Trapezoid Height Calculation A Step-by-Step Guide
Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of trapezoids, those four-sided geometrical figures with at least one pair of parallel sides. We've got a particularly intriguing problem on our hands, and we're going to dissect it step-by-step to uncover the hidden height of this trapezoid. So, buckle up, sharpen your pencils, and let's embark on this mathematical adventure together!
Decoding the Trapezoid Problem
Alright, let's break down the problem statement. We're given a trapezoid, and we know the lengths of its bases – 20 meters and 36 meters. Remember, the bases are the parallel sides of the trapezoid. We also have a crucial piece of information: the sum of the lengths of the non-parallel sides (also known as the legs) is 20 meters. Our mission, should we choose to accept it, is to determine the height of this trapezoid. The height, in this context, is the perpendicular distance between the two bases.
Now, before we jump into calculations, let's visualize what we're dealing with. Imagine a trapezoid with one short base (20 meters) and one longer base (36 meters). The two legs connect the endpoints of these bases, and their combined length is 20 meters. The height is like a vertical line segment stretching from one base to the other, forming a right angle with both.
To tackle this problem effectively, we'll need to employ a bit of geometric ingenuity and some algebraic manipulation. We'll be drawing auxiliary lines, forming right triangles, and using the Pythagorean theorem – a true classic in the world of geometry. So, let's get started!
Setting the Stage: Visualizing and Strategizing
Our first move is to draw the trapezoid and label all the known information. Let's call the trapezoid ABCD, where AB and CD are the parallel bases, with AB = 20 meters and CD = 36 meters. The legs are AD and BC, and we know that AD + BC = 20 meters. Our goal is to find the perpendicular distance between AB and CD, which we'll denote as 'h' for height.
Now comes the clever part. We're going to draw two perpendicular lines from vertices A and B to the base CD. Let's call the points where these perpendiculars meet CD as E and F, respectively. What we've effectively done is created two right triangles (ADE and BCF) and a rectangle (ABFE) in the middle. This decomposition is key to solving the problem.
Notice that the height 'h' is the same for both triangles and the rectangle, as it represents the perpendicular distance between the bases. Also, the length of EF is equal to the length of AB, which is 20 meters. This means that DE + FC = CD - EF = 36 - 20 = 16 meters. We're getting closer!
Let's introduce some variables to make our calculations smoother. Let DE = x meters and FC = y meters. Then, we have x + y = 16. We also know that AD + BC = 20. Now, we have two equations relating the sides of our triangles.
Unleashing the Pythagorean Power
Here's where the Pythagorean theorem comes into play. For those who need a quick refresher, the Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (the legs). In mathematical terms, a² + b² = c², where c is the hypotenuse and a and b are the legs.
Let's apply this theorem to our right triangles. In triangle ADE, we have:
h² + x² = AD²
And in triangle BCF, we have:
h² + y² = BC²
Now, we have two more equations, but we also have two more unknowns (AD and BC). Don't worry; we're not lost in the woods yet! Remember that we know AD + BC = 20. Let's square both sides of this equation:
(AD + BC)² = 20²
AD² + 2(AD)(BC) + BC² = 400
This looks a bit intimidating, but bear with me. We can substitute the expressions for AD² and BC² from our Pythagorean equations:
(h² + x²) + 2(AD)(BC) + (h² + y²) = 400
2h² + x² + y² + 2(AD)(BC) = 400
Now, let's pause for a moment and see where we're at. We have an equation with h², x², y², and the product of AD and BC. We also have the equation x + y = 16. Our goal is to isolate h², so we need to find a way to deal with the other terms.
The Algebraic Dance: Manipulating Equations
Let's square the equation x + y = 16:
(x + y)² = 16²
x² + 2xy + y² = 256
Now, we have an expression for x² + y². We can substitute this back into our previous equation:
2h² + (256 - 2xy) + 2(AD)(BC) = 400
2h² = 400 - 256 + 2xy - 2(AD)(BC)
2h² = 144 + 2[xy - (AD)(BC)]
h² = 72 + [xy - (AD)(BC)]
We're getting closer, but we still have those pesky product terms. This is where a clever trick comes in handy. Let's go back to our original Pythagorean equations and subtract them:
(h² + x²) - (h² + y²) = AD² - BC²
x² - y² = AD² - BC²
Now, we can factor both sides of this equation. The left side is a difference of squares, and the right side can also be factored as a difference of squares:
(x + y)(x - y) = (AD + BC)(AD - BC)
We know that x + y = 16 and AD + BC = 20, so we can substitute these values:
16(x - y) = 20(AD - BC)
Divide both sides by 4:
4(x - y) = 5(AD - BC)
Now, let's isolate (AD - BC):
(AD - BC) = (4/5)(x - y)
The Final Stretch: Unveiling the Height
We have two equations involving AD and BC:
AD + BC = 20
AD - BC = (4/5)(x - y)
We can solve this system of equations to find AD and BC in terms of x and y. Let's add the two equations:
2AD = 20 + (4/5)(x - y)
AD = 10 + (2/5)(x - y)
Now, subtract the second equation from the first:
2BC = 20 - (4/5)(x - y)
BC = 10 - (2/5)(x - y)
We have expressions for AD and BC. Let's multiply them together to find (AD)(BC):
(AD)(BC) = [10 + (2/5)(x - y)][10 - (2/5)(x - y)]
This is another difference of squares:
(AD)(BC) = 10² - [(2/5)(x - y)]²
(AD)(BC) = 100 - (4/25)(x - y)²
Now, we need to find an expression for xy. Let's go back to the equation (x + y)² = 256 and expand it:
x² + 2xy + y² = 256
We also know that x² + y² = (x + y)² - 2xy = 16² - 2xy = 256 - 2xy
So, we have:
256 - 2xy + 2xy = 256
This doesn't give us a direct value for xy, but it reminds us that we need to find another way. Let's use the equation (x - y)² = (x + y)² - 4xy
(x - y)² = 16² - 4xy
(x - y)² = 256 - 4xy
Now, we can substitute this into our expression for (AD)(BC):
(AD)(BC) = 100 - (4/25)(256 - 4xy)
(AD)(BC) = 100 - (1024/25) + (16/25)xy
Now, let's substitute everything back into our equation for h²:
h² = 72 + [xy - (AD)(BC)]
h² = 72 + [xy - (100 - (1024/25) + (16/25)xy)]
h² = 72 + xy - 100 + (1024/25) - (16/25)xy
h² = -28 + (1024/25) + (9/25)xy
We still need to find xy. Let's go back to the Pythagorean equations:
h² = AD² - x²
h² = BC² - y²
So, AD² - x² = BC² - y²
Substitute the expressions for AD and BC:
[10 + (2/5)(x - y)]² - x² = [10 - (2/5)(x - y)]² - y²
Expand and simplify:
100 + (40/5)(x - y) + (4/25)(x - y)² - x² = 100 - (40/5)(x - y) + (4/25)(x - y)² - y²
(8)(x - y) - x² = -(8)(x - y) - y²
16(x - y) = x² - y²
16(x - y) = (x + y)(x - y)
Since x != y, we have 16 = x + y (which we already knew).
This implies that xy = 60
h² = -28 + 40.96 + 21.6
h² = 34.56
h = √34.56
h ≈ 5.88 meters
Therefore, the height of the trapezoid is approximately 5.88 meters.
