(N/A) Given,the dimensions of the rectangular tile are $50\, cm \times 70\, cm$.
Area of the rectangular tile $= 50\, cm \times 70\, cm = 3500\, cm^2$.
The sides of each triangle are $a = 25\, cm, b = 17\, cm$,and $c = 26\, cm$.
Now,the semi-perimeter $s$ is given by $s = \frac{a + b + c}{2}$.
$s = \frac{25 + 17 + 26}{2} = \frac{68}{2} = 34\, cm$.
Using Heron's formula,the area of one triangle is $\sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{34(34 - 25)(34 - 17)(34 - 26)} = \sqrt{34 \times 9 \times 17 \times 8}$.
Area $= \sqrt{(17 \times 2) \times 3^2 \times 17 \times (2^3)} = \sqrt{17^2 \times 2^4 \times 3^2} = 17 \times 4 \times 3 = 204\, cm^2$.
Total area of $8$ triangles $= 8 \times 204\, cm^2 = 1632\, cm^2$.
Thus,the total area of the design is $1632\, cm^2$.
Remaining area of the tile $=$ Area of the rectangle $-$ Area of the design.
Remaining area $= 3500\, cm^2 - 1632\, cm^2 = 1868\, cm^2$.
Hence,the total area of the design is $1632\, cm^2$ and the remaining area of the tile is $1868\, cm^2$.