Rose is fishing from a small boat. A fish swimming at the same depth as the hook at the end of her fishing line is 8 meters away
1 answer:
The question is incomplete, as the depth of the hook isn't given in the question. However, using a proposed depth for the hook, we could give a stepwise solution to the exercise.
Answer:
Kindly check explanation
Step-by-step explanation:
If given a hook depth of 6 meters, and a distance of 8 m(fishing line from hook) at the the Same depth, then the distance between rose and the fish, x can be calculated, using the relation.
Hypotenus² = opposite² + adjacent²
(refer to attached picture)
x² = 6² + 8²
x² = 36 +. 64
x² = 100
x = sqrt(100)
x = 10 meters
You can use the actual depth value to obtain the actual solution to your question.
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This problem can be completed in 2 ways. Both are acceptable.
Option 1:
This is an isosceles trapezoid that can be divided into a rectangle and two congruent triangles.
The area of the rectangle is the base times the height.
The area of one of the triangles is half the base times the height.
The other triangle must have that area too.
The area is 56 square centimeters.
Option 2:
We can use the area formula for the trapezoid.
Where
is the length of the shorter base
and
is the length of the longer base
and
is the height.
The length of the shorter base is 9.
The length of the longer base is 9+5+5, or 19.
The height is 4.
Same answer. The area is 56 square centimeters.
Both options are two acceptable ways the problem can be tackled.
Option 1:
This is an isosceles trapezoid that can be divided into a rectangle and two congruent triangles.
The area of the rectangle is the base times the height.
The area of one of the triangles is half the base times the height.
The other triangle must have that area too.
The area is 56 square centimeters.
Option 2:
We can use the area formula for the trapezoid.
Where
and
and
The length of the shorter base is 9.
The length of the longer base is 9+5+5, or 19.
The height is 4.
Same answer. The area is 56 square centimeters.
Both options are two acceptable ways the problem can be tackled.