Question

Pls ans the question!!!!!!!

Answer 1

Answer:

formula of square area side×side

Explanation:

base of toy=7cm

height of toy=8cm

square area of toy

=side×side

=7×8cm

=56cm

Answer 2

Answer:

Question :

A toy is in the form of cone mounted on a hemisphere the radius of the base and the height of the cone or 7 cm and 8 cm respectively find the surface area of the toy.

Given :

• → Radius of cone = 7 cm
• → Height of cone = 8 cm
• → Radius of hemisphere = 7 cm
• → Height of hemisphere = 8 cm

To Find :

• → Surface area of cone
• → Surface area of hemisphere
• → Surface area of toy

Using Formulas :

$\star\:\small{\underline{\boxed{\sf{l=\sqrt{r^2+h^2}}}}}$

$\star\:\small{\underline{\boxed{\sf{Surface \: area \: of \: cone = \pi r l}}}}$

$\star\:\small{\underline{\boxed{\sf{Surface \: area \: of \: hemisphere = 2\pi r^2}}}}$

$\star\:\small{\underline{\boxed{\sf{Surface \: area_{(toy)} = Surface \: area_{(Cone)} + Surface \: area_{(Hemisphere)}}}}}$

Solution :

★ Fistly, finding the slant height of cone for find surface area of cone :

$\longrightarrow\:\:{\sf{l=\sqrt{r^2+h^2}}}$

$\longrightarrow\:\:{\sf{l=\sqrt{(7)^2 + (8)^2}}}$

$\longrightarrow\:\:{\sf{l=\sqrt{(7 \times 7)+ (8 \times 8)}}}$

$\longrightarrow\:\:{\sf{l=\sqrt{(49)+ (64)}}}$

$\longrightarrow\:\:{\sf{l=\sqrt{49+ 64}}}$

$\longrightarrow\:\:{\sf{l=\sqrt{113}}}$

$\longrightarrow\:\:{\sf{\underline{\underline{l \approx 10.63 \: cm}}}}$

∴ The slent height of cone is 10.63 cm.

$\rule{300}{1.5}$

★ Finding the surface area of cone by substituting the values in the formula :

$\longrightarrow \: \: {\sf{Surface \: area \: of \: cone = \pi r l}}$

${\longrightarrow \: \: {\sf{Surface \: area \: of \: cone = \dfrac{22}{7} \times 7 \times 10.63}}}$

${\longrightarrow \: \: {\sf{Surface \: area \: of \: cone = \dfrac{22}{\cancel{7}} \times \cancel{7}\times 10.63}}}$

${\longrightarrow \: \: {\sf{Surface \: area \: of \: cone =22 \times 10.63}}}$

${\longrightarrow \: \: {\sf{\underline{\underline{Surface \: area \: of \: cone \approx 233.86 \: {cm}^{2}}}}}}$

∴ The surface area of cone is 233.86 cm².

$\rule{300}{1.5}$

★ Finding the surface area of hemisphere by substituting the values in the formula :

${\longrightarrow\: \: {\sf{Surface \: area \: of \: hemisphere = 2\pi r^2}}}$

${\longrightarrow\: \: {\sf{Surface \: area \: of \: hemisphere = 2 \times \dfrac{22}{7} \times (7)^2}}}$

${\longrightarrow\: \: {\sf{Surface \: area \: of \: hemisphere = 2 \times \dfrac{22}{7} \times 7 \times 7}}}$

${\longrightarrow\: \: {\sf{Surface \: area \: of \: hemisphere = 2 \times \dfrac{22}{\cancel{7}} \times \cancel{7} \times 7}}}$

${\longrightarrow\: \: {\sf{Surface \: area \: of \: hemisphere = 2 \times 22 \times 7}}}$

${\longrightarrow\: \: {\sf{\underline{\underline{Surface \: area \: of \: hemisphere = 308 \: {cm}^{2} }}}}}$

∴ The surface area of hemisphere is 308 cm².

$\rule{300}{1.5}$

★ Now, finding the surface area of toy by substituting the values in the formula :

${\longrightarrow \: \:{\sf{Surface \: area_{(toy)} = Surface \: area_{(Cone)} + Surface \: area_{(Hemisphere)}}}}$

${\longrightarrow \: \:{\sf{Surface \: area_{(toy)} = 233.86 + 308}}}$

${\longrightarrow \: \:{\sf{\underline{\underline{Surface \: area_{(toy)} = 541.86 \: {cm}^{2}}}}}}$

∴ The surface area of toy is 541.86 cm².

Learn More :

$\boxed{\begin{minipage}{6 cm}\bigstar \:\underline{\textbf{Formulae Related to Cone :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:Curved \: Surface \: Area = \pi rl\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:TSA = Area\:of\:Base + CSA=\pi r^2+\pi rl\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\dfrac{1}{3}\pi r^2h\\ \\{\textcircled{\footnotesize\textsf{5}}} \: \:Slant \: Height=\sqrt{r^2 + h^2}\end{minipage}}$

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$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

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