Answer: 6 m

Step-by-step explanation:

Given,

Perimeter of the floor = 250m.

Since the hall is rectangular, the length and width of the hall will be equal to the length and width of 1 wall.

Thus, we can say that the AREA of the four walls COMBINED will be the lateral surface area of the cuboid (also in hint).

We know that

Lateral Surface area of Cuboid (room if taken along with walls is cuboid shaped) = 2(l+b)*h

But, perimeter = 2(l+b)

Thus,

Lateral surface area of the room = 250*h (....................................[1])

We see that in order to equate h, we need lateral surface area

We can obtain the area of 4 walls by so -

Total cost = 15000 for x meter^2

But cost of m^2 = 10 rs

Hence, Area of the four walls = 15000/10 = 1500 m^2

Therefore,

From [1],

250*h = 1500

or h = 1500/250 = 6 m.

We conculde that the hight of the hall is 6 m.

Hope it helped!

Verified answer

Answer:

[tex]\sf\:\boxed{\bf\:Height\:of\:hall\:is\: 6 \: m \: } \\ [/tex]

Step-by-step explanation:

Let assume that l, b and h represents the length, breadth and height of a rectangular hall respectively.

Given that, The floor of a rectangular hall has a perimeter 250 m.

[tex]\implies\bf\: 2(l + b) = 250 - - - (1)\\ [/tex]

Further given that, the cost of painting the four walls at the rate of Rs 10 per m² is Rs 15000.

So,

[tex]\sf\: Area\:of\:four\:walls = \dfrac{15000}{10} \: {m}^{2} \\ [/tex]

[tex]\sf\: Area\:of\:four\:walls = 1500 \: {m}^{2} \\ [/tex]

[tex]\sf\: 2(l + b) \times h = 1500 \: \\ [/tex]

On substituting the value from equation (1), we get

[tex]\sf\: 250 \times h = 1500 \: \\ [/tex]

[tex]\implies\sf\:h = 6 \: m \\ [/tex]

Hence,

[tex]\implies\sf\:\boxed{\bf\:Height\:of\:hall\:is\: 6 \: m \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information:

[tex] \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}[/tex]