Approximately 15.62 feet.
It depends on what the 7' and 10' refers to. They could be adjacent sides of a rectangle or a parallelogram. If the latter, more information is required before the question can be answered.
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
10 ft x 20 ft room = 200 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.
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.
It depends on what the 7' and 10' refers to. They could be adjacent sides of a rectangle or a parallelogram. If the latter, more information is required before the question can be answered.
Across a Crowded Room was created in 1984-10.
10' x 10' = 100'2
10 X 10 = 100
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.
100 sq. ft.
A room that is 110 square feet typically measures about 10 feet by 11 feet.
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
10 feet*10 feet=100 square feet