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r-ruslan [8.4K]

3 years ago

5

2. Brian is filling a conic container with water. He has the container half full. The radius of the container is 5 inches and th

e height is 20 inches. (a) What is the current volume of the water? Show your work and explain your steps. (b) If Brian wants to transfer the water to a cylinder with a radius of 5 inches, and fill it completely, what height would the cylinder have to be? Show your work and explain your steps.

1 answer:

MakcuM [25]3 years ago

5 0

Answer:

current volume of the water = 785.714 in²

Height of new container = 10 inches

Step-by-step explanation:

Radius of the container = 5 inches

Height of the container = 20 inches

Volume of the container = πR²H = (22/7) × 5² × 20 = 1571.428571428 in²

Since the container is half filled with water, volume of the water = 1571.428571428 ÷ 2 = 785.714 in²

b. Volume of cylinder = πR²H = volume of water from first container = 785.714 in²

Hence 785.714 = (22/7) × 5² × H

H = (785.714) ÷ [(22/7) × 25] = 10 inches

Height of new container = 10 inches

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<h2>○=> <u>Correct options</u> :</h2><h2>□ $\color{plum}\tt\bold{(C) \: SR = 27}$ </h2><h2>□ $\color{plum}\tt\bold{(D) \: QR = 54 }$ </h2><h3><u>Steps to derive the correct options</u> :</h3>

Since two sides and one included angle is equal in △PQS and △PRS, we can conclude that they are congruent under the SAS congruence criterion.

Which means :

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▪︎QS = RS

Given :

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Measure of segment RS = 4n+11

Thus :

$= \tt6n + 3 = 4n + 11$

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Measure of segment QS :

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$= \tt6 \times 4 + 3$

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$\color{plum}\tt \: \bold{QS = 6n + 3 = \tt\bold{27}}$

Thus, measure of QS = 27

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$=\tt 4n + 11$

$=\tt 4 × 4 + 11$

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$\color{plum}\tt \: \bold{RS = 4n + 11 = \tt\bold{27}}$

Measure of QR :

$=\tt 27+27$

$\color{plum}\tt \: \bold{QR = 27+27 \tt\bold{=54}}$

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▪︎QR = 54

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▪︎(C) SR = 27

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