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Assuming 15b means the lengths of each side of the square base and 12l means the length of the triangle from the top to the bottom (note that I'm not assuming 12l is the height of the pyramid) the SA (surface area) will be:
15^2 + 15x12x4/2 = 225 + 360 = 585 units squared (whatever your units are)
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If the perimeter of a rectangle is 150 inches and the length is 3 inches longer than five times the width what is the length?
Given: P = 150 and L = (5W + 3) Solve the 'L' equation for 'W': (L-3) = 5W ===> W = (L-3)/5 Perimeter = 2L + 2W = 2L + 2[ (L-3)/5 ] = 2L + (2L-6)/5 5 x perimeter = 10L + (2L-6) = 12L - 6 Perimeter = 150 = (12L-6)/5 ===> 750 = (12L - 6) ===> 756 = 12L ===> L = 756/12 = 63=============================== Check: L = 63, W = 12, Perimeter = 2(L+W) = 2(63+12) = 2(75) = 150. Yay !
A figure with an area of 6 and a perimeter of 12?
The figure could be a rectangle with length 4.73 and width 1.27(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 6 : Then W = 6/L Perimeter = 2 (L + W) = 12 : substituting for W then :- 2(L + 6/L) = 12 : 2L + 12/L = 12 : 2L2 - 12L + 12 = 0 Solving for a quadratic equation gives L = (12 ± √48) ÷ 4 = 3 ± √3 = 3 ± 1.73 (2dp)
