Question

three equal cubes are placed adjjacently 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​

Answer 1

Answer:

Let the side of the cube be 'x' cm

when placed end to end , the length of the cuboid is '3x' cm

height=breadth='x' cm

T.S.A of cuboid = 2(lb+lh+bh)=2(3x^2 + 3x^2 + x^2)=14 x^2  sq.cm.

sum of T.S.A of cubes = 3*6*x*x=18 x^2

ratio = 14:18=7:9

please mark as brainliest.

Answer 2

Answer: 7a sq : 9a sq

Step-by-step explanation:

We know that the formula of T.S.A of a cube is, 6a sq.

So when the three equal cubes are joined to form a cuboid, the length of the cuboid becomes,

a + a + a = 3a

The breadth of the cube becomes, 1a

And the height of the cuboid will be, 1a

Also we know the formula of T. S. A for the cuboid,

= 2 ( lb + bh + hl )

Therefore the T. S. A of the cuboid after doing the calculations becomes,

= 2 ( 3a × a + a × a + a × 3a )

= 2 ( 7a sq. )

= 14 a sq.

Now total surface area of three cubes will be, 6a sq + 6 a sq. + 6 a sq.

= 18 a sq.

Now ratio of cuboid by cube,

= 14 a sq : 18 a sq.

= 7 a sq. : 9 a sq.