the csa of cylindrical pillar is 264m sq. and it's volume is 924mcube . Find the diameter and height of the pillar..
Answers
264 =2×22/7×R×H
264×7/2×22 =R×H
42=R×H
VOLUME OF CYLINDER =PIE ×R^2×H
924 =PIE×R×R×H
924 =22/7×R×42
924×7/22×42 =R
7=R
CSA OF CYLINDER =2×PIE ×R×H
264=2×22/7×7×H
264=44×H
4=h
Answer:
Diameter of the cylindrical pillar is 14 m and the height of the cylindrical pillar is 6 m.
Step-by-step explanation:
Given :-
CSA of a cylindrical pillar is 264 m² and volume is 924 m³.
To find :-
Diameter and height of the cylindrical pillar.
Solution :-
Let the radius of the cylindrical pillar be r m and the height of the cylindrical pillar be h m.
Formula used :
★{\boxed{\sf{CSA\: of\: cylinder=2\pi\:rh}}}
CSAofcylinder=2πrh
★ {\boxed{\sf{Volume\:of\: cylinder=\pi\:r^2h}}}
Volumeofcylinder=πr
2
h
According to the 1st condition,
CSA of the cylindrical pillar is 264 m².
\begin{gathered}\to \sf \: 2\pi \: rh \: = 264 \\ \\ \to \sf \: 2 \times \dfrac{22}{7} \times rh = 264 \\ \\ \to \sf \: rh = 264 \times \dfrac{1}{2} \times \dfrac{7}{22} \\ \\ \to \sf \: rh = 42.............(i)\end{gathered}
→2πrh=264
→2×
7
22
×rh=264
→rh=264×
2
1
×
22
7
→rh=42.............(i)
According to the 2nd condition,
Volume of the cylindrical pillar is 924 m³.
\begin{gathered}\to \sf \: \pi {r}^{2} h = 924 \\ \\ \to \sf \: \dfrac{22}{7} \times rh \times r = 924 \\ \\ \to \sf \: \dfrac{22}{7} \times 42 \times r = 924 \\ \\ \to \sf \: 22 \times 6 \times r \: = 924 \\ \\ \to \sf \: 132r \: = 924 \\ \\ \to \sf \: r \: = 7\end{gathered}
→πr
2
h=924
→
7
22
×rh×r=924
→
7
22
×42×r=924
→22×6×r=924
→132r=924
→r=7
Radius of the cylindrical pillar is 7 m.
Diameter = 2×7 = 14 m.
Now put r = 7 in eq(i).
rh = 42
→ 7×h = 42
→ h = 42/7
→ h = 6
Height of the cylindrical pillar is 6 m.