Treasure Hunt: Finding the Values of Right Angle Triangles
This final weeks course asks us to find a treasure with two pieces of a map. Now this may not be a common use of the Pythagorean Theorem to solve the distances for a right angled triangle but it is a fun exercise to find the values of the right angle triangle.
Buried treasure: Ahmed has half of a treasure map,which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x?
So, if you walk x paces north, then 2x+4 paces east, you have moved roughly east northeast to a distance of 2x+6 paces. This is a right angle, with 2x+6 on the hypotenuse or line c, so we must solve using the Pythagorean Theorem:
a² + b² = c²
add in the values
(x)² + (2x+4)² = (2x+6)²
multiply inside the parenthesis x² + 4x² + 16x + 16 = 4x² + 24x + 36
subtract 4x² + 24x + 36 from both sides x² - 8x - 20 = 0
factor the quadratic equation
(x -10)(x + 2) = 0
use zero factor property to solve
X – 10 = 0 or x + 2 = 0
creating a compound equation
x = 10 or x = -2
the answer cannot be – 2
x = 10
Now we will plug in the value and solve: x paces north and 2x + 4 paces east or 10 paces north and 2(10) + 4 = 24 paces east of Leaning Rock. Or 2x + 6 paces northeast or 2(10) + 6 = 26 paces northeast from the rock to reach the buried treasure.
In this exercise we learned how to find the value of a right angle triangle with one given point and two variables based off that point. I personally learned to take my time with a written out problem like this, as at first was a bit frustrated and confused with it. Overall this course MAT 221 has helped me quite a bit to refresh on my Algebra skills. Thank you for...
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