A cone with the same base but a height 3 times taller than the given cylinder exists. What is the volume of each figure? Express your answers in terms of pi.
Volume of the cylinder =
Volume of the cone =

Pls i need help

Answer:

Step-by-step explanation:

131.95

The volume of cylinder and cone are [tex]36\pi[/tex] cubic unit.

It is given by formula, [tex]V=\pi r^{2}h[/tex]

Where r is base radius and h is height of cylinder.

Given that,  [tex]r=3,h=4[/tex]

Substitute values in above formula,

         [tex]V=\pi*3^{2} *4=36\pi[/tex]

Volume of the cone is,

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

Given that, [tex]r=3,h=3*4=12[/tex]

Substitute values in above formula.

    [tex]V=\frac{1}{3}\pi*3^{2}*12=36\pi[/tex]

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1.
Given points B(−12,2), T(−8,10), A(−7,12), F(−6,10), and G(−2,2), which of the following proves that △BAG~△TAF?


2.
Given points A(2,1), B(1,4), C(−1,10), D(8,13), and E(4,5), which of the following proves that △ABE~△ACD?
Answers:

By the Distance Formula,AB=310‾‾‾√,AC=10‾‾‾√, AE=65‾√, and AD=25‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SAS ∼.


By the Distance Formula,AB=310‾‾‾√,AC=10‾‾‾√, AE=65‾√, and AD=25‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SSS ∼.

By the Distance Formula,AB=10‾‾‾√,AC=310‾‾‾√, AE=25‾√, and AD=65‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SSS ∼.

By the Distance Formula,AB=10‾‾‾√,AC=310‾‾‾√, AE=25‾√, and AD=65‾√. Since, ABAC=AEAD=13, and ∠A≅∠A by the Reflexive Property, △ABE∼△ACD by SAS ∼.



3.
Given points M(1,2), N(4,4), P(5,−1), J(−7,−1), K(−1,3), and L(1,−7), which of the following proves that △MNP≈△JKL?
Answers:

By the Distance Formula,MN=13‾‾‾√,NP=26‾‾‾√, and PM=5.Also, JK=213‾‾‾√,KL=226‾‾‾√, and LJ=10.Therefore, MNJK=NPKL=PMLJ=12, andtherefore, △MNP∼△JKL by SAS ∼.

By the Distance Formula,MN=13‾‾‾√,NP=26‾‾‾√, and PM=5.Also, JK=213‾‾‾√,KL=226‾‾‾√, and LJ=10.Therefore, MNJK=NPKL=PMLJ=12, andtherefore, △MNP∼△JKL by SSS ∼.

By the Distance Formula,MN=13,NP=26, and PM=25.Also, JK=52,KL=104, and LJ=100.Therefore, MNJK=NPKL=PMLJ=14, andtherefore, △MNP∼△JKL by SSS ∼.

By the Distance Formula,MN=13,NP=26, and PM=25.Also, JK=52,KL=104, and LJ=100.Therefore, MNJK=NPKL=PMLJ=14, andtherefore, △MNP∼△JKL by SAS ∼.



4.
Given points M(−5,5), R(−2,6), I(−2,4), B(2,−2), A(8,0), and G(8,−4), which of the following proves that △MRI~△BAG?
Answers:

By the Distance Formula,MR=10‾‾‾√,RI=2, and IM=10‾‾‾√.Also, BA=210‾‾‾√,AG=4, and GB=210‾‾‾√.Therefore, MRBA=RIAG=IMGB=10√210√=12,and therefore, △MRI~△BAG by SSS ~.

By the Distance Formula,MR=10,RI=2, and IM=10.Also, BA=40,AG=8, and GB=40.Therefore, MRBA=RIAG=IMGB=1040=14,and therefore, △MRI∼△BAG by SSS ∼.

By the Distance Formula,MR=2,RI=10, and IM=10.Also, BA=8,AG=40, and GB=40.Therefore, MRBA=RIAG=IMGB=1040=14,and therefore, △MRI∼△BAG by SAS ∼.

By the Distance Formula,MR=10‾‾‾√,RI=2, and IM=10‾‾‾√.Also, BA=210‾‾‾√,AG=4, and GB=210‾‾‾√.Therefore, MRBA=RIAG=IMGB=10√210√=12,and therefore, △MRI∼△BAG by SAS ∼.



5.
Given points L(−2,−2), A(−6,−3), U(−14,−5), X(4,−5), and P(0,−3), which of the following proves that △XUL~△PAL?

By the Distance Formula,LX=317‾‾‾√,PL=17‾‾‾√, LA=5‾√, and LU=35‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SSS ∼.

By the Distance Formula,LX=35‾√,PL=5‾√, LA=17‾‾‾√, and LU=317‾‾‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SAS ∼.

By the Distance Formula,LX=35‾√,PL=5‾√, LA=17‾‾‾√, and LU=317‾‾‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SSS ∼.

By the Distance Formula,LX=317‾‾‾√,PL=17‾‾‾√, LA=5‾√, and LU=35‾√. Since, LXPL=LULA=31=3, and ∠L≅∠L by the Reflexive Property, △XUL∼△PAL by SAS ∼.

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