Question

heena hold a japnese fan in her hand as shown in the figure. it is shaped like a sector of a circle and made of a thin material such as paper or feather. the inner and outer radii are 6cm and 10cm respectively. the fan has three colours i.e.pink, blue and black​

Answer 1

Answer:

The dimensions of equal strips are 25 cm, 25 cm and 14 cm. 

Then area of each strip=s(s−a)(s−b)(s−3)=32(32−25)(32−25)(32−14)=32×7×7×18=168

Then total the area of each type of paper needed=168×5=840 cm2

Step-by-step explanation:

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Answer 2

Note: The given question is incomplete as the question should be as shown in the question image.

Given: inner and outer radii are 6cm and 10cm respectively.

Part (i):

Find the area of the region having blue colour.

Area of the region having blue colour$\[ = \frac{{80}}{{360}} \times \pi {\left( {10} \right)^2} - \frac{{80}}{{360}} \times \pi {\left( 6 \right)^2}\]$

                                                               $\[\begin{array}{l} = \frac{{80}}{{360}} \times \frac{{22}}{7} \times \left[ {{{10}^2} - {6^2}} \right]\\ = \frac{{80}}{{360}} \times \frac{{22}}{7} \times \left[ {\left( {10 + 6} \right)\left( {10 - 6} \right)} \right]\end{array}\]$

                                                               $\[\begin{array}{l} = \frac{{80}}{{360}} \times \frac{{22}}{7} \times 16 \times 4\\ = \frac{{112640}}{{2520}}\\ = 44.69c{m^2}\end{array}\]$

Part (ii):

Find the area of the region having black colour.

Area of the region having blue colour$\[ = \frac{{40}}{{360}} \times \pi {\left( {10} \right)^2} - \frac{{40}}{{360}} \times \pi {\left( 6 \right)^2}\]$

                                                              $\[\begin{array}{l} = \frac{{40}}{{360}} \times \frac{{22}}{7} \times \left[ {{{10}^2} - {6^2}} \right]\\ = \frac{{40}}{{360}} \times \frac{{22}}{7} \times \left[ {\left( {10 + 6} \right)\left( {10 - 6} \right)} \right]\end{array}\]$

                                                              $\[ = \frac{{40}}{{360}} \times \frac{{22}}{7} \times 16 \times 4\]$

                                                              $\[\begin{array}{l} = \frac{{56320}}{{2520}}\\ = 22.34c{m^2}\end{array}\]$

Part (iii):

Draw the required figure.

Find the perimeter of the region having pink colour.

Perimeter of the region having

pink colour$\[ = 2\left( {10 - 6} \right)\]$ $+$ length of the arc having radius 6 cm $+$length of the

                                            arc having radius 10 cm.

                 $\[ = 8 + \frac{{30}}{{360}} \times 2 \times \frac{{22}}{7} \times 6 + \frac{{30}}{{360}} \times 2 \times \frac{{22}}{7} \times 10\]$

                 $\[ = 8 + \frac{{7920}}{{2520}} + \frac{{13200}}{{2520}}\]$

                 $\[\begin{array}{l} = 8 + 31.42 + 5.238\\ = 16.38cm\end{array}\]$

Part (iv):

Find the total angle of region having 6 cm radius/ or the shaded region.

$\[{80^ \circ } + {40^ \circ } + {30^ \circ } + {150^ \circ } = {150^ \circ }\]$

Find the area of region having 6 cm radius.

$\[ = \frac{{150}}{{360}} \times \frac{{22}}{7} \times 6 \times 6\]$

$\[ = \frac{{11800}}{{2520}}\]$

$\[ = 47.14c{m^2}\]$

Part (v):

Observe that the given shaded region has an angle of 150 degrees.

Therefore, it represents the minor sector.

Hence, the correct answer is option (a). i.e., minor sector.