Question

the depth of the pit.
7. Find the height of the cylinder whose total surface area is 660 cm and diameter is 10 cm.
8. The ratio of radius and height of a cylinder is 5:7 and its volume is 550 cm'. Find its radius and heie​

Answer 1

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$\underline{\underline{\bf \green {Questions\::}}}$

• Find the height of the cylinder whose total surface area is 660 cm² and diameter is 10 cm.

• The ratio of radius and height of a cylinder is 5:7 and it's volume is 550 cm³. Find it's radius and height.

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$\underline{\underline{\bf \green {Given}}}\begin{cases} & \qquad \underline{\sf \red{ In \:1^{st} \:question\::}} \\ & \sf{\bullet\:Total\:surface\;area_{(Cylinder)} = \bf{660\;cm^{2}}} \\ & \sf{\bullet\:Diameter_{(Cylinder)}\:=\:\bf{10\:cm}} \\ & \qquad \underline{\sf \red { In\:2^{nd}\:question\::}} \\ & \sf{\bullet\:Ratio\:of\:radius_{(Cylinder)}\:and\:height_{(Cylinder)}\:=\:\bf{5:7}} \\ & \sf{\bullet\:Volume_{(Cylinder)}\:=\:\bf{550\:cm^{3}}}\end{cases}$

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$\underline{\underline{\bf \green {To\:Find}}}\begin{cases} & \qquad \underline{\sf \red{ In \:1^{st} \:question\::}} \\ & \sf{\bullet\:Height_{(Cylinder)} ? } \\ & \qquad \underline{\sf \red { In\:2^{nd}\:question\::}} \\ & \sf{\bullet\:Radius_{(Cylinder)}\:and\:height_{(Cylinder)} ? }\end{cases}$

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$\underline{\underline{\bf \green {Solving\:1^{st}\:question\::}}}$

$\qquad \qquad\tiny \: {\underline{\rm{\red\bigstar\:Formula \:that \:we \:will \:use \::}}}$

$\orange\bigstar\:\boxed{\boxed{\sf \blue{Total\:surface\:area_{(Cylinder)}\:=\:\bf{2\pi r (h\:+\:r)}}}}$

$\qquad \qquad\tiny \: {\underline{\rm{\green\bigstar\:Values\:that\;we\;have\::}}}$

(i) Total surface area = 660 cm²

(ii) Radius = d/2 = 10/2 = 5 cm

$\qquad \qquad\tiny \dag \: {\underline{\frak{Putting\:all\;values\;in\;the\:formula\::}}}$

$:\implies\:\sf 660\:=\:2\:\times\:\dfrac{22}{7}\:\times\:5\:\times\:(h\:+\:5)$

$:\implies\:\sf 660\:=\:\dfrac{2\:\times\:22\:\times\:5}{7}\:\times\:(h\:+\:5)$

$:\implies\:\sf 660\:=\:\dfrac{220}{7}\:\times\:(h\:+\:5)$

$:\implies\:\sf 660\:\times\:\dfrac{7}{220}\:=\:h\:+\:5$

$:\implies\:\sf \cancel{660}\:\times\:\dfrac{7}{\cancel{220}}\:=\:h\:+\:5$

$:\implies\:\sf 3\:\times\:7\:=\:h\:+\:5$

$:\implies\:\sf 21\:=\:h\:+\:5$

$:\implies\:\sf h\:=\:21\:-\:5$

$:\implies\:\sf h\:=\:\underline{\bf{16}}$

$\underline{\boxed {\frak {\therefore \pink {Height_{(Cylinder)}\:\leadsto\:16\:cm}}}}\:\purple\bigstar$

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$\underline{\underline{\bf \green {Solving\:2^{nd}\:question\::}}}$

$\qquad \qquad\tiny \: {\underline{\rm{\red\bigstar\:Formula \:that \:we \:will \:use \::}}}$

$\orange\bigstar\:\boxed{\boxed{\sf \blue{Volume_{(Cylinder)}\:=\:\bf{\pi r^{2} h}}}}$

$\qquad \qquad\tiny \: {\underline{\rm{\green\bigstar\:Values\:that\;we\;have\::}}}$

(i) Volume = 550 cm³

(ii) Ratio of radius and height = 5:7

⛳ Let radius and height = 5x and 7x ⛳

$\qquad \qquad\tiny \dag \: {\underline{\frak{Putting\:all\;values\;in\;the\:formula\::}}}$

$:\longmapsto\:\sf 550\:=\:\dfrac{22}{7}\:\times\:(5x)^{2}\:\times\:7x$

$:\longmapsto\:\sf 550\:=\:\dfrac{22}{7}\:\times\:25x^{2}\:\times\:7x$

$:\longmapsto\:\sf 550\:=\:\dfrac{22}{7}\:\times\:175x^{3}$

$:\longmapsto\:\sf 550\:=\:\dfrac{22}{7}\:\times\:175\:\times\:x^{3}$

$:\longmapsto\:\sf \dfrac{550\:\times\:7}{22\:\times\:175} = x^{3}$

$:\longmapsto\:\sf \dfrac{\cancel{550}\:\times\:\cancel{7}}{\cancel{22}\:\times\:\cancel{175}} = x^{3}$

$:\longmapsto\:\sf \dfrac{25}{25} = x^{3}$

$:\longmapsto\:\sf \dfrac{\cancel{25}}{\cancel{25}} = x^{3}$

$:\longmapsto\:\sf 1 = x^{3}$

$:\longmapsto\:\sf x = \underline{\bf{\sqrt[3]{1}}}$

$\underline{\boxed {\frak {\therefore \pink {Radius_{(Cylinder)}\:height_{(Cylinder)}\:\leadsto\:5\sqrt[3]{1}\:cm\:and\:7\sqrt[3]{1}\:cm}}}}\:\purple\bigstar$

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Answer 2

Answer:

r:h=5:7

V=550cm²

d=?

V=πr²h

550=22/7×5y×5y×7y

550×7÷22=25y²×7y

1=y³

y=1cm

⇒r =5cm

d =2r

  =2×5

  =10cm