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
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
I'll keep that under advisement