Q.1 A round thali has 2 inbuilt triangular sections for serving vegetables and a separate semi-circular area for keeping rice or chapati. If radius of thali is 21 𝑐𝑚, find the area of the thali that is shaded in the figure :

 

(OR)

 

Q.2 A toy is in the form of a cylinder of diameter $2\sqrt{2}$ 𝑚 and height 3. 5 𝑚 surmounted by a cone whose vertical angle is 90°. Fund total surface area of the toy.

We have been given the radius of the thali as 21 𝑐𝑚 . We can find the area of the shaded region as follows: 

By the diagram, we can infer that,

 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒 − 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶

Area of semicircle= ${\mathrm{\pi r}}^2$ , where r is the radius of the semicircle.

$\begin{array}{r}\Rightarrow \text{ Area of semicircle }=\frac{22}{7}\times \frac{1}{2}\times {21}^2\\ \Rightarrow \text{ Area of semicircle }=693{\mathrm{cm}}^2\\ \text{ Now, Area of triangle }=\frac{1}{2}\times \text{ base }\times \text{ height }\end{array}$

Here, the base will be the diameter (2 × 21 𝑐𝑚) and the height will be the radius.

$\begin{array}{r}=\frac{1}{2}\times 42\mathrm{cm}\times 21\mathrm{cm}\\ =441{\mathrm{cm}}^2\end{array}$

⇒ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = (693 − 441) 𝑠𝑞. 𝑐𝑚 

Hence, 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 is equal to 252 𝑠𝑞. 𝑐𝑚

(OR)

We have been given a toy cylinder of diameter $2\sqrt{2}$ 𝑚 and height 3. 5 𝑚. 

By Pythagoras Theorem we know that the sum of the squares on the legs of a right triangle is equal to the square of the hypotenuse. We can see that ∠𝐶 = 90°

$\Rightarrow A{B}^2=A{C}^2+B{C}^2$

Let the slant height of the cone 𝐴𝐶 be 𝑥.

$\begin{array}{r}\Rightarrow A{B}^2=x^2+x^2[AC=BC]\\ \Rightarrow (2\sqrt{2}{)}^2=2x^2\\ \Rightarrow x=4m\\ \Rightarrow x=2m\end{array}$

$\begin{array}{r}\text{ Radius of the cylinder }=\frac{\text{ diameter }}{2}\\ \Rightarrow r=\frac{2\sqrt{2}}{2}\\ \Rightarrow r=\sqrt{2}m\end{array}$

Now, total surface area of toy = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 (𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝑐𝑜𝑛𝑒 + 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑏𝑜𝑡𝑡𝑜𝑚) 

We know that, Surface area of cylinder = 2π𝑟ℎ 

Surface area of cone = π𝑟𝑙

$\begin{array}{r}\text{ Area of bottom of cylinder }=\pi r^2\\ \Rightarrow \text{ Surface area }=2\pi rh+\pi rl+\pi r^2\\ =\pi r(2h+l+r)\end{array}$

$\begin{array}{r}=\pi \times \sqrt{2}\times (2\times 3.5)+2+\sqrt{2})\\ =\pi [2+9\sqrt{2}] m^2\end{array}$

Hence, the total surface area of the toy is $\pi [2+9\sqrt{2}] m^2$