The Murderous Maths Research Lab proudly presents...
This is possibly the most exciting page on the whole internet. Or the dullest. It depends on you.
Here's one that we photographed earlier.
Usually if you want to find the area of a trapezium you use the simple little formula.
In the formula, a and c are the lengths of the two parallel sides, and "h" is the height of the trapezium.
This is probably all you ever need to know about the area of trapeziums, so if you want to leave now, that's fine. Goodbye! You now continue at your own risk. So far it's all been simple BUT... ...we were asked if we could work out the area if we knew the lengths of the four sides, but we DIDN'T know the height! Obviously the usual formula wasn't going to help much, so with the help of some of our regular visitors to this site, we came up with another formula. The first sensible suggestion came from our Singapore friend Hu Yi Jie. He suggested splitting up the trapezium up into three bits and then attacking the problem like this:
where s=½(b+d+ac) And when you swap every "s" in the main formula with ½(b+d+ac) and boil it down you get our formula for the area of a trapezium...
Of course, we've only shown you Hu Yi Jie's EASY version here. We had already worked out another way of solving the problem which we think is even easier. But then we would wouldn't we?
We tested it on a trapezium with sides a=30, b=10, c=9 and d=17. This conveniently makes a trapezium of height 8, and it breaks down into a rectangle measuring 9x8 (area=72), and two right angled triangles of 6810 (area=24) and 81517 (area=60). The trapezium area is 72+24+60=156 which is what the formula gives. Oooooh.... we're so pleased with ourselves. The PYTHAGORAS FORMULAMany thanks to Timothy Ti'en Ern Ang  yet another young genius from Singapore!

Just in case you're wondering, this web page had been online for 12 years before Tim sent us this. The amazing thing is that Tim is ten years old, so the rest of this page has been on the planet two years longer than Tim has! 
And thanks to Kevin Wai who noticed we'd made a mistake when we first put this online! Kevin was seven years old.
If any babies find mistakes, do please get in touch. 
You'll find out how algebra and equations and all this stuff works in THE PHANTOM X. 
For the first triangle, Pythagoras says: b^{2} = h^{2} + z^{2}
For the second triangle: d^{2} = h^{2} + (acz)^{2}
We'd like to know what z is, so as we have two simultaneous equations, we can subtract the second equation from the first equation to get rid of h. (Some people enjoy this sort of thing.)
b^{2}  d^{2} = z^{2}  (acz)^{2}
Woohoo! If we multiply the ugly thing at the end out we get
(acz)^{2} = (a^{2}  ac az ca +c^{2} + cz  za +zc + z^{2})
which simplifies to
(a^{2} +c^{2} +z^{2}  2ac 2az + 2cz )
We now put this in place of the (acz)^{2}, not forgetting that there was a minus sign in front of the bracket...
b^{2}  d^{2} = z^{2}  a^{2} c^{2} z^{2} + 2ac +2az  2cz
Hey! There's a +z^{2} and a z^{2} so they disappear...
b^{2}  d^{2} =  a^{2} c^{2} + 2ac +2az  2cz
We now get everything without a z to the same side.
b^{2}  d^{2} + a^{2} 2ac + c^{2} = 2az  2cz
And now we conveniently notice that a^{2} 2ac + c^{2} factorises to (ac)^{2}, and the other side factorises too.
b^{2}  d^{2} + (ac)^{2} = 2z(ac)
Divide through by 2(ac), swap the sides over and we've got a value for z. BINGO!
z = [b^{2}  d^{2} + (ac)^{2} ]/2(ac)
At this point it's helpful to work out what z^{2} will be:
z^{2} = [b^{2}  d^{2} + (ac)^{2} ]^{2}/[2(ac)]^{2}
All we do now is go back to our first Pythagoras triangle equation and twiddle it round so we get h^{2} = b^{2} z^{2} and then put in the z^{2} value we just worked out.
h^{2} = b^{2}  [b^{2}  d^{2} + (ac)^{2} ]^{2}/[2(ac)]^{2}
Take a square root of both sides and we get...
Awesome.
TOUGHER FORMULAS! 
JENNY'S AREA OF TRAPEZIUM = ½(a+c)*b*sin{cos^{1}[(d^{2}(ac)^{2}b^{2})/(2(ac)b]}
CARL'S AREA OF TRAPEZIUM = (a+c)(½bd(sin(cos^{1}(b^{2}+d^{2}(ac)^{2}))))/4(ac)
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