,G) = sqrt((5-1)^2 + (5-1)^2) = 5.6569

The other distances are d(G,H) and d(H,E).

So, if we choose the point H = (-3,1), we have:

d(G,H) = sqrt((1+3)^2 + (1-1)^2) = 4

d(H,E) = sqrt((-3-0)^2 + (1-5)^2) = 5

Total fencing = 5 + 5.6569 + 4 + 5 = 19.6569 < 20

This point can be chosen.

If we take H = (-3,-1):

d(G,H) = sqrt((1+3)^2 + (1+1)^2) = 4.4721

d(H,E) = sqrt((-3-0)^2 + (-1-5)^2) = 6.7082

Total fencing = 5 + 5.6569 + 4.4721 + 6.7082 = 21.8372 > 20

This point can't be chosen.

If we take H = (-5,1):

d(G,H) = sqrt((1+5)^2 + (1-1)^2) = 6

d(H,E) = sqrt((-5-0)^2 + (1-5)^2) =6.4031

Total fencing = 5 + 5.6569 + 6 + 6.4031 = 23.06 > 20

This point can't be chosen.

If we take H = (-5,-1):

d(G,H) = sqrt((1+5)^2 + (1+1)^2) = 6.3246

d(H,E) = sqrt((-5-0)^2 + (-1-5)^2) = 7.8102

Total fencing = 5 + 5.6569 + 6.3246 + 7.8102 = 24.7917 > 20

This point can't be chosen.

So the point Alex should place point H is (-3,1)

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

the answer is b