The diagonal of a rectangle 8 feet by 12 feet is √208 feet, or about 14 feet 5.06 inches.
The diagonal measurement of an 8 ft square is: 11.31 feet.
I assume that you refer to a 12 ft x 12 ft 8 in square. The length of the diagonal is determined easily suing Pythororas's therem: Diagonal = sqrt[122 + (128/12)2] feet = sqrt(304.4... ) ft = 17.448 ft approx.
the square root of 128 or 8 times the square root of 2.
24 sq. ft. * * * * * The above answer must use some exotic units for measuring distance since 12 ft*8 feet is greater than 24 sq ft, and any of 12 feet*8 inches 12 inches*8 feet 12 inches*8 inches is smaller than 24 sq feet. It is not possible to give a more substantive answer since the measurement units are not given in the question.
Convert both units to feet and multiply: 12 in = 1 ft → 2 ft × 8 in = 2 ft × 8 ÷ 12 ft = 4/3 sq ft = 1⅓ sq ft ≈ 1.33 sq ft
The diagonal measurement of an 8 ft square is: 11.31 feet.
I assume that you refer to a 12 ft x 12 ft 8 in square. The length of the diagonal is determined easily suing Pythororas's therem: Diagonal = sqrt[122 + (128/12)2] feet = sqrt(304.4... ) ft = 17.448 ft approx.
Don't tell me you are in class!
Around 17.8. I did this with cossin, but check the answer with a calculator
17.8885'
The diagonal of an 8 ft by 10 ft square, by Pythagoras, would be sqrt(82 + 102) = sqrt(64+100) = sqrt(164) = 12.806 ft (to 5 sig figs)
The square has a diagonal measurement of: 1.13 km
(Diagonal)2 = (15)2 + (8)2 = (225) + (64) = 289Diagonal = sqrt(289) = 17
8 ft.
24 ft. wide, 8 ft. high, 4 ft. at the bottom- This is the standard measurement for a soccer net.
If you mean a rectangle then its diagonal using Pythagoras; theorem is 2 times the square root of 41 or about 12.806 feet rounded to 3 decimal places.
the square root of 128 or 8 times the square root of 2.