Question

There is a bamboo 10 feet high, the upper end of which, being broken, reaches the ground 5 feet from the stem. find the height of the break.The height at which the break occurs is ___. (ft, ft^2, ft^3)(Type an integer or decimal rounded to the nearest hundredth)

Answer 1

The height at which the bamboo stick is broken can be found using the Pythagorean theorem. By setting up the equation, solving for the height 'x', and calculating the value, the height of the break is approximately 8.66 feet.

Step-by-step explanation:

To find the height of the break in the bamboo, we can use the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this problem, the broken bamboo forms a right triangle with the ground and the remaining part of the bamboo standing upright.

The hypotenuse is the original height of the bamboo, which is 10 feet. The horizontal distance from the base to where the tip of the bamboo hits the ground is 5 feet. Let's call the height at which the bamboo has broken 'x' feet. So, the unbroken part of the bamboo extending from the break to the top is (10 - x) feet, representing the vertical side of the triangle.

Applying the Pythagorean theorem:

(x)^2 + (5 feet)^2 = (10 feet)^2
(x)^2 + 25 = 100
(x)^2 = 100 - 25
(x)^2 = 75
x = √75
x ≈ 8.66 feet

Therefore, the height at which the break occurs is approximately 8.66 feet.

Answer 2

Let x be the height where the bamboo breaks.

The breaking represents a right triangle, wherein x represents the legs and the base is equal to 5 feet (ground to stem braking distance). We can use the Pythagorean theorem to solve for the value of x, such that

`\begin{gathered} c^2=a^2+b^2 \\ c=\sqrt[]{x^2+5^2} \\ c=\sqrt[]{x^2+25} \end{gathered}`

The total height of the bamboo is equal to the value of x plus the value of c. Hence, we can have the equation

`\begin{gathered} x+c=10 \\ x+\sqrt[]{x^2+25}=10 \end{gathered}`

Solve for x, we get

`\begin{gathered} \sqrt[]{x^2+25}=10-x \\ x^2+25=(10-x)^2 \\ \cancel{x^2}+25=100-20x+\cancel{x^2} \\ 20x=100-25 \\ 20x=75 \\ x=(75)/(20) \\ x=3.75 \end{gathered}`

Therefore, the break occurs at 3.75 ft.

Answer: 3.75 ft