Using Pythagoras' theorem the length of the diagonal is 20 feet
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
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
13 57/64 "
Use Pythagoras' theorem: 52+72 = 74 and the square root of 74 is 8.602325267 Answer: 8.6 units to one decimal place
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
15 and 20 inches because these dimensions comply with Pythagoras' theorem and the area of the rectangle.
13 57/64 "
Use Pythagoras' theorem: 52+72 = 74 and the square root of 74 is 8.602325267 Answer: 8.6 units to one decimal place
You forgot to put in the length of the rectangle's perimeter.
Half a rectangle is a right angled triangle. Pythagoras : Triangle whose hypotenuse is 5 and one side is 1 unit greater than the third must be a 3-4-5 triangle, so the rectangle is 3 cm x 4 cm
The length of the diagonal is not enough to determine the dimensions of the rectangle. Take any length W such that 0<W<sqrt(132/2) that is, 0<W<9.192 cm. And let L = sqrt(132 - W2) cm. Then 9.192<L<13 so that any combination of W and L is unique. There are infinitely many possible values for W and so infinitely many different combinations for W and L. That is, infinitely many rectangles whose width is W cm and length is L cm. And, since, L2 = 132 - W2, L2 + W2 = 132 so the diagonal is 13 cm.
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A rectangle with width 4 can have any length!
The square's diagonal is 11.314 cm
The area of rectangle is : 152.0