A Very Nice Geometry Problem | Find the area of the triangle | 2 Different Methods to Solve

[@gandelve]

3 steps easy. Drop a perpendicular from A to meet BC at E. 13^2 = h^2 + (x+h)^2 = h^2 + x^2 + 2xh + h^2 10^2 = h^2 + (x-h)^2 = h^2 + x^2 - 2xh +h^2 Subract these equations 160 -100 = 69 = 4xh The area of ABC is 2xh/2 = xh = 69/ 4

[@aysemaysem3042]

Thank you very much

[@nathancc2526]

Appolonius theorem will work ryt? And then find area by herons formula

[@sunilmathews4439]

Nice.. it would be better, if you could explain where this will be used in practical life in the lecture.

[@gandelve]

It's basic practice. Imagine you were building a bridge. In the middle of the river you could build a pylon to support two halves (which were the same size because of industrial pre-fabrication). And you would support the bridge with a 23m cable from a point which was off-center. And so on. If you can't solve this problem, you will never be an engineer.

[@dickroadnight]

You would not need the area to build a bridge. It could be a surveying or building problem… you need the area for paint, concrete etc.

[@alokranjan4149]

Very beautiful practical application 👍👍

[@rachidmsmdi6433]

69:4

[@dickroadnight]

The area of a triangle is half the base times the height… x is half the base, and the height is y/sqrt2. You got xy at 16:42, and the area is x.y/sqrt2 = 69/4.

[@yakupbuyankara5903]

69/4.