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2 answers:
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Answer:
The distance you are from the base of plateau =
Step-by-step explanation:
Given:
Height of plateau = 70 m
Angle of elevation to the top of plateau = 35°
To find the distance you are from the base of plateau.
We will construct a triangle ABC to model the given situation. The triangle would be a right triangle for which we know an angle and its opposite side. We need to find the adjacent side of the triangle.
We will apply trigonometric ratio to find the adjacent side.
where represents the angle of reference.
Plugging in the values from the triangle.
Multiplying both sides by AC.
Dividing both sides by
∴
The distance you are from the base of plateau =
Let
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²