Diagonal = 18.867 feet.
Approximately 15.62 feet.
The diagonal of a rectangle with sides measuring 7 feet and 10 feet can be calculated using the Pythagorean theorem. The square of the diagonal is equal to the sum of the squares of the two sides, so in this case, the diagonal would be √(7^2 + 10^2) = √(49 + 100) = √149 = approximately 12.21 feet.
A box is a 3-dimensional object and only two measures are given so it is not clear what the question means. Across the base of the box, the diagonal is 10*sqrt(2) = 14.142 approx.
10 feet*10 feet=100 square feet
The diagonal is 10 feet. The one side is 8 feet and the other side is 6 feet (by Pythagoras 82 + 62 = 102)). The perimeter is thus 28 feet.
Diagonal = 18.867 feet.
Approximately 15.62 feet.
Assuming you are talking about a rectangular area, the diagonal would be found using the Pythagorean Theorem. 5^2 + 10^2 = d^2, so 125 = d^2, then take square root of both sides. This means the diagonal is approximately 11.18 feet. It is exactly 5√5 feet.
The diagonal of a rectangle with sides measuring 7 feet and 10 feet can be calculated using the Pythagorean theorem. The square of the diagonal is equal to the sum of the squares of the two sides, so in this case, the diagonal would be √(7^2 + 10^2) = √(49 + 100) = √149 = approximately 12.21 feet.
Across a Crowded Room was created in 1984-10.
10' x 10' = 100'2
A box is a 3-dimensional object and only two measures are given so it is not clear what the question means. Across the base of the box, the diagonal is 10*sqrt(2) = 14.142 approx.
10 X 10 = 100
100 sq. ft.
Area of Rhombus = length of first diagonal x length of second diagonal / 2 / means divided by So for your problem: Area of Rhombus = 10 feet x 14 feet / 2 = 70 square feet
A room that is 110 square feet typically measures about 10 feet by 11 feet.