Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answers (1)

best_answer

The required surface area of the toy is given by :

 Area of toy  =   Surface area of hemisphere   +   Surface area of the cone

Firstly consider the hemisphere :

  The surface area of a hemisphere is   =\ 2 \pi r^2

or                                                =\ 2 \times \frac{22}{7}\times \left ( 3.5 \right )^2

or                                                 =\ 77\ cm^2

Now for cone we have :

                                 The surface area of a cone   =\ \pi rl

Thus we need to calculate slant of the cone.

                We know that  :

                                                  l^2\ =\ h^2\ +\ r^2

or                                                     =\ 12^2\ +\ 3.5^2

or                                                    =\ \frac{625}{4}

or                                               l\ =\ \frac{25}{2}\ =\ 12.5\ cm

Thus surface area of a cone     =\ \pi rl

or                                            =\ \frac{22}{7}\times 3.5\times 12.5

or                                            =\ 137.5\ cm^2

Hence the total surface area of toy  =   =\ 77\ +\ 137.5\ =\ 214.5\ cm^2

Posted by

Devendra Khairwa

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10 board exams from February 15 to March 18; download date sheet 2025 PDF
anam-khan

CBSE Date Sheet 2025 Live: CBSE Class 10, 12 board exam timetable soon at cbse.gov.in; practical dates
anam-khan

CBSE Date Sheet 2025 Live: CBSE to announce Class 10, 12 board exam dates soon at cbse.gov.in; sample paper

Latest Question