Multiple choice

A tree $AC$ is broken over by over by wind from $B. D$ is the point where the top of the broken tree touches the ground and $BD$ makes an angle of $\displaystyle 45^{\circ}$ with the ground, if the distance between he base of the tree and the point $D = 10$ m. What is the height of the tree ?

  1. $20$ m
  2. $\displaystyle 10(1+\sqrt{2})$ m
  3. $\displaystyle 10\sqrt{2}$ m
  4. $\displaystyle 20\sqrt{2}$ m
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B Correct answer
Explanation

Let the tree break at height h. The top touches the ground at distance 10m. The broken part forms the hypotenuse of a 45-45-90 triangle. Hypotenuse = 10 / cos(45) = 10 * sqrt(2). The height of the tree is the vertical part (10 * tan(45) = 10) plus the broken part (10 * sqrt(2)). Total = 10 + 10 * sqrt(2) = 10(1 + sqrt(2)).

AI explanation

In the right triangle formed by the broken part of the tree, the horizontal distance is 10 m and the angle with the ground is 45 degrees. Using the tangent ratio, the vertical height of the break is 10 times tan(45), which equals 10 m. The length of the broken fallen part is the hypotenuse, calculated as 10 divided by sin(45), which equals 10 times sqrt(2) m. The total height of the tree is the sum of the standing part and the fallen part, resulting in 10 plus 10 times sqrt(2) m, or 10(1 + sqrt(2)) m.

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