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nobillionaireNobley
Find the volume and the lateral area of a frustum of a right circular cone whose radii are 4 and 8 cm, and slant height is 6 cm.
[tex]h= \sqrt{s^2-(R_1-R_2)^2} \\ = \sqrt{6^2-(4-8)^2} \\ = \sqrt{36-16} \\ = \sqrt{20} [/tex]
[tex]Volume= \frac{1}{3} \pi h(R_1^2+R_1R_2+R_2^2) \\ = \frac{1}{3} \pi \times \sqrt{20} (4^2+4 \times 8+8^2) \\ = \frac{1}{3} \pi \sqrt{20} (16+32+64) \\ = \frac{1}{3} \pi \sqrt{20} (112) \\ =524.5cm^3[/tex]
Lateral area = Total surface area - area of base - area of top
[tex]Lateral \ area= \pi (R_1+R_2)s \\ = \pi (4+8) \times 6 \\ =12 \pi \times 6 \\ =72 \pi \\ =226.2cm^2[/tex]
calculista

the complete answers in the attached figure

Part 1) we have

[tex] r=4cm\\ R=8 cm\\ L=6cm [/tex]

Find the height h

[tex] h^{2}=L^{2} -(R -r)^{2}\\ h^{2}=6^{2} -(8-4)^{2}\\ h^{2}=36-16\\ h=\sqrt{20} cm [/tex]

Find the volume

[tex] V=\frac{1}{3}\pi[R^{2} +r^{2} +Rr]h\\\\ V=\frac{1}{3}\pi[8^{2} +4^{2} +8*4]\sqrt{20}\\ \\ V=\frac{1}{3}\pi[112]\sqrt{20}\\ \\ V=524.52 cm^{3} [/tex]

Find the lateral area

[tex] LA=\pi (R+r)L\\ LA=\pi *(8+4)*6\\ LA=226.19 cm^{2} [/tex]

the answer Part 1) is

a) the volume is equal to [tex] 524.52 cm^{3} [/tex]

b) The Lateral area is equal to [tex] 226.19 cm^{2} [/tex]

Part 2) we have

[tex] r=4ft\\ R=5 ft\\ h=100 ft [/tex]

Find the slant height L

[tex] L^{2}=h^{2}+(R -r)^{2}\\ L^{2}=100^{2} +(5-4)^{2}\\ L^{2}=10000+1\\ L=\sqrt{10001} ft [/tex]

Find the lateral area

[tex] LA=\pi (R+r)L\\ LA=\pi *(5+4)*\sqrt{10001}\\ LA=2,827.57 ft^{2} [/tex]

the answer part 2) is

a) The Lateral area is equal to [tex] 2,827.57 ft^{2} [/tex]

Part 3) we have

[tex] V=52\pi ft^{3} \\ h=3ft\\ R=3r [/tex]

Step 1

Find the values of R and r

[tex] [/tex][tex] V=\frac{1}{3}\pi[R^{2} +r^{2} +Rr]h [/tex]

substitute [tex] R=3r [/tex] in the formula above

[tex] V=\frac{1}{3}\pi[(3r)^{2} +r^{2} +(3r)*r]*3 [/tex]

[tex] V=\frac{1}{3}\pi[7r)^{2}]*3 [/tex]

[tex] V=[tex] 52\pi [/tex]

[tex] 52\pi =\pi [7r^{2} ]\\ r^{2} =\frac{52}{7} \\ \\ r=2.73 ft [/tex]

[tex] R=3*2.73\\ R=8.19 ft [/tex]

Step 2

Find the slant height L

[tex] L^{2}=h^{2}+(R -r)^{2}\\ L^{2}=3^{2} +(8.19-2.73)^{2}\\ L^{2}=38.81\\ L=6.23 ft [/tex]

Step 3

Find the lateral area

[tex] LA=\pi (R+r)L\\ LA=\pi *(8.19+2.73)*6.23 LA=213.73 ft^{2} [/tex]

the answer Part 3) is

a) The lateral area is equal to [tex] 213.73 ft^{2} [/tex]

Part 4) we have

[tex] r=15 in\\ R=33 in\\ h=24 in [/tex]

Find the slant height L

[tex] L^{2}=h^{2}+(R -r)^{2}\\ L^{2}=24^{2} +(33-15)^{2}\\ L^{2}=576+324\\ L=30 in [/tex]

Find the lateral area

[tex] LA=\pi (R+r)L\\ LA=\pi *(33+15)*30\\ LA=4,523.89 in^{2} [/tex]

