Case study:
The school administration planned to renovate the common room and asked Rohan to calculate its dimensions so that the new furniture could be arranged properly. The common room is cuboidal in shape.

After collecting the necessary data, Rohan found that :

The volume of the common room is 144 cubic units.

The total surface area of the common room is 192 sq. units.

The height of the room is 3 units.

Based on the above information answer the following:

(a) Form equations for volume and total surface area assuming x and y as the length and breadth.

(b) Solve the equations for finding the dimensions of the common room.

(c) While construction is going on, what length of the biggest iron rod can be put in the room?

Given,

Volume of the common room = 144 cubic units.

Total surface area of common room = 192 sq. units

Height of the room = 3 units.

(a) Let, length = x units and breadth = y units.

Volume of cuboid = l × b × h

144 = x × y × 3

3xy = 144

xy = $\dfrac{144}{3}$

xy = 48 ………(1)

By formula,

Total surface area of cuboid = 2(lb + bh + hl)

192 = 2(x × y + y × 3 + 3 × x)

192 = 2(xy + 3y + 3x)

xy + 3y + 3x = $\dfrac{192}{2}$

xy + 3y + 3x = 96……..(2)

Hence, equation for volume : 3xy = 144 and for TSA = 2(xy + 3y + 3x) = 192.

(b) Substituting the value of equation (1) in equation (2), we get :

48 + 3y + 3x = 96

3x + 3y = 96 - 48

3x + 3y = 48

3(x + y) = 48

x + y = $\dfrac{48}{3}$

x + y = 16

x = 16 - y ………(3)

Substituting the value of x from equation (3) in equation (1) :

(16 - y) × y = 48

16y - y² = 48

y² - 16y + 48 = 0

y² - 12y - 4y + 48 =0

y(y - 12) - 4(y - 12) = 0

(y - 12) (y - 4) = 0

y = 12 or y = 4

If y = 12, then

x = 16 - y = 16 - 12 = 4.

If y = 4, then

x = 16 - y = 16 - 4 = 12.

Since length is longer than breadth,

∴ x = 12 and y = 4

Hence, x = 12 and y = 4.

(c) Length of the biggest rod that can be put in a room = Diagonal of room

Diagonal of room (cuboid) = $\sqrt{l^2 + b^2 + h^2}$

$\Rightarrow d = \sqrt{12^2 + 4^2 + 3^2} \\[1em] \Rightarrow d = \sqrt{144 + 16 + 9} \\[1em] \Rightarrow d = \sqrt{169} \\[1em] \Rightarrow d = 13 \text{ units}.$

Hence, length of the biggest iron rod can be put in the room = 13 units.