Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), and G (6, 1). He has 16 units of fencing. where could alex place point H so that he does not have to buy more fencing?
Answers
Given :- Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), and G (6, 1). He has 16 units of fencing.
To Find :- where could alex place point H so that he does not have to buy more fencing ?
A) (0 , 1)
B) (0 , -2)
C) (1 , 1)
D) (1 , -2)
Answer :-
Let coordinates of H are (x , y) .
Using distance formula we get,
→ EF = √{(3 - 1)² + (5 - 5)²} = √(4 + 0) = 2 units.
→ FG = √{6 - 3)² + (1 - 5)²} = √(9 + 16) = 5 units .
→ GH = √{(x - 6)² + (y - 1)²}
→ HE = √{(x - 1)² + (y - 5)²}
since,
→ EF + FG + GH + HE = 16 units
→ 2 + 5 + √{(x - 6)² + (y - 1)²} + √{(x - 1)² + (y - 5)²} = 16
→ √{(x - 6)² + (y - 1)²} + √{(x - 1)² + (y - 5)²} = 9 units .
using options now we get,
A) (0, 1)
→ √(36) + √(1 + 25) = 6 + √26 ≠ 9 units.
B) (0, - 2)
→ √(36 + 9) + √(1 + 49) = √45 + √50 ≠ 9 units .
C) (1 , 1)
→ √(25 + 0) + √(0 + 16) = 5 + 4 = 9 = 9 units .
Therefore, Alex should place Point H at (1, 1) .
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