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In the given figure, from a cuboidal solid metallic block, of dimensions 15 cm x 10 cm x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.
Answers
Step-by-step explanation:
Given :-
Measurements of a cuboid are 15 cm, 10 cm and 5cm
A cylindrical hole of diameter 7cm is drilled out from that .
To find:-
Find the surface area of the remaining block.
Solution:-
Length of the cuboid = 15 cm
Breadth of the cuboid = 10 cm
Height of the cuboid = 5 cm
Total surface area of a cuboid =
2(lb+bh+hl ) sq.units
We have ,
l = 15 cm
b = 10 cm
h = 5 cm
Total surface area of the given cuboid
=> 2(15×10+10×5+5×15) sq.cm
=> 2(150+50+75 ) sq.cm
=> 2(275 ) sq.cm
=> 550 sq.cm
Diameter of the drilling out cylindrical part = 7 cm
Radius = Diameter/2
Radius of the Cylindrical part = 7/2 cm
Curved Surface Area of the cylinder = 2πrh sq.units
We have , r = 7/2 cm
h = 5 cm
CSA of the Cylindrical part
= 2×(22/7)×(7/2)×(5) sq.cm
=> (2×22×7×5)/(7×2) sq.cm
=> 22×5 sq.cm
=> 110 sq.cm
The part has two circular parts at its end
Area of a circle = πr^2 sq.units
=> (22/7)×(7/2)^2 sq.cm
=>(22/7)×(7/2)×(7/2) sq.cm
=> (22×7×7)/(7×2×2) sq.cm
=> (11×7)/2 sq.cm
=> 772/ sq.cm
Area of two circular ends= 2×(77/2) sq.cm
=> 77 sq.cm
Now ,
The surface area of the remaining block after the circular hole is drilled out
= Total surface area of the cuboidical block + Curved Surface Area of the cylindrical part - areas of two circular ends
=> 550 sq.cm + 110 sq.cm -77 sq.cm
=> 660 sq.cm -77sq.cm
=>583 sq.cm
Answer:-
The surface area of the remaining block after the circular hole is drilled out is 583 sq.cm
Used formulae:-
- Total surface area of a cuboid =
- 2(lb+bh+hl ) sq.units
- Area of a circle = πr^2 sq.units
- Curved Surface Area of the cylinder = 2πrh sq.units
- Radius = Diameter/2
- r=radius
- h =height
- b=breadth
- l=length
- π=22/7
Answer:
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Step-by-step explanation: