Let
x,y, and z the long sides of the triangle
we know that
x+y+z=170 ft------> equation 1
x/y=25/14----> y=14x/25------> equation 2
y/z=14/12-----> equation 3
x/z=25/12----> z=12x/25------> equation 4
substitute equation 2 and equation 4 in equation 1
x+[14x/25]+[12x/25]=170------> multiply by 25 both sides
25x+14x+12x=4250
51x=4250
x=4250/51
x=250/3
y=14x/25------> y=(250/3)*(14/25)----> y=140/3
z=12y/14-----> (140/3)*12/14----> z=40
Using Heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)
where s is the semiperimeter
s=170/2-----> s=85 ft
Area=√85*[85-250/3]*[85-140/3]*[85-40]
Area=9.22*[1.67]*[38.33]*[45]------> Area=26558.21 ft²
the answer is
26558.21 ft²
Answer:
6y+3xy
Step-by-step explanation:
You can combine the y's together to get this answer
<span>This is a right triangle problem. c^2 = a^2 + b^2
Let b = 7 mi (Rosa's distance from home)
Let a = the east leg (Juan's distance from home
(a+1) = c (hypotenuse, distance between them)
(a+1)^2 = a^2 + 7^2
FOIL (a+1)(a+1)
a^2 + 2a + 1 = a^2 + 49
Combine like terms
a^2 - a^2 + 2a = 49 - 1
2a = 48
a =
<span>a = 24 mi, Juan's distance from home
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Answer:
The triangle would be Obtuse
