Question

find the volume of a cone it the radius of its base is 1.5cm and its perpendicular height is 5cm​

Answer 1

Answer:

Volume of the cone is 11.79 cm³ (approx).

Step-by-step explanation:

Given :-

• The radius of the base of the cone is 1.5 cm.
• Its perpendicular height is 5 cm.

To find :-

• Volume of the cone.

Solution :-

• Radius = 1.5 cm
• Height = 5 cm

Formula used :

${\boxed{\sf{Volume\:of\:cone=\dfrac{1}{3}\pi\:r^2h}}}$

• r = radius
• h = height

Volume of the cone,

= ⅓ πr²h

= [⅓ × (22/7) ×1.5×1.5×5 ] cm³

= (11×15)/14 cm³

= 165/14 cm³

= 11.79 cm³

Therefore, the volume of the cone is 11.79 cm³(approx).

_______________

Additional information :-

• Volume of cylinder = πr²h
• T.S.A of cylinder = 2πrh + 2πr²
• Volume of cone = ⅓ πr²h
• C.S.A of cone = πrl
• T.S.A of cone = πrl + πr²
• Volume of cuboid = l × b × h
• C.S.A of cuboid = 2(l + b)h
• T.S.A of cuboid = 2(lb + bh + lh)
• C.S.A of cube = 4a²
• T.S.A of cube = 6a²
• Volume of cube = a³
• Volume of sphere = 4/3πr³
• Surface area of sphere = 4πr²
• Volume of hemisphere = ⅔ πr³
• C.S.A of hemisphere = 2πr²
• T.S.A of hemisphere = 3πr²

Answer 2

Given :

$\bullet\:\:\textsf{Radius \: of \: the \: cone = \textbf{1.5 cm}}$

$\bullet\:\:\textsf{Height of the cone = \textbf{5 cm}}$

Solution:

$:\implies \sf Volume \: of \: cone = \dfrac{1}{3} \pi {r}^{2} h$

$:\implies \sf Volume \: of \: cone = \dfrac{1}{3} \times \dfrac{22}{7} \times (1.5)^{2} \times 5$

$:\implies \sf Volume \: of \: cone = \dfrac{1}{3} \times \dfrac{22}{7} \times 1.5 \times 1.5 \times 5$

$:\implies \sf Volume \: of \: cone = \dfrac{22}{7} \times 0.5 \times 1.5 \times 5$

$:\implies \sf Volume \: of \: cone = \dfrac{22}{7} \times 3.75$

$:\implies \underline{ \boxed{\sf Volume \: of \: cone = 11.785 \: cm^{3}}}$

$\therefore\:\underline{\textsf{The volume of cone is \textbf{11.785}} \: \sf {cm}^{3}}.$

_________________

$\boxed{\underline{\underline{\bigstar \: \bf\:Extra\:Brainly\:knowlegde\:\bigstar}}}$

$\boxed{\bigstar{\sf \ Cylinder :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cylinder= \pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ cylinder= 2\pi r h\\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ cylinder= 2\pi r (h+r)$

$\boxed{\bigstar{\sf \ Cone :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cone= \dfrac{1}{3}\pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Cone = \pi r l \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Cone = \pi r (l+r) \\ \\ \\ \sf {\textcircled{\footnotesize4}} Slant \ Height \ of \ cone (l)= \sqrt{r^2+h^2}$

$\boxed{\bigstar{\sf \ Hemisphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Hemisphere= \dfrac{2}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Hemisphere = 2 \pi r^2 \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Hemisphere = 3 \pi r^2$

$\boxed{\bigstar{\sf \ Sphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Sphere= \dfrac{4}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Surface\ Area \ of \ Sphere = 4 \pi r^2$