a tent is in the form of right circular cylinder surrounded by a cone the diameter of a cylinder is 24 M. height of the cylinder is the portion is 11 M. while the vertex of the cone is 16 m above the ground. find the area of the Canvas required for the tent.
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Diameter of cylindrical part of the tent = 24M
Therefore,
Radius of cylindrical part of the tent = D/2 = 24/2 = 12 M
Total height of the tent = 16 M
and the height of cylindrical part of the tent = 11 M
Therefore,
Height of conical part of the tent = 16-11 = 5M
Radius of the conical part of the tent = 12 M
Slant height of the conical part of the tent (L) = (H)² + (R)² = (5)² + (12)² = 25 + 144 = ✓169 = 13M
CSA of cone = πRL
= 22/7×12×13 = 490.28M²
TSA of cylindrical part of the tent = 2πr(r+h)
PUT VALUES AND SOLVE THIS YOU WILL GET THE TSA OF CYLINDRICAL PART OF THE TENT.
THEREFORE,
CANVAS REQUIRED TO MAKE THE TENT = CSA OF CONICAL PART OF THE TENT+ TSA OF CYLINDRICAL PART OF THE TENT .
PUT VALUES AND SOLVE U WILL GET THE ANSWER
Therefore,
Radius of cylindrical part of the tent = D/2 = 24/2 = 12 M
Total height of the tent = 16 M
and the height of cylindrical part of the tent = 11 M
Therefore,
Height of conical part of the tent = 16-11 = 5M
Radius of the conical part of the tent = 12 M
Slant height of the conical part of the tent (L) = (H)² + (R)² = (5)² + (12)² = 25 + 144 = ✓169 = 13M
CSA of cone = πRL
= 22/7×12×13 = 490.28M²
TSA of cylindrical part of the tent = 2πr(r+h)
PUT VALUES AND SOLVE THIS YOU WILL GET THE TSA OF CYLINDRICAL PART OF THE TENT.
THEREFORE,
CANVAS REQUIRED TO MAKE THE TENT = CSA OF CONICAL PART OF THE TENT+ TSA OF CYLINDRICAL PART OF THE TENT .
PUT VALUES AND SOLVE U WILL GET THE ANSWER
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