Approximate the change in the lateral surface area (excluding the area of the base) of a right circular cone of fixed height of h=6 m when its radius decreases from r= 10m to r= 9.9 m

Answer:The change in the lateral surface area is approximate [tex]6.32\ m^{2}[/tex]Step-by-step explanation:we know thatThe lateral surface area of the cone is equal to[tex]LA=\pi r l[/tex]wherer is the radius of the basel is the slant heightPart 1 we have [tex]r=10\ m, h=6\ m[/tex]Calculate the slant height l (applying the Pythagoras Theorem)[tex]l^{2}=r^{2}+h^{2}[/tex]substitute the values[tex]l^{2}=10^{2}+6^{2}[/tex][tex]l^{2}=136[/tex][tex]l=\sqrt{136}\ m[/tex]Find the lateral area of the cone[tex]LA=\pi (10)(\sqrt{136})[/tex][tex]LA=10\pi \sqrt{136}\ m^{2}[/tex]Part 2 we have [tex]r=9.9\ m, h=6\ m[/tex]Calculate the slant height l (applying the Pythagoras Theorem)[tex]l^{2}=r^{2}+h^{2}[/tex]substitute the values[tex]l^{2}=9.9^{2}+6^{2}[/tex][tex]l^{2}=134.01[/tex][tex]l=\sqrt{134.01}\ m[/tex]Find the lateral area of the cone[tex]LA=\pi (9.9)(\sqrt{134.01})[/tex][tex]LA=9.9\pi \sqrt{134.01}\ m^{2}[/tex]Part 3Find  the change in the lateral surface area[tex]10\pi \sqrt{136}-9.9\pi \sqrt{134.01}[/tex]assume [tex]\pi =3.14[/tex][tex]10(3.14)\sqrt{136}-9.9(3.14)\sqrt{134.01}=6.32\ m^{2}[/tex]