you can draw this triangle ... as the sides follow the triangle inequality , i.e. , sum of two sides is always greater than third side. also if you let 20 be base , the area is 96
Yes this is the simple answer. But when you double the area (96+96=192) and divide it by height (9.8) the answer should be 20. In case there is an error which should be rectified Use the formula, 1/4 ((4a^2*c^2- (a^2*c^2-b^)^2.
Since 12-16-20 is 4 times 3-4-5 and that is a right triangle 3*3+4*4=5*5 with area 6, we get that the area is 16*6 = 60+36=96. Thus the heights in this triangle are 12,16 and 96*2/20=9,6 and not 9,8 as falsly noted.
Brilliantly presented solutions as always.
genius solution, Richard! never even thought of it at first glance!
In Richard's style, "and we're done!"
This guy is legendary. SO COOL DUDE! keep up ! ;)
Awesome explanation as always. Richard, do you have any tips for an 8th grader for the AMC 10? I'm doing it tomorrow :)
Sanjay Suresh at least you got the last 5 questions
PhCuber05 wdym
I had trouble keeping up with this one.
What web can i find out them
Nice vid
Can you draw a triangle of sides 20, 16 and 12 meters. Also show that the area is 96. The height is 9.8 meters.
you can draw this triangle ... as the sides follow the triangle inequality , i.e. , sum of two sides is always greater than third side. also if you let 20 be base , the area is 96
Yes this is the simple answer. But when you double the area (96+96=192) and divide it by height (9.8) the answer should be 20. In case there is an error which should be rectified Use the formula, 1/4 ((4a^2*c^2- (a^2*c^2-b^)^2.
Isn't this just an alternative for a 3-4-5 right triangle?
Since 12-16-20 is 4 times 3-4-5 and that is a right triangle 3*3+4*4=5*5 with area 6,
we get that the area is 16*6 = 60+36=96.
Thus the heights in this triangle are 12,16 and 96*2/20=9,6 and not 9,8 as falsly noted.
10:04
*vsauce music intensifies*
one one one one lol