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Sauron [17]
3 years ago
7

Help with my geometry. please?

Mathematics
1 answer:
Jet001 [13]3 years ago
4 0

if indeed the QT || RS, that simply means that QT is the midsegment of ∡PRS, but that also means that the corresponding sides of the small triangle ∡PQT and the larger one ∡PRS are in a proportion, meaning


\bf \cfrac{PQ}{PR}=\cfrac{PT}{PS}~\hspace{5em}\cfrac{18}{18+38}=\cfrac{36}{36+80}\implies \cfrac{18}{56}=\cfrac{36}{116}\implies \cfrac{9}{28}\ne \cfrac{9}{29}~~\bigotimes

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  • Volume of the cylinder = 1540 cu. cm

\\

Step-by-step explanation:

Given:

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  • Height of the cylinder = 10 cm

\\

To Find:

  • Curved surface area
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\\

Solution:

\\

Using formula:

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{CSA  \: of  \: cylinder = 2\pi rh}}}}}}  \:  \star \\  \\

<em>Substituting the required values: </em>

\\

\dashrightarrow \:  \:  \sf CSA {(cylinder)} = 2 \times \dfrac{22}{7} \times 7 \times  10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 2 \times 22 \times 10 \\  \\ \\   \dashrightarrow \:  \:  \sf CSA {(cylinder)} = 44 \times 10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 440 \:  {cm}^{2}  \\ \\ \\

Now,

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{Volume {(cylinder)}=  \pi {r}^{2} h}}}}}}  \:  \star \\  \\ \\

<em>Substituting the required values, </em>

\\

\dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \dfrac{22}{7}  \times  {(7)}^{2}  \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \frac{22}{7}  \times 49  \times 10 \\ \\ \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 22 \times 7 \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 1540 {cm}^{3}  \\ \\ \\

Hence,

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