calendarBack
Quant

/

Geometry

/

Mensuration
ALL MODULES

CAT 2025 Lesson : Mensuration - Cylinder & Prism

bookmarked

3.3 Right Circular Cylinder

A right circular cylinder is a circular base raised to a certain height. This is very often referred to as just cylinder in the entrance tests.
In the adjacent figure,
r is the radius of the base.

h is the height of the cylinder.

Volume of cylinder = πr2h\pi r^2 h
Curved surface area (CSA) = 2πrh2\pi r h
Flat Surface Area (FSA) = 2πr22\pi r ^2
Total Surface Area (TSA) = 2πr(h+r)2\pi r (h + r)

Example 8

A cylindrical jar contains water upto a height of 24 cm. If the water in this jar is transferred to another cylindrical jar whose base radius is double that of the earlier jar, then what is the height of the water (in cm) in the new jar?

Solution

Let the radius of the first and second jars be x\bm{x} cm and 2x\bm{2x} cm respectively. Let the height of the second jar be yy cm.

Volume of water in the first container =πr2h=π×x2×24=24πx2= \pi r^{2} h = \pi \times x^{2} \times 24 = 24 \pi x^{2}

Volume of water in the second container =π×(2x)2×y=4πx2y= \pi \times (2x)^{2} \times y = 4 \pi x^{2} y

As the two volumes are equal,
24πx2=4πx2y24 \pi x^{2} = 4 \pi x^{2} y
⇒ y=6y = 6 cm

Answer: 6

Example 9

If the ratio of the diameters of 2 cylinders and the ratio of their volumes are 2 : 3 and 3 : 4 respectively, then what is the ratio of their heights?

Solution

As diameter is twice the radius, the ratio of radii will be the same as the ratio of the diameters.

Let the radii of the 2 cylinders be 2rr and 3rr. Let h1h_{1} and h2h_{2} be their respective heights.

Ratio of volumes =π(2r)2h1π(3r)2h2=34 = \dfrac {\pi (2r)^{2}h_{1}}{\pi (3r)^{2}h_{2}} = \dfrac{3}{4} ⇒ 4h19h2=34 \dfrac{4h_{1}}{9h_{2}} = \dfrac{3}{4}

⇒ h1h2=2716\dfrac{h_{1}}{h_{2}} = \dfrac{27}{16}

Answer: 27 : 16

3.4 Right Prism

Like cylinders, the bottom and top of a right prism are identical. However, while the cylinder has a circular base, the right prism has a polygonal base.

By definition, a right prism has a polygonal base raised to a certain height. The polygon can be any nn-sided figure - triangle, quadrilateral, pentagon, hexagon, etc.

In the adjacent figure,
h\bm{h} is the height of the right prism.

Volume of Right Prism = Area of Base ×\times hh

Lateral Surface Area (LSA) = Perimeter of Base ×\times hh
Total Surface Area (TSA) = LSA + (2 ×\times Area of Base)

Example 10

A solid right triangular prism, with the base being an equilateral triangle, is melted and recast into a solid right hexagonal prism, where the base is a regular hexagon (with no material wastage). If the heights of the two right prisms are the same, then what is the ratio of the perimeters of the bases of the triangular prism and the hexagonal prism?
(1) 1:21 : \sqrt{2}       (2) 3:1\sqrt{3} : 1        (3) 6:1\sqrt{6} : 1        (4) 3:2\sqrt{3} : \sqrt{2}

Solution

Let the length of the side of the equilateral triangle in the base of the first prism be a1a_1 and that of the regular hexagon in the second prism be a2a_2 . Let the heights of the two prisms be hh.

Volume of Equilateral triangular Prism =34×a12×h = \dfrac {\sqrt{3}}{4} \times a_{1}^{2} \times h

Volume of Regular Hexagonal Prism =6×34×a22×h = 6 \times \dfrac {\sqrt{3}}{4} \times a_{2}^{2} \times h

As one prism is recast into the other, the above volumes are equal.

34×a12×h\dfrac {\sqrt{3}}{4} \times a_{1}^{2} \times h = 6×34×a22×h 6 \times \dfrac {\sqrt{3}}{4} \times a_{2}^{2} \times h

⇒ a12a22=6\dfrac {a_{1}^{2}}{a_{2}^{2}} = 6 ⇒ a1a2=6\dfrac {a_{1}}{a_{2}} = \sqrt{6}

Ratio of Perimeters =3a16a2=12×a1a2=12×6=32= \dfrac{3a_{1}}{6a_{2}} = \dfrac{1}{2} \times \dfrac{a_1}{a_2} = \dfrac{1}{2} \times \sqrt{6} = \dfrac{\sqrt{3}}{\sqrt{2}}

Answer: (4) 3:2\sqrt{3} : \sqrt{2}

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock