NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 solutions Ex 13.1

As per the question:

The radius of the conical part \( =3.5 \mathrm{~cm} \)
And,
Radius of the hemispherical part \( =3.5 \mathrm{~cm} \)
Height of hemispherical part \( = \) Radius \( (\mathrm{r}) \)
\(=3.5 \mathrm{~cm}\)
\(=\frac{7}{2} \mathrm{~cm}\)
Height of conical part \( (\mathrm{h})=15.5-3.5=12 \mathrm{~cm} \)
for cone we know that,
\(\mathrm{r}^{2}+\mathrm{h}^{2}=\mathrm{l}^{2}\)
where \( r \) is the radius of base, \(l\) is the slant height and \( h \) is the height of cone.
Slant height of the conical part
\(l=\sqrt{r^{2}+h^{2}}\)
\(l=\sqrt{\left(\frac{7}{2}\right)^{2}+(12)^{2}}\)
\(l=\sqrt{\frac{49}{4}+144}\)
\(l=\sqrt{\frac{49+576}{4}}\)
\(l=\sqrt{\frac{625}{4}}\)
\(l=\frac{25}{2}\)
Total surface area of toy \( = \) CSA of conical part \(+\) CSA of hemispherical part
\(=\pi r^{1}+2 \pi r^{2}\)
\(=\left(\frac{22}{7} \times \frac{7}{2} \times \frac{25}{2}\right)+\left(2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)\)
\(=\left(\frac{275}{2}\right)+(11 \times 7)\)
\(=137.5+77\)
\(=214.5 \mathrm{~cm}^{2}\)