Answer:
The surface area of Design A is smaller than the surface area of Design B.
The area of Design A is 94.65% of the Design B.
Step-by-step explanation:
The area of a right cone is given by the sum of the circle area of the base and the lateral area:
[tex] A = \pi r^{2} + \pi rL [/tex] (1)
Where:
r: is the radius
L: is the slant height
The slant height is related to the height and to the radius by Pitagoras:
[tex] L^{2} = H^{2} + r^{2} [/tex]
[tex] L = \sqrt{H^{2} + r^{2}} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] A = \pi r^{2} + \pi r(\sqrt{H^{2} + r^{2}}) [/tex]
Now, let's find the area of the two cases.
Design A: height that is double the diameter of the base, H= 2D = 4r
[tex] A_{1} = \pi r^{2} + \pi r(\sqrt{(4r)^{2} + r^{2}}) = \pi r^{2}(1+ \sqrt{17}) [/tex]
The volume of the cone is:
[tex]V = \frac{1}{3}\pi r^{2}H[/tex]
We can find "r":
[tex] V = \frac{1}{3}\pi r^{2}(4r) = \frac{4}{3}\pi r^{3} [/tex]
[tex] r = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3*1000}{4\pi}} = 6.20 cm [/tex]
The area is:
[tex] A_{1} = \pi (6.20)^{2}(1+ \sqrt{17}) = 618.7 cm^{2} [/tex]
Design B: height that is triple the diameter of the base, H = 3D = 6r
The radius is:
[tex] r = \sqrt[3]{\frac{3V}{6\pi}} = \sqrt[3]{\frac{3*1000}{6\pi}} = 5.42 cm [/tex]
The area is:
[tex]A_{2} = \pi r^{2} + \pi r(\sqrt{(6r)^{2} + r^{2}}) = \pi r^{2}(1 + \sqrt{37}) = \pi (5.42)^{2}(1 + \sqrt{37}) = 653.7 cm^{2}[/tex]
Hence, the surface area of Design A is smaller than the surface area of Design B.
The percent of the surface area of Design A is less than Design B by:
[tex] \% A = \frac{618.7 cm^{2}}{653.7 cm^{2}}\times 100 = 94.65 \% [/tex]
Therefore, the area of Design A is 94.65% of the Design B.
I hope it helps you!
