Question

Two different right cones are being considered by a design team to hold one liter (1000

Answer 1

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:

$A = \pi r^{2} + \pi rL$   (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:

$L^{2} = H^{2} + r^{2}$  

$L = \sqrt{H^{2} + r^{2}}$    (2)

By entering equation (2) into (1) we have:

$A = \pi r^{2} + \pi r(\sqrt{H^{2} + r^{2}})$  

Now, let's find the area of the two cases.

Design A: height that is double the diameter of the base, H= 2D = 4r

$A_{1} = \pi r^{2} + \pi r(\sqrt{(4r)^{2} + r^{2}}) = \pi r^{2}(1+ \sqrt{17})$  

The volume of the cone is:

$V = \frac{1}{3}\pi r^{2}H$

We can find "r":

$V = \frac{1}{3}\pi r^{2}(4r) = \frac{4}{3}\pi r^{3}$

$r = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3*1000}{4\pi}} = 6.20 cm$

The area is:

$A_{1} = \pi (6.20)^{2}(1+ \sqrt{17}) = 618.7 cm^{2}$  

Design B: height that is triple the diameter of the base, H = 3D = 6r                  

The radius is:

$r = \sqrt[3]{\frac{3V}{6\pi}} = \sqrt[3]{\frac{3*1000}{6\pi}} = 5.42 cm$  

The area is:

$A_{2} = \pi r^{2} + \pi r(\sqrt{(6r)^{2} + r^{2}}) = \pi r^{2}(1 + \sqrt{37}) = \pi (5.42)^{2}(1 + \sqrt{37}) = 653.7 cm^{2}$  

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:

$\% A = \frac{618.7 cm^{2}}{653.7 cm^{2}}\times 100 = 94.65 \%$

Therefore, the area of Design A is 94.65% of the Design B.

                         

I hope it helps you!