Lighting hit a tree three-eighths of the distance up the trunk from the ground and broke the tree so that its top landed 40 feet from the base of the trunk. How tall was the originally tree?
The wording is somewhat ambiguous. The way I read it says that the part that fell off must be 5/8 of the tree
so (5/8)x = 40
5x = 320
x = 64
or
Is the part that broke off still attached and when bent over lands 40 feet from the tree?
then (3x/8)^2 + 40^2 = (5x/8)2
9x^2/64 + 1600 = 25x^2/64
(16/64)x^2 = 1600
take √ of both sides
(1/2)x = 40
x = 80
To solve this problem, we need to set up an equation and use a bit of algebra to find the height of the original tree. Let's break down the problem step by step:
1. Let's assume that the height of the entire tree is 'h' feet.
2. The lightning hit the tree at a point three-eighths of the distance up the trunk from the ground. This point can be represented as (3/8)h feet.
3. The top of the tree fell 40 feet from the base of the trunk. Since the lightning hit the tree at (3/8)h feet, the remaining height of the tree after the lightning strike would be (h - (3/8)h) = (5/8)h feet.
4. We know that the remaining height of the tree, (5/8)h feet, is equal to 40 feet. So we can set up the equation: (5/8)h = 40.
5. To solve for 'h', we will multiply both sides of the equation by (8/5): h = 40 * (8/5) = 64.
Therefore, the original height of the tree was 64 feet.