Raju wants to paint a cuboidal box which is 20 cm in length , 15 cm in breadth and 6 cm in depth. Find the area to be painted if the box is open at the top.
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Answer:
ɢɪᴠᴇɴ:-
- Length (l) = 20 cm
- Breadth (b) = 15 cm
- Height (h) = 6 cm
ᴛᴏ ғɪɴᴅ :-
Area of Box open at the top
sᴏʟᴜᴛɪᴏɴ :-
We know that,
Area of Cuboid(Box) = 2(lb + lh + bh)
➮ 2(lb + lh + bh)
➮ 2(20 × 15 + 20 × 6 + 15 × 6)
➮ 2(300 + 120 + 90)
➮ 2(510)
➮ 1020 cm²
Now,
Required area of Box = Area of box - Area of Top
➮ Required area = 1020 - (l × b)
➮ 1020 - (20 × 15)
➮ 1020 - 300
➮ 720 cm²
Hence,
Area of Box to be painted is 720 cm²
TO MORE INFORMATION :-
- Total Surface Area of a Cuboid (TSA) = 2 (lb + bh + lh) square units
- Lateral Surface Area of a cuboid (LSA) = 2h (l+b) square units
- Volume of the cuboid (v) =lbh
Answered by
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Height = 6cm, Breadth = 15cm and Length = 20cm
Formula for painting a cuboidal box with its top open:
TSA of a cuboid - area of rectangle =
2(lb + bh + hl) - lb
=> 2( 20x15 + 15x6 + 6x20 ) - 20x15
=> 2 ( 300 + 90 + 120 ) - 300
=> 2 ( 510 ) - 300
=> 1020 - 300
=> 720 cm^2
The area to be painted if the box is open at the top will be 720 cm^2.
Please mark as brainliest
Formula for painting a cuboidal box with its top open:
TSA of a cuboid - area of rectangle =
2(lb + bh + hl) - lb
=> 2( 20x15 + 15x6 + 6x20 ) - 20x15
=> 2 ( 300 + 90 + 120 ) - 300
=> 2 ( 510 ) - 300
=> 1020 - 300
=> 720 cm^2
The area to be painted if the box is open at the top will be 720 cm^2.
Please mark as brainliest
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