Jump to

- GST
- Banking
- Shares and Dividends
- Quadratic Equations in One Variable
- Factorization
- Ratio and Proportion
- Matrices
- Arithmetic and Geometric Progression
- Reflection
- Section Formula
- Equation of Straight Line
- Similarity
- Locus
- Circles
- Constructions
- Mensuration
- Trigonometric Identities
- Trigonometric Tables
- Heights and Distances

A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.

Answer:

**Solution:**

As the hall comprises of cylinder and hemisphere so its capacity or volume will be the sum of volume of cylinder and hemisphere.

Let the radius of the hemisphere be r.

Inner diameter = 2r

Given greatest height equal to inner diameter.

So total height of the hall = 2r

Height of the hemispherical part = r

Height of cylindrical area, h = 2r-r = r

Capacity of the hall = Volume of cylindrical area + volume of hemispherical area

=π r^{2}h + (2/3)π r^{3}

=π r^{3}+ (2/3)π r^{3} [∵h = r]

=π r^{3} (1+2/3)

=π r^{3} (3+2)/3

= (5/3)π r^{3}

Given capacity of hall = 48510 m^{3}

(5/3)π r^{3}= 48510

(5/3)×(22/7)r^{3}= 48510

r^{3} = 48510×3×7/(22×5)

r^{3} = 9261

Taking cube root on both sides,

r = 21

Area of the floor = r^{2}

= (22/7)×21^{2}

= 22×21×3

= 1386 m^{2}

**Hence the Area of the floor is 1386 m ^{2} .**

Related Questions

Was This helpful?

Exercises

Chapters

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2022 © Quality Tutorials Pvt Ltd All rights reserved