Trapezium Area

Notice that we don't know which sides of the trapezoid are parallel. And we have to figure that out.

From this point on, we're going to call our trapezoid $ABCD$ with $AB=a$ , $BC=d$ , $CD=b$ and $DA=c$ . Without loss of generality, we can say $b>a$ . And we're going to say that $AB||CD$ .

Now let's draw a line $BE$ parallel to $AD$ and say $BE$ intersects $CD$ at $E$ . Since we said earlier that $CD>AB$ , we can ensure that the point $E$ will be strictly on the line segment $CD$ .

Now our trapezoid $ABCD$ is divided into a parallelogram $ABED$ and a triangle $BEC$ .

And,

$AB=DE=a$ .

$AD=BE=c$ ,

$BC=d$ ,

$CE=b-a$ .

In order for $\triangle BEC$ to have a positive area, its sides must fulfill the triangle inequality. That means the sum of any two sides of $\triangle BEC$ must be greater than the other side.

So, we must have:

$c+(b-a)>d \Rightarrow b-a>d-c$

$c+d>b-a$

$d+(b-a)> c \Rightarrow b-a> c-d$ .

All these inequalities must hold if $\triangle BEC$ is non-degenerate.

Letting $d>c$ [WLOG], these inequalities can be written like this:

$d-c<b-a<c+d$ .

Now we're going to find out the possible cases for $a$ and $b$ . With the constraint of $a<b$ , it gives us $6$ cases to consider:

$\begin{array}{c|c} a & b & c & d & d-c & b-a & c+d & \text{does it fulfill the requirements?}\\ \hline 6 & 8 & 10 & 12 & 2 & 2 & 22 & \text{no} \\ \hline 6 & 10 & 8 & 12 & 4 & 4 & 20 & \text{no}\\ \hline 6 & 12 & 8 & 10 & 2 & 6 & 18 & \text{yes}\\ \hline 8 & 10 & 6 & 12 & 6 & 2 & 18 & \text{no}\\ \hline 8 & 12 & 6 & 10 & 4 & 4 & 16 & \text{no}\\ \hline 8 & 12 & 6 & 8 & 2 & 2 & 14 & \text{no}\\ \end{array}$

We see that there is only one case that satisfies all the requirements. Luckily for us, it turns out that in this case $\triangle BEC$ is a right triangle with side lengths $6$ , $8$ and $10$ . [If it was not a right triangle, we'd have to work a bit more to find the trapezoid height]

So the height of our trapezoid is simply $BE=c=8$ .

And its area= $\frac{1}{2} \times (AB+CD) \times BE$ $=\frac{1}{2} \times(a+b) \times c$

$=\frac{1}{2} \times(6+12) \times 8=72$ .

Note that all non-degenerate trapezoids with congruent but not necessarily corresponding side lengths have the same area. So any non-degenerate trapezoid with side lengths $6, 8, 10, 12$ will satisfy for the purpose of our calculation.

Construct a right triangle $ABC$ with $AB = 8, BC = 6, AC = 10$ and $BC$ is horizontal. Draw $DC$ such that $DC = 10$ and $AD$ is parallel to $BC$ . Since $ACD$ is isosceles, the median drawn from $C$ creates a rectangle $ABCM$ . Thus, $AM = BC = 6$ and $AD = 12$ , giving us a non-degenerate trapezoid with side lengths $6, 8, 10, 12$ .

The area of triangle $ABC$ is $(\frac {1}{2})(6)(8) = 24$ and, using Heron's formula, the area of triangle $ACD$ is $\sqrt {(16)(6)(6)(4)} = 48$ . Thus, the area of trapezoid $ABCD$ is $24 + 48 = 72$ .

Let the trapezoid be $EFLK$ , with $EF \| LK$ the bases (parallel sides). Let $a = EF$ , $b = LK$ , $c = EK$ , and $d = FL$ ; without loss of generality, assume $a > b$ and $c \geq d$ .

Let $G$ be a point on $EF$ such that $EGLK$ is a parallelogram; then $GL = c$ , $EG = b$ , and $GF = a - b$ . Apply the triangle inequality in triangle $GFL$ : this results in the condition $GL < FL + FG\ \ \ \therefore\ \ \ c < d + a - b \ \ \ \therefore\ \ \ a + d > b + c.$

Of the 24 possible assignments of $6, 8, 10, 12$ to $a, b, c, d$ , there is only one that satisfies the three conditions $a > b$ , $c \geq d$ , and $a + d > b + c$ . This is $a = 12,\ b = 6,\ c = 10,\ d = 8.$

If the height of the parallelogram is $h$ , we must have $c^2 = h^2 + x^2$ and $d^2 = h^2 + y^2$ , with $x + y = a - b = 6$ , with $y$ possibly negative.

Now $(c+d)(c-d) = c^2 - d^2 = (h^2 + x^2) - (h^2 + y^2) = x^2 - y^2 = (x+y)(x-y),$ so that $x - y = \frac{(c+d)(c-d)}{x+y} = \frac{18\cdot 2}{6} = 6,$ from which it follows that $x = 6$ and $y = 0$ . It is easy to see that the height of the parallelogram is $h = 6$ . Finally, $A = \tfrac12h(a+b) = \tfrac12\cdot 6\cdot (12+6) = \boxed{72}.$

According to Pythagoras' Theorem, 6,8,10 is one of the Pythagorean Triples .

We let the angle between 6, 8 equals to 90 degrees. Then we will get 10 as the hypotenuse.

One of the side of the trapezoid equals to 10, so we let this side and the hypotenuse of the triangle (6,8,10) form a isosceles triangle, with a base 12.

So, we will get a trapezoid with sides 8,6,10,12 accordingly.

The area of the right-angle triangle is 6 \times 8 \times \frac{1}{2} = 24 square units

We can use Pythagoras' Theorem to find out the height of the isosceles triangle with sides 10, 10 and 12. Draw a perpendicular line on the base. It's also the median of the triangle. Then we get { \frac{12}{2} }^2 + {height}^2 = {10}^2 . By solving the equation, we get the height as 8 units.Therefore, the area of the isosceles triangle = 12 \times 8 \times \frac{1}{2} = 48 square units.

Hence, the area of the trapezoid = Area of Right-angle Triangle + Area of Isosceles Triangle = 24 + 48 = 72 square units.

Let $a$ and $b$ be the parallel sides of the trapezoid, $b$ and $c$ be the remaining 2 sides of the trapezoid.

Let $a>b$ and $c>d$

Since the trapezoid is $non-degenerate$

therefore, from the formula

$Area\ of\ trapezoid\ = \frac{a+b}{|a-b|} \sqrt((-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d))$

We have the inequalities,

$a+c>b+d$

$a+d>b+c$

Since $a+c$ and $a+d \neq 18$ , hence by logic when $a=6,8,10$ , the values of $b,c,d$ contradict the inequalities above. The only possible values are $a=12$ , $b=6$ , $c=10$ , $d=8$

Let $a=x+y+6$ , $h$ as the height of the trapezoid and where by applying the $Pythagoras's\ Theorem$ , we have

$8^{2}\ - x^{2}\ = h^{2}$

$10^{2}\ - y^{2}\ = h^{2}$

$x+y=6$

Solving the simultaneous equations, we have $x = 0$ , $y = 6$ , hence we can conclude that the height, $h$ of the trapezoid is $d = 8$

$Area\ of\ trapezoid\ = \frac{1}{2} (a +b)(d) = 72$

@Calvin Lin we have for the existence of trapezium, |a-b|>|d-c| where a,b are parallel sides,from the numbers we can see only (12,6) can be the parallel sides

Call the distance between the parallel side $x$ . (What else, I ask you?)

Now we can choose pairs of possible parallel sides from the four given. One of these will be longer than the other. There will be right triangles formed (possibly degenerate) at the two ends of the trapezium between the parallel sides. We can find quadratic equations in $x$ now for each possible pair of sides and attempt to solve them.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

from itertools import combinations
from sympy import *

sides=[6,8,10,12]
var('x')

for choice in combinations(sides, 2):
    small, large=choice
    A,B=list(set(sides)-set(choice))
    eqn='%(large)s-(sqrt(%(A)s**2-x**2)+sqrt(%(B)s**2-x**2))-%(small)s' %locals()
    #~ print eqn
    solns=solve(sympify(eqn))
    if solns:
        print '*** solutions'
        print 'parallels', small, large
        print 'others', A,B
        for soln in solns:
            print soln, N(sqrt(A**2-soln**2)), N(sqrt(B**2-soln**2))

If I tried to do this 'manually' I'd be at it for days and I would still get the wrong answer. Someone with insight would have got this instantly.

1
2
3
4
5

*** solutions
parallels 6 12
others 8 10
-8 0 6.00000000000000
8 0 6.00000000000000

Because we're not given the order of the side lengths or which sides are parallel, we can arrange them any way. The easiest way is with the parallel sides being 6 and 12. This creates a right trapezoid of height 8. The area is then ((6 + 12) / 2) * 8 = 72