Question

Answer the 4 th one
on paper in good handwritting​

Answer 1

Answer:

Dimensions of the box= 8m*7.5m*6m

Surface area of cuboidal box= 2(lb+bh+hl)

= 2(8*7.5+7.5*6+6*8)

= 2(60+45+48)

= 2(153)

= 306

But Harshit did not paint the bottom

So, Surface area of cuboid-Area of bottom=The total area painted.

=(306) - (8*7.5)

=306-60

=246

Therefore, the area painted by Harshit is 246 square metres.

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Answer 2

» Question :

Harshit Painted the outer surface of a cuboidal box of dimensions 8m × 7.5 m × 6 m .

Find the surface area Painted by him , if he didn't paint the bottom of the box.

» To Find :

The surface area of the part or the four walls painted by Harshit.

» Given :

• Length of the Cuboid = 8 m

• Breadth of the Cuboid = 7.5 m

• Height of the Cuboid = 6m

» We Know :

Area of Four walls :

$\sf{\underline{\boxed{A = 2(l + b)h}}}$

Where ,

• A = Area of four walls.

• l = Length of the Cuboid.

• b = Breadth of the Cuboid.

• h = Height of the Cuboid.

» Concept :

Here ,we have Taken the area of four walls as ,

According to the question , it says that the bottom part was not painted and the ceiling is an exception.

We have taken the dimensions as length , breadth and height for ease of calculation.

» Solution :

Given values :

• Length of the Cuboid = 8 m

• Breadth of the Cuboid = 7.5 m

• Height of the Cuboid = 6m

Using the formula ,and substituting the values in it ,we get :

$\sf{\underline{\boxed{A = 2(l + b)h}}}$

$\sf{\Rightarrow A = 2 \times (8 + 7.5) \times 6}$

$\sf{\Rightarrow A = 2 \times 15.5 \times 6}$

$\sf{\Rightarrow A = 31 \times 6}$

$\sf{\Rightarrow A = 186 m^{2}}$

Hence ,the area which is painted by the Harshit is 186 m².

» Additional information :

• Surface area of a Cuboid = 2(lb + bh + lh)

• Volume of a Cuboid = lbh

• Curved surface area of a Cube = 4(a)²

• Curved surface area of a Cube = 4(a)²

• Total surface area of a Cube = 6(a)²