Question

*21 સેમી વ્યાસ અને 12સેમી તિર્યક ઊંચાઈના શંકુનો પાયો અર્ધગોળાકાર છે, તો કુલ સપાટીનું પૃષ્ઠ્ફળ __________ છે.*

1️⃣ 1098 ચો સેમી
2️⃣ 1089 ચો સેમી
3️⃣ 1079 ચો સેમી
4️⃣ 1099 ચો સેમી​

Answer 1

Answer:

Given: A cone with hemispherical surface.

Slant Height (l)=12 cm

diameter (d)=21 cm

Total surface area= Total surface Area

of cone + Total surface area of Hemisphere

Total surface area of cone =πre+πr  

2

 

But since we have a hemispherical surface as be x

Total surface are of cone will be =πrl

Total  surface area of hemisphere =3πr  

2

 

but since the base of cone hemisphere is some and total surface area will not consider any be x area

∴ Total surface area of hemisphere in this case =2πr  

2

 

Total surface area = Total surface area of cone+ Hemisphere  

putting value of l=12 cm and r=d/2=21/2=10.5 cm

∴π(12)(  

2

21

​  

)+2π(  

2

21

​  

)  

2

 

=1088.01 cm  

2

Answer 2

$\huge \colorbox{orange}{✎...Answer}$

4. 1099 યો સેમી