It can't.
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The circumference of a 24cm circle is: 75.4 cm(diameter x pi = circumference).
Anything greater than 24 cm. P = 2*L + 2*W, since the length is fixed at 24cm we have: 2*(24cm) +2*W > 96cm 2W > 96 - 48cm 2W > 48cm, or Width > 24cm. So any width greater than 24cm will make the perimeter >96cm. Obviously a width of 24cm wouldn't work since a polygon with all four sides equal would be a square and not a rectangle.
Since an isosceles triangle can be represented by two right triangles back to back, you can utilize the pythagorean theorum to solve this example. Specifically: 18cm/2 = 9cm = 1 leg of right triangle (A) 24cm = hypotenuse of right triangle (C) A squared + B squared = C squared Altitude = B = square root of (C squared - A squared) = approximately 19.875
A = 1/2 BH A = 24 = 1/2 BH 24 * 2 = BH 48 = BH H = B - 2 48 = B(B - 2) B2 - 2B = 48 B2 - 2B - 48 = 0 (B - 8)(B + 6) = 0 B = 8 or -6 Since the base cannot be negative, the base is 8 and the height is 6 (8 - 2)
v=LWH (length x witdhx height) .16m x .24m x 1.4m = 0.06144 m2