135.15 sq. cm
165.8 sq cm
The area of a trapezoid is 1/2 times the sum of the bases * the height. In this case, we have the area of 65 = 1/2 * (13+13) * height. Solving for height, we have 65 * 2 / 26, so h = 5. If the two bases of a quadrilateral are of the same length, it is not a parallelogram, but a rectangle.
6
8
It is 20 units.
135.15 sq. cm
165.8 sq cm
The area of a trapezoid is 1/2 times the sum of the bases * the height. In this case, we have the area of 65 = 1/2 * (13+13) * height. Solving for height, we have 65 * 2 / 26, so h = 5. If the two bases of a quadrilateral are of the same length, it is not a parallelogram, but a rectangle.
6
Using Pythagoras' theorem the length of the hypotenuse is 13 units
Isosceles trapezoid ABCD has an area of 276 If AD = 13 inches and DE = 12 inches, find AB.
8
Assuming the height is 13 inches (not 13 feet or something else), Area = 0.5*(12+14)*13 = 169 sq inches.
13 feet
It is the square root of 169 which is 13 units of length
Area = Length * Width = 13 ft * 6 ft = 13*6 sq ft = 78 sq ft