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Three Equal Cubes Are Placed Adjacently in a Row. Find the Ratio of Total Surface Area of the New Cuboid to that of the Sum of the Surface Areas of the Three Cubes. - CBSE Class 9 - Mathematics

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Question

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

Solution

Length of new cuboid= 3a 
Breadth of cuboid=a 
Height of new cuboid= a 

The total surface area of new cuboid

`⇒(TSA)_1= 2[lb+bh+hl]`

`⇒(TSA)_1=2[3axxa+axxa+3axxa]`

`⇒(TSA)= 14 a^2`

Total surface area of three cubes

`⇒(TSA)_2=3xx6a^2=18a^2`

`∴(TSA)_2/(TSA_2)=(14a^2)/(18a^2)= 7/9`

`Ratio is 7:9`

 

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APPEARS IN

 RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 18: Surface Areas and Volume of a Cuboid and Cube
Ex.18.10 | Q: 6 | Page no. 14
Solution Three Equal Cubes Are Placed Adjacently in a Row. Find the Ratio of Total Surface Area of the New Cuboid to that of the Sum of the Surface Areas of the Three Cubes. Concept: Surface Area of a Cuboid and a Cube.
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