How about we take the area of tha triangle which is (1/2)* 10 *x make it twice for a diagonal cuts a rectangle in two congruent triangles and then equate it to 6*8.(area of the rectangle)

I solved it with y+z=10 -> y=10-z y²+x²=36 z²+x²=64 Put y in II (10-z)²+x²=36 100-20z+z²+x²=36 x²= -64+20z-z² Put x² in III z²+(-64+20z-z²)=64 z²-64+20z-z²=64 20z=128 z=6,4 That means y=3,6 (I) Solving x with I 3,6²+x²=36 x²=23,04 x=4,8

This format is a bit weird... With long pauses

@StickScience

3 жыл бұрын

I'll keep that under advisement