Question

The given figure is a combined solid made up of a cylinder and a
cone. The radius of the base of the solid object is 7cm, length of
the cylindrical part is 35 cm, and the total length of the solid
object is 59cm. Find the total surface area of the solid object
(Ans: 2244cm)
​

Answer 1

$\sf\small\underline\purple{Given:-}$

$\sf{\implies Radius\:_{(solid\: object)}=7cm}$

$\sf{\implies Length\:_{(solid\: object)}=59cm}$

$\sf{\implies Height\:_{(Cylinder)}=35cm}$

$\sf{\implies Height\:_{(Cone)}=59-35=24cm}$

$\sf\small\underline\purple{To\: Find:-}$

$\sf{\implies T.S.A\:_{(solid\: object)}=?}$

$\sf\small\underline\purple{Solution:-}$

As per the given figure we have to find out the solid surface area area of object. Simply by applying formula to calculate the TSA of the object.

$\sf\small\underline\purple{Formula\:Used:-}$

$\bf{\implies T.S.A\:_{(cylinder)}=2\pi\:r\:h+\pi\:r^2}$

• Only one base use for Cylinder

$\bf{\implies T.S.A\:_{(cylinder)}=\pi\:r(2h+r)}$

$\tt{\implies \dfrac{22}{7}*7(2*35+7)}$

$\tt{\implies 22*(70+7)}$

$\tt{\implies 22*77}$

$\tt{\implies 1694cm^2}$

$\bf{\implies Slant\: height\:_{(cone)}=Radius^2+height^2}$

$\tt{\implies L^2=H^2+R^2}$

$\tt{\implies L^2= 24^2+7^2}$

$\tt{\implies L^2=576+49}$

$\tt{\implies L^2=625}$

$\tt{\implies L=\sqrt{625}}$

$\tt{\implies L=25cm}$

$\bf{\implies C.S.A\:_{(cone)}=\pi\:r\:L}$

$\tt{\implies \dfrac{22}{7}*7*25}$

$\tt{\implies 22*25}$

$\tt{\implies 550cm^2}$

$\sf\large{Hence,}$

$\tt{\implies T.S.A\:_{(solid\: object)}=T.S.A\:_{(cylinder)}+C.S.A\:_{(cone)}}$

$\tt{\implies T.S.A\:_{(solid\: object)}=1694+550}$

$\bf{\implies T.S.A\:_{(solid\: object)}=2244cm^2}$

Answer 2

Given : The radius of the base of the solid object is 7cm , length or Height of the cylindrical part is 35 cm , and the total length of the solid object is 59cm .

Exigency To Find : The total surface area of the solid object .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀☆ Finding Total Surface Area [ T.SA ] of solid object :

As, We know that ,

⠀⠀⠀⠀⠀⠀If we have to find Total Surface Area of any object that is formed by joining two objects so , we should have to add their Curved surface area & there is solid so we to add there area of Base to .

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\: \bf \:Formula\:for\: T.S.A \:\:of\:solid\:object\::$

$\qquad \dag\:\:\bigg\lgroup \sf{T.S.A _{(Solid \:Object\:)} \:\:= \:\: C.S.A \:_{(Cylinder)} + C.S.A \:_{(Cone)} + Area_{(Circular\:Base \:)} }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀Here , C.S.A is the Curved Surface Area , T.S.A is the Total Surface Area ,

$\qquad \qquad \leadsto \:\:\bf C.S.A \:_{(Cylinder)} \:=\: 2 \pi r h\: sq.unit \:\:$

$\qquad \qquad \leadsto \:\:\bf C.S.A \:_{(Cone)} \:=\: \pi r l\: sq.unit \:\:$

$\qquad \qquad \leadsto \:\:\bf Area \:_{(Circe)} \:=\: \pi r^2\: sq.unit \:\:$

⠀⠀⠀⠀⠀⠀Here , r is the Radius , h is the Height , l is the slant Height of cone & $\pi \:= \: \dfrac{22}{7} \:$.

