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Starting with an equilateral triangle of side 2, dropping a perpendicular from one vertex to the opposite base creates two equal right angled triangles with hypotenuse of length 2, base length 1 and height of length √(22 - 12) = √3 which is the longer leg of the 30-60-90 triangle. Thus the ratio of longer_leg : hypotenuse is √3 : 2
Base and height
base*height
the base area is = PI * radius^2 = PI*16 the area of the curved surface is = 2*PI*radius*height = 2*PI*4*h ( h - is the height ) = 8*PI*h the ratio base area : curved surface = 134 PI * 16 : 8 * PI *h ( PI*16 )/( 8 * PI * h) = 134 2/h = 134 h = ( 1/67 )//
As area_of_parallelogram = base x height if they are both doubled then: new_area = (2 x base) x (2 x height) = 4 x (base x height) = 4 x area_of_parallelogram Thus, if the base and height of a parallelogram are [both] doubled, the area is quadrupled.