13 57/64 "
Using Pythagoras' theorem the length of the diagonal is 20 feet
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
Using Pythagoras, the diagonal is sqrt(122 + 162) = sqrt(144 +256) = sqrt(400) = 20 feet.
I can and will. So if you put the diagonal through a rectangle you create a right angled triangle of which you can use pythagoras theorem. x2 = 152 + 62 x2= 261 x = 16.16m
A rectangle with width 4 can have any length!
Using Pythagoras' theorem the length of the diagonal is 20 feet
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
Using Pythagoras, the diagonal is sqrt(122 + 162) = sqrt(144 +256) = sqrt(400) = 20 feet.
The diagonal of a rectangle does not provide enough information to determine the length of the rectangle. Let L be any real number such that 32/sqrt(2) < L < 32. let B = sqrt(32^2 - L^2) Since L < 32 the above square root exists, and since L > 32/sqrt(2), B < L. So the rectangle with sides of L and B will have a diagonal of 32 inches. But L is any of an infinite number of possible real numbers. Therefore there are infinitely many possible solutions.
I can and will. So if you put the diagonal through a rectangle you create a right angled triangle of which you can use pythagoras theorem. x2 = 152 + 62 x2= 261 x = 16.16m
the width of the rectangle is 12.....because area= lenght x width so if the lenght is 5 inches and the area is 60....u divide 60 divided by 5 = 12
A rectangle with width 4 can have any length!
Area = L*W = 15 * 2 = 30 square inches.
The square's diagonal is 11.314 cm
a rectangle as a longer side because there's 48 inches for a rectangle but there's only 45 inches for a triangle. a rectangle as a longer side because there's 48 inches for a rectangle but there's only 45 inches for a triangle.
Length + width = half of perimeter ie 36.7 in. If length is 4.7 in then width is 32 in.
The area of rectangle is : 152.0