Question

A cubical rectangular bar has the dimensions with the ratio 5 : 4 : 3. Its volume is 7500. What is the surface area of the bar?Ans. 2350

Answer 1

Heya,

We know that,

Volume=l×b×h

7500=5x×4x×3x

7500=60x

x=7500/60

x=125

5x=125×5=625

4x=125×4=500

3x=125×3=375

length=625

breadth=500

height=375

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

=2(25*20+20*15+25*15)

=【2350 sq.units】Answer..

HOPE IT HELPS:)

Answer 2

Given: A cubical rectangular bar has the dimensions with the ratio 5 : 4 : 3. Its volume is 7500.

To find: The surface area of the bar

Solution: The ratio of dimensions= 5:4:3

Let the length(l) be 3x, breadth(b) be 4x and height(h) be 3x.

The volume of a cubical rectangular bar is given as= length × breadth × height

Using the formula, volume of the bar

= l × b × h

= 5x × 4x × 3x

$= 60 {x}^{3}$

But volume is given as 7500.

Therefore,

$60 {x}^{3} = 7500$

$= > {x}^{3} = \frac{7500}{60}$

$= > {x}^{3} = 125$

$= > x = \sqrt[3]{125}$

$= > x = 5$

Length of the bar

= 5x

= 5× 5

= 25 units

Breadth of the bar

= 4x

= 4× 5

= 20 units

Height of the bar

= 3x

= 3× 5

= 15 units

Surface area is given by the formula

= 2 ( lb + bh + hl)

= 2 (25×20 + 20×15+ 15×25)

= 2 × 1175

= 2350

Therefore, the surface area of the cubical rectangular bar is $2350 \: ({units})^{2}$.