Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
The width of the rectangle is 13 meters. The perimeter of the rectangle is 50 meters, and its length is 12 meters. The 50 meters is 2 times the length plus 2 times the width. With a length of 12 meters, twice that is 24 meters. That leaves 50 meters - 24 meters for twice the width. And 50 - 24 = 26, which is twice the width. The 26 meters divided by 2 = 13 meters, which is the width of the rectangle.
perimeter 24 so length + width half of that ie 12. Length 3 times width must be 9 and width 3.
The perimeter of a rectangle = 2 x (length + width) = 2 x (12 + 8) = 2 x 20 = 40 inches
The equation for a rectangle is h(height) x w(width) = area . By manipulating the equation we can show that width=area/length. So 132/12=11. This means the width of the rectangle is 11.
The diagonal is 20.
The diagonal length is about 18.44 inches.
d = 13
if a rectangle has width of 5 and diagonal with lenght of 13, what is the area of the rectangle? Use Pythagoras' theorem to find the length of the rectangle which will be 12 5*12 = 60 square units
13 units. Use Pythagoras.
13 57/64 "
Width is 16.
If the length of a rectangle is 12 and the width of the rectangle is 16, by the Pythagorean theorem we know that one diagonal is 20 units long. You can draw two diagonals within a rectangle, so the length of both diagonals together is 20+20 = 40 units.
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
The area of the rectangle is 72 square feet
12 squft
diagonal is 13 inch length of the rectangle of 12 and 5 inches sides