A strong cone of sweep r and tallness h is set over a strong chamber having same base range and stature as that of a cone. The complete surface space of the consolidated strong is πr[√(r2 + h2 +3r + 2h].
A strong cone of sweep r and tallness h is set over a strong chamber having same base range and stature as that of a cone. The complete surface space of the consolidated strong is πr[√(r2 + h2 +3r + 2h].

False

Clarification:

At the point when a strong cone is put over a strong chamber of same base range, the foundation of cone and top of the chamber won’t be shrouded in absolute surface region.

Since the stature of cone and chamber is same,

We get,

All out surface space of cone\[=\text{ }\pi rl\text{ }+\text{ }\pi r2\] , where r = base sweep and l = incline tallness

Complete surface space of shape framed = Total surface space of cone + Total Surface space of chamber – 2(Area of base)

NCERT Exemplar Class 10 Maths Chapter 12 Ex. 12.2 Question 3

Complete surface space of chamber\[=\text{ }2\pi rh\text{ }+\text{ }2\pi r2h\] , where r = base span and h = tallness

\[=\text{ }\pi r\left( r\text{ }+\text{ }l \right)\text{ }+\text{ }\left( 2\pi rh\text{ }+\text{ }2\pi r2 \right)\text{ }\text{ }2\left( \pi r2 \right)\]

\[=\text{ }\pi r2\text{ }+\text{ }\pi rl\text{ }+\text{ }2\pi rh\text{ }+\text{ }2\pi r2\text{ }\text{ }2\pi r2\]

\[=\text{ }\pi r\left( r\text{ }+\text{ }l\text{ }+\text{ }h \right)\]

NCERT Exemplar Class 10 Maths Chapter 12 Ex. 12.2 Question 3-1