Math, asked by lakshyagupta19361, 10 months ago

The external dimensions of an open iron box are 20 cm, 1-
ensions of an open iron box are 20 cm, 15 cm and 12 cm. If the thickness ofironis 1 cm, find
the volume of the iron used.

Answers

Answered by RvChaudharY50
100

Correct Question :-- The external dimensions of an open iron box are 20 cm, 15 cm and 12 cm. If the thickness of iron is 1 cm, findthe volume of the iron used. ?

Concept and Formula :---

→ when box is opened , that means it is opened from upper part, Hence, when finding internal sides of cuboidal box , we use :--

=> internal Length = (External Length or actual length - 2*Thickness)

=> internal Breadth = (External Breadth or actual Breadth - 2*Thickness)

=> internal Height = ( External Height - Thickness) { as box is opened . } .

→ Volume of box = length * Breadth * Height .

→ Volume of iron used = (External volume - inner volume ) .

__________________________

Solution :--

External dimensions of the rectangular box = 20cm*15cm*12cm

→external volume of the box = 3600cm³

Now, since the thickness of the box is 1 cm and box is open.

→ internal length = (20-1*2)cm= 20-2 = 18cm

→ internal breadth = 15-1*2=15-2=13 cm

→ internal height = 12 -1 = 11 cm [since box is open]

_______________________

So, the internal dimensions of the rectangular box are = 18cm*13cm*11cm

→ internal volume of the box = 2574cm³

So, the volume of the iron = (External volume of the box - internal volume of the box).

→ Volume of the iron = 3600 - 2574

→ Volume of the iron = 1026cm³ .

Hence, volume of iron used to make box will be 1026cm³...

Answered by Anju12335
0

Answer:

Step-by-step explanation:

Similar questions