Question

A Christmas tree is supported by a wire that is 4 meters longer than the height of the tree. The wire is anchored at a point whose distance from the base of the tree is 28 meters shorter than the height of the tree. What is the height of the tree?

Answer 1

The scenario would be represented by a given right angle triangle as shown below

Looking at the triangle,

BC = x represents the height of the tree

AC = x + 4 represents the length of the wire

AB = x - 28 represents the distance from the base of the tree to the wire.

We would solve for x by applying pythagorean theorem which states that

Hypotenuse^2 = shorter leg^2 + longer leg^2

hypotenuse = x + 4

shorterleg = x - 28

Longer leg = x

Thus, we have

`\begin{gathered} (x+4)^2=(x-28)^2+x^2 \\ (x+4)(x+4)=(x-28)(x-28)+x^2 \\ x^2+4x+4x+16=x^2-28x-28x+784+x^2 \\ \text{Collecting like terms, we have} \\ x^2-x^2-x^2\text{ +4x + 4x + 28x + 28x + 16 - 784 = 0 } \\ x^2+64x\text{ - 768 = 0} \end{gathered}`

We would solve the quadratic equation by applying the general formula for solving quadratic equations which is expressed as

`\begin{gathered} x\text{ = }\frac{-\text{ b}\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{From the equation, } \\ a\text{ = 1, b = 64, c = - 768} \\ x\text{ = }\frac{-\text{ 64 }\pm\sqrt[]{64^2-4(1*-768)}}{2*1} \\ x\text{ = }\frac{-64\pm\sqrt[]{4096\text{ + 3072}}}{2} \\ x\text{ = }\frac{-\text{ 64 }\pm84.66}{2} \\ x\text{ = }(20.66)/(2)\text{ or x = }\frac{-\text{ 148.66}}{2} \\ x\text{ = 10.3 or x = } \end{gathered}`