Question

the total surface area of cylinder is 40 /pi
sq. cm. and its height is 5.5cm. Find the radius of the base.​

Answer 1

Step-by-step explanation:

total surface area of a cylinder is 40π cm^ 2

2pi * r * (r + h) = 40Pi

2r * (r + 5.5) = 40

2r ^ 2 + 11r = 40

2r ^ 2 + 11r - 40 = O

2r ^ 2 + 16r - 5r - 40 = 0

2r * (r + 8) - 5(r + 8) = 0

(2r - 5)(r + 8) = o

from here

r can be 5/2 = 2.5

and

r can be - 8

radius = 2.5 cm

but it is not possible as Radius can't be negative

HOPE IT HELPS YOU

Answer 2

Answer:

AnsWer:

$\bold{\underline{\boxed{\red{\sf{Radius\:of\:the\:base\:=\:2.5cm}}}}}$

$\bold{\underline{\underline{\large{\mathfrak{StEp\:by\:stEp\:explanation:}}}}}$

$\bold{\underline{\underline{\red{\tt{GiVeN:}}}}}$

$Total surface area of a cylinder is \tt{40\:\pi\:cm^2}40πcm </p><p>2</p><p> </p><p>Height = 5.4 cm</p><p>\bold{\underline{\underline{\red{\tt{To\:FiNd:}}}}}$

$Base radius.</p><p>\bold{\underline{\underline{\red{\tt{SoLuTioN:}}}}} </p><p>SoLuTioN:$

Formula :

$\bold{\large{\red{\boxed{\tt{TSA_{cylinder}\:=\:2\pi\:r(h+r)}}}}} </p><p>TSA$

cylinder

=2πr(h+r)

The values of,

TSA = 40π cm²

π = 22/7

h = 5.4 cm

Block in the values,

$\longrightarrow⟶ \tt{40\:\pi\:=\:2\:\pi\:r\:(5.4\:+r)}40π=2πr(5.4+r)$

$\longrightarrow⟶ \tt{40\:\pi\:=\:5.4(2\pi\:r)\:+\:r\:(2\:\pi\:r)}40π=5.4(2πr)+r(2πr)$

$\longrightarrow⟶ \tt{40\:\pi\:=\:10.8\:\pi\:r\:+\:2\:\pi\:\:r^2}40π=10.8πr+2πr </p><p>2$

$\longrightarrow⟶ \tt{2\:\pi\:r^2\:+\:10.8\:\pi\:r\:=\:40\:\pi}2πr </p><p>2</p><p> +10.8πr=40π$

$\longrightarrow⟶ \tt{2\:\pi\:r^2\:+\:10.8\:\pi\:r\:-\:40\:\pi\:=0}2πr </p><p>2</p><p> +10.8πr−40π=0$

Solve using factorization method,

$\longrightarrow⟶ \tt{2\:\pi\:r^2\:+\:10.8\:\times\:3.14\:r\:-\:40\:\pi\:=0}2πr </p><p>2</p><p> +10.8×3.14r−40π=0$

$\longrightarrow⟶ \tt{2\:\pi\:r^2\:33.92\:r\:-\:40\:\pi\:=0}2πr </p><p>2</p><p> 33.92r−40π=0$

$\longrightarrow⟶ \tt{2\:\times\:3.14\:r^2+\:33.92\:r\:-\:40\:\times\:3.14\:=0}=2×3.14r </p><p>2</p><p> +33.92r−40×3.14=0=$

$\longrightarrow⟶ \tt{6.28r^2\:+\:33.92\:r\:-125.6\:=\:0}6.28r </p><p>2</p><p> +33.92r−125.6=0$

$\longrightarrow⟶ \tt{r\:(6.28\:r\:+\:33.92)\:-\:125.6\:=0}r(6.28r+33.92)−125.6=0$

$\longrightarrow⟶ \tt{6.28r^2\:+\:33.92r\:-\:125.6=0}6.28r </p><p>2</p><p> +33.92r−125.6=0$

$\longrightarrow⟶ \tt{2\:\pi\:(r\:+\:2.7)^2\:-\:171.46\:=0}2π(r+2.7) </p><p>2</p><p> −171.46=0$

$\longrightarrow⟶ \tt{2(3.14r^2\:+16.96r\:-62.83\:=\:0}2(3.14r </p><p>2</p><p> +16.96r−62.83=0</p><p>$

$\longrightarrow⟶ \tt{6.28\:r^2\:+\:33.92r\:-\:125.6\:=0}6.28r </p><p>2</p><p> +33.92r−125.6=0$

$\longrightarrow⟶ \tt{r\:=\:-7.92\:\:OR\:\:r\:=\:2.5}r=−7.92ORr=2.5$

Radius cannot be negative.

•°• r = - 7.92 is not acceptable.

$\tt{\therefore{Radius\:of\:the\:base\:=\:2.5cm}}∴Radiusofthebase=2.5cm$