A flag in the shape of a right triangle is hung over the side of a building as shown below. The total weight of the flag is 250 pounds and it has uniform density. a = 15 and b = 39.

Answer:(a) The density of flag is [tex]\dfrac{25}{27}[/tex] Pounds per square foot(b) The weight of strip is [tex]\dfrac{20}{9}(15-h)\Delta h[/tex] Pounds(c) The work is [tex]\dfrac{20}{9}(15-h)h\Delta h[/tex] foot-pounds(d) The exact work by roof is 1250 foot-poundsStep-by-step explanation:(a) We are given a flag in the shape of a right triangle. The total weight of flag is 250 pounds and Uniform density. Base of the flag [tex]=\sqrt{b^2-a^2}[/tex]                           [tex]=\sqrt{39^2-15^2}=36[/tex]Area of the flag [tex]=\dfrac{1}{2}\times 36\times 15 = 270[/tex]Weight of flag = 250 pounds [tex]Density =\dfrac{Weight}{Area}=\dfrac{250}{270}=\dfrac{25}{27}[/tex]Hence, The density of flag is [tex]\dfrac{25}{27}[/tex](b) Weight of strip which is h feet below the roof. Length of strip [tex]=\dfrac{36}{15}(15-h)[/tex]Width of strip [tex]\Delta h[/tex]Weight = Density x area             [tex]=\dfrac{25}{27}\times \dfrac{36}{15}(15-h)[/tex]             [tex]=\dfrac{20}{9}(15-h)\Delta h[/tex]Hence, The weight of strip is [tex]\dfrac{20}{9}(15-h)\Delta h[/tex](c) Work slice to move h feet above to the roofwork[tex]=Weight\times displacement[/tex]        [tex]=\dfrac{20}{9}(15-h)\Delta h\times h[/tex]        [tex]=\dfrac{20}{9}(15-h)h\Delta h[/tex]Hence, The work is [tex]\dfrac{20}{9}(15-h)h\Delta h[/tex] foot-pounds(d) Exact work on the roof by hanging flag[tex]W=\int_0^{15}\dfrac{20}{9}(15-h)hd h[/tex][tex]W=\dfrac{20}{9}(\dfrac{15h^2}{2}-\dfrac{h^3}{3})|_0^{15}[/tex][tex]W=1250-0[/tex][tex]W=1250[/tex]Hence, The exact work by roof is 1250 foot-pounds