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3 years ago

15

If the radius of a cone is 5 in and the height is 12 in what is the volume of the cone

2 answers:

expeople1 [14]3 years ago

7 0

In form of pi it is 100pi. In simplified form it is 314.15927

Sav [38]3 years ago

4 0

Answer:

Volume of cone = 314 in³

Step-by-step explanation:

Radius of the cone (r) = 5 in

Height of the cone (h) = 12 in

$Voume \: of \: the \: cone = \frac{1}{3} \pi {r}^{2} h \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{3} \times \pi \times {(5)}^{2} \times 12\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{3} \times \pi \times 25 \times 12\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{3} \times \pi \times 300 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \pi \times 100\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 314 \: {in}^{3}$

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From the graph:

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In parallelogram TQRS

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then;

Area of parallelogram TQRS = $QR \cdot PT$

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Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

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the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

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