A chips wrapper is 4.5 cm long and 5.9 cm wide. Its area upto significant figures will be?
Answer: 26.55 cm^2
Explanation
To find the area of the chip wrapper, you multiply the length by the width:
Area = Length x Width
= 4.5 cm x 5.9 cm
= 26.55 cm^2
So, the area of the chip wrapper is 26.55 cm^2.
This question appeared in
Subjects (4 times)
EVERYDAY SCIENCE (2 times)
GENERAL SCIENCE (2 times)
Related MCQs
- A room is 74 cm long and 61 cm wide. Aliza wants to carpet the area 2 cm wide all around outside the room. Find the area of the carpeting part of the room?
- An 82 cm long and 75 cm wide rectangle is enclosed by a strip outside it that is 4 cm wide. Find the area of the strip?
- A room is 7.8 m long and 6.5 m wide. Aliza wants to carpet 2.3 m wide all round inside the room. Find the area of the carpeting part of the room?
- A student added three figures 72.1, 3.32 and 0.003. The correct answer regrading the rules of the addition of the significant figures will be ______?
- A rectangle WXYZ is 45 m long and 35 m wide. A path is constructed towards the inner side of the rectangle that is 3 m wide. Find the area of the path?
- A rectangle ABCD is 20 cm long and 10 cm wide. There is a border towards the inner side of the rectangle that is 2 cm wide. Find the area of the border?
- A garden is 75 cm long and 82 cm wide. A 5 cm wide road is made all around the inside of the garden. Find the area of the road?
- A rectangle WXYZ is 60 m long and 40 m wide. A path is constructed towards the inner side of the rectangle that is 2.5 wide. The area of the path is?
- A boy is cutting a rectangular lawn 40 ft long and 30 ft wide. He has cut all of it except for a rectangle that is 20 ft long and 15ft wide. The fractional part of the lawn remains uncut--------.
- Portrait in art gallery is 7 feet wide and 9 feet long, if its frame has a width of 2 inches. What is the ratio of area of the frame to the area of portrait?
