हिंदी

Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). - Mathematics

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प्रश्न

Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

उत्तर

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + l b

= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2

= [2(10 + 7.5) + 12] m2

= 47 m2

Therefore, 47 m2 tarpaulin will be required.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Surface Area and Volumes - Exercise 13.1 [पृष्ठ २१३]

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एनसीईआरटी Mathematics [English] Class 9
अध्याय 13 Surface Area and Volumes
Exercise 13.1 | Q 8 | पृष्ठ २१३

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