Answer
Verified
393.9k+ views
Hint:Here, all the five regions are in a circle shape. The area of the gold region is calculated by the area of a circle, which is given by $A = \pi {r^2}$, where r is the radius of the circle. While the formula of area of a ring is used to find out the areas of other four regions i.e., red, blue, black and white, which is given by $\pi \left( {{R^2} - {r^2}} \right)$, where $R$ is the radius of outer ring and $r$ is the radius of inner ring.Using this formula we can calculate the area of each of the five scoring regions.
Complete step-by-step answer:
Given, Diameter of gold region= $21$$cm$
Therefore, Radius of gold region =$\dfrac{{21}}{2}$$cm$ $ = 10.5$$cm$
Area of gold region= $\pi {r^2}$
$ = \pi {\left( {10.5} \right)^2} = \dfrac{{22}}{7} \times 110.25 = 346.5$$c{m^2}$
Hence, Area of gold region= $346.5c{m^2}$
Now, the radius of the red region is calculated by adding the width of the band in the radius of the previous region.
Radius of red region= Radius of gold region+ width of band
$ = 10.5 + 10.5 = 21cm$
Also, the area of red region is calculated by the formula of area of a ring, which is given by $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring .
The radius of a particular region is the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.
Area of red region= \[