Question

Look at this cone:,,

Answer 1

• Slant height=l=3ft
• Radius=r=2ft

We know

$\boxed{\sf \star TSA_{(Cone)}=\pi r(r+\ell)}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=2 \dfrac{22}{7}\times 2(2+3)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{44}{7}(5)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{220}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=31.4ft^2$

Now

• New slant height =2(3)=6cm
• New radius=2(2)=4cm

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{22}{7}\times 4(4+6)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{88}{7}(10)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{880}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=125.7cm^2$

So

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{125.7}{31.4}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cone)}}:{TSA_{(Old\:Cone)}}=4:1}}}$