Question

A room is 30m long, 24m broad and 18m high. Find: (a) length of the longest rod that can be placed in the room. (b) its total surface area.(c) its volume.

Answer 1

Hey!!!
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●For room
l=30m
b=24m
h=18m
●Longest rod that can be placed in the room(diagonal)=42.42m
$\sqrt{ {l}^{2} + {b}^{2} + {h}^{2} } = \sqrt{ {30}^{2} + {24}^{2} + {18}^{2} } = \sqrt{1800}$
●TSA=2(lb+bh+lh)=2{(30×24)+(24×18)+(30×18)}=2 (720+432+540)=2×1692=3384cm^2
●Volume=l×b×h=(30×24×18)=12960cm^3
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Hope it helps!!! :)

Answer 2

Answer:

a) Length of the longest rod that can be placed in the room is 42.43 m

b) Total surface area of rectangle is 3384 m²

c) Volume of cuboid is 12960 m³.

Step-by-step explanation:

A room is 30 m long, 24 m broad and 18  m high

Given length l = 30 m

breadth b = 24 m and

height h = 18 m

a) Longest rod that can be placed in a room is nothing but its diagonal.

Length of diagonal of a cuboid = $\sqrt{(l^2+b^2+h^2)}$

Substitute the values,

Length of diagonal of a cuboid = $\sqrt{((30)^2+(24)^2+(18)^2)}$

Length of diagonal of a cuboid = $\sqrt{(900+576+324)}$

Length of diagonal of a cuboid = $42.43$ meters.

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

Putting values,

Total surface area of rectangle = 2(30 × 24+ 24 × 18 + 18× 30 )

Total surface area of rectangle = 2(720+ 432+ 540 )

Total surface area of rectangle = 2( 1692 ) = 3384 m²

c)  Volume of cuboid = l × b × h

Volume of cuboid = 30 × 24 × 18 = 12960 m³.