Answer:

[tex]\boxed{\bf\:Cost \: of \: polishing \: the \: tiles \: is \: Rs \: 705.60 \: }\\ [/tex]

Step-by-step explanation:

Given that, A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm.

So, Let's first evaluate the area of one triangular tile, whose sides are 9 cm, 28 cm and 35 cm.

Let assume that a = 9 cm, b = 28 cm, c = 35 cm

Now, Consider Semi-perimeter, s

[tex]\sf\: s = \dfrac{a + b + c}{2} \\ [/tex]

[tex]\sf\: s = \dfrac{9 + 28 + 35}{2} \\ [/tex]

[tex]\sf\: s = \dfrac{72}{2} \\ [/tex]

[tex]\implies\sf\:s = 36 \: cm \\ [/tex]

Now,

[tex]\sf\: Area\:of\:one\:triangular\:tile \\ [/tex]

[tex]\sf\: = \: \sqrt{s(s - a)(s - b)(s - c)} \\ [/tex]

[tex]\sf\: = \: \sqrt{36(36 - 9)(36 - 28)(36 - 35)} \\ [/tex]

[tex]\sf\: = \: \sqrt{36(27)(8)(1)} \\ [/tex]

[tex]\sf\: = \: \sqrt{(6)(6)(3)(3)(3)(2)(2)(2)} \\ [/tex]

[tex]\sf\: = \: 6 \times 3 \times 2 \sqrt{6} \\ [/tex]

[tex]\sf\: = \: 36 \sqrt{6} \: {cm}^{2} \\ [/tex]

Thus,

[tex]\sf\: Area\:of\:one\:triangular\:tile = 36 \sqrt{6} \: {cm}^{2} \\ [/tex]

So,

[tex]\sf\: Area\:of\:16\:triangular\:tiles = 36 \sqrt{6} \times 16 = 576 \sqrt{6} \: {cm}^{2} \\ [/tex]

[tex]\implies\sf\: Area\:of\:16\:triangular\:tiles = 576 \times 2.45 = 1411.20 \: {cm}^{2} \\ [/tex]

Now, Further given that

[tex]\sf\: Cost \: of \: polishing \: 1 \: {cm}^{2} = 50 \: p \\ [/tex]

[tex]\sf\: Cost \: of \: polishing \: 1411.20\: {cm}^{2} =1411.20 \times 50 = 70560 \: p \\ [/tex]

Thus,

[tex]\implies\sf\: Cost \: of \: polishing \: 1411.20\: {cm}^{2} = Rs \: 705.60 \: \\ [/tex]

Hence,

[tex]\implies\sf\: \boxed{\bf\:Cost \: of \: polishing \: the \: tiles \: is \: Rs \: 705.60 \: }\\ [/tex]