⠀⠀⠀⠀⠀⠀Finding Height of Cone :

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\:\bf Formula\:for\:\:Height\:of\:cone \:: \\\\ \qquad \dag\:\:\bigg\lgroup \sf{ Height_{(Cone)} \: =\: Height _{(Solid \:Object \:)} \: - \: Height \:_{(Cylinder)} \: }\bigg\rgroup \\\\\\ \qquad:\implies \sf Height_{(Cone)} \: =\: Height _{(Solid \:Object)} \: - \: Height \:_{(Cylinder)} \\\\\qquad:\implies \sf Height_{(Cone)} \: =\: 59\: - \: 35\\\\ \qquad:\implies \sf Height_{(Cone)} \: =\: 24\\\qquad:\implies \bf Height_{(Cone)} \: =\: 24\:cm$

⠀⠀⠀⠀⠀⠀Finding Slant Height of Cone :

$\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\:\bf Formula\:for\:Slant\:Height\:of\:cone \:: \\\\\qquad \dag\:\:\bigg\lgroup \sf{ (Slant\:Height\:)^2 \: =\: (Radius)^2 \:+ \:(Height)^2 \: }\bigg\rgroup \\\\\qquad \quad \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\qquad:\implies \sf (Slant\:Height\:)^2 \: =\: (Radius)^2 \:+ \:(Height)^2 \\\\\qquad:\implies \sf (Slant\:Height\:)^2 \: =\: (7)^2 \:+ \:(24)^2 \\\\\qquad:\implies \sf (Slant\:Height\:)^2 \: =\: 576 \:+ \:49 \\\\\qquad:\implies \sf (Slant\:Height\:)^2 \: =\: 625 \\\\\qquad:\implies \sf Slant\:Height\:\: =\: \sqrt {625} \\\\\qquad:\implies \sf Slant\:Height\:\: =\: 25 \\\\\qquad:\implies \bf Slant\:Height_{(Cone)}\:\: =\: 25 \: cm$

Now ,

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\: \bf \:Formula\:for\: T.S.A \:\:of\:solid\:object\::$

$\qquad \dag\:\:\bigg\lgroup \sf{T.S.A _{(Solid \:Object\:)} \:\:= \:\: C.S.A \:_{(Cylinder)} + C.S.A \:_{(Cone)}+ Area_{(Circular\:Base \:)}}\bigg\rgroup$

$\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: C.S.A \:_{(Cylinder)} + C.S.A \:_{(Cone)}+ Area_{(Circular\:Base \:)}$

$\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \pi r h \:+\pi r l+ \pi r^2$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \pi r h \:+\pi r l + \pi (r)^2\\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \pi \times 7 \times 35 \:+\pi \times 7 \times 25 + \pi \times (7)^2 \\\\ \qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \pi \times 7 \times 35 \:+\pi \times 7 \times 25 + \pi \times 49 \\\\ \qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \times \dfrac{22}{7} \times 7 \times 35 \:+\dfrac{22}{7} \times 7 \times 25 + \dfrac{22}{7} \times 49\\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \times \dfrac{22}{\cancel {7}} \times \cancel {7} \times 35 \:+\dfrac{22}{\cancel {7}} \times \cancel {7} \times 25+ \dfrac{22}{\cancel {7}} \times \cancel {49} \\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2 \times 22\times 35 \:+ 22 \times 25 + 22 \times 7 \\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:) }\:\:= \:\: 44 \times 35 \:+ 22 \times 25 + 154 \\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:) }\:\:= \:\: 44 \times 35 \:+ 550 + 154\\\\\qquad:\implies \sf T.S.A _{(Solid \:Object\:) }\:\:= \:\: 1540 \:+ 550+ 154\\\\ \qquad:\implies \sf T.S.A _{(Solid \:Object\:) }\:\:= \:\: 2090 + 154\\\\ \qquad:\implies \sf T.S.A _{(Solid \:Object\:) }\:\:= \:\: 2244\\\\ \qquad:\implies \sf T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2244 \: cm^2 \:\\\\ \qquad :\implies \frak{\underline{\purple{\: T.S.A _{(Solid \:Object\:)} \:\:= \:\: 2244 \: cm^2}} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Total \:Surface \:Area \:of\:Solid\:object \:is\:\bf{2244\:cm^2}}}}$