Question

The area of the floor of a cuboidal party hall is 80
m
2
. If the area of two walls of the hall is 135
m
2
and 108
m
2
, what is the height of the hall(in m)?

Answer 1

Step-by-step explanation:

The perimeter of Floor is given as 2(l+b)=250m

The height of walls h=6m

The Lateral surface area is 2h(l+b)=6×250

1500m

2

Cost of 1m

2

is 8

Cost of 1500m

2

is 8×1500

12000Rs.

Answer 2

Given:

The area of the floor of a cuboidal party hall=80$m^{2}$

The area of two walls of the hall=135$m^{2}$ and 108$m^{2}$

To find:

The height of the hall

Solution:

The height of the hall is 13.5 m.

We can find the height by following the given process-

We know that the hall is cuboidal.

So, the floor and the walls are rectangular.

Let the length of the hall be L, breadth be B, and height be H.

The area of the floor of the hall is the product of its length and breadth.

The area of the floor=length of the hall×breadth of the hall

80=L×B (1)

One of the walls of the hall is a product of its breadth and height and the other wall is the product of its length and height.

So, the area of one wall=breadth of the hall×height of the hall

135=B×H (2)

Similarly, the area of the other wall=Length of the hall×height of the hall

108=L×H (3)

Using the three equations, we will find out the height of the hall.

From (1), L=80/B

Substituting this in (3),

H×80/B=108

80H/B=108

80H/108=B

Putting the value of B in (2),

80H/108×H=135

$H^{2}$ = 135×108/80

$H^{2}$ =182.25

H=13.5

Therefore, the height of the hall is 13.5 m.