Find the volume

[tex] V=\frac{1}{3}\pi[R^{2} +r^{2} +Rr]h\\\\ V=\frac{1}{3}\pi[33^{2} +15^{2} +33*15]24\\ \\ V=\frac{1}{3}\pi[112]24\\ \\ V=142.83 in^{3} [/tex]

the answer is

a) The lateral area is equal to [tex] 4,523.89 in^{2} [/tex]

b) the volume is equal to [tex] 142.83 in^{3} [/tex]

Part 5) we have

[tex] r=5 cm\\ h=8√3 cm [/tex]

Step 1

Find the value of (R-r)

[tex] tan 60=\sqrt{3} [/tex]

[tex] tan 60=\frac{(R-r)}{8\sqrt{3}} \\\\ R-r= \sqrt{3} *8\sqrt{3} \\ R-r=24 cm\\ R=24+r\\ R=24+5\\ R=29 cm [/tex]

Step 2

Find the value of slant height L

[tex] L^{2}=h^{2}+(R -r)^{2}\\ L^{2}=(8\sqrt{3})^{2}+(24-5)^{2}\\ L^{2}=192+361\\ L=23.52 cm [/tex]

Step 3

Find the lateral area

[tex] LA=\pi (R+r)L\\ LA=\pi *(24+5)*23.52\\ LA=2,142.82 cm^{2} [/tex]

Step 4

Find the total area

total area=lateral area+area of the top+area of the bottom

Area of the top

[tex] r=5 cm\\ A=\pi *r^{2} \\ A=\pi *25\\ A=78.54 cm^{2} [/tex]

Area of the bottom

[tex] r=24 cm\\ A=\pi *r^{2} \\ A=\pi *576\\ A=1,809.56 cm^{2} [/tex]

Total surface area

[tex] SA=2,142.82+78.54+1,809.56\\ SA=4,030.92 cm^{2} [/tex]

the answer is

a) The total surface area is [tex] 4,030.92 cm^{2} [/tex]

Part 6)

Part a) Find the volume of the water tank

we have

[tex] r=4 ft\\ R=6 ft\\ h=8 ft [/tex]

Step 1

Find the volume

[tex] V=\frac{1}{3}\pi[R^{2} +r^{2} +Rr]h\\\\ V=\frac{1}{3}\pi[6^{2} +4^{2} +6*4]8\\ \\ V=\frac{1}{3}\pi[76]8\\ \\ V=636.70 ft^{3} [/tex]

the answer Part a) is [tex] 636.70 ft^{3} [/tex]

Part b) Find the volume of the wetted part of the tank if the depth of the water is 5 ft

by proportion find the radius R of the upper side for h=5 ft

[tex] \frac{(R1-r)}{8} =\frac{(R2-r)}{5} \\\\ \frac{(6-4)}{8} =\frac{(R2-4)}{5}\\ \\(R2-4)= 1.25\\ R2=4+1.25\\ R2=5.25 ft [/tex]

Find the volume for [tex] R2=5.25 ft [/tex]

[tex] V=\frac{1}{3}\pi[R^{2} +r^{2} +Rr]h\\\\ V=\frac{1}{3}\pi[5.25^{2} +4^{2} +5.25*4]5\\ \\ V=\frac{1}{3}\pi[64.56]5\\ \\ V=338.05 ft^{3} [/tex]

the answer Part b) is [tex] 338.05 ft^{3} [/tex]

Part 7) we have

[tex] SA=435\pi cm^{2} \\ A1=144\pi cm^{2}\\ A2=81\pi cm^{2} [/tex]

Step 1

Find the value of R and the value of r

[tex] A1=\pi *R^{2} \\ 144\pi =\pi *R^{2}\\ R=12 cm [/tex]

[tex] A2=\pi *r^{2} \\ 81\pi =\pi *r^{2}\\ r=9 cm [/tex]

Step 2

Find the value of lateral area

[tex] LA=SA-A1-A2\\ LA=435\pi -144\pi -81\pi \\ LA=210\pi cm^{2} [/tex]

Step 3

Find the slant height

[tex] LA=\pi (R+r)L\\\\ L=\frac{LA}{\pi(R+r)} \\ \\ L=\frac{210\pi}{\pi(12+9)} \\ \\ L=10 cm [/tex]

Find the altitude of the frustum

[tex] h^{2} =L^{2} -(R-r)^{2} \\ h^{2} =10^{2} -(12-9)^{2}\\ h^{2}=91\\ h=9.54 cm [/tex]

the answer Part a) is

the slant height is [tex] 10 cm [/tex]

the answer Part b) is

the altitude of the frustum is [tex] 9.54 cm [/tex]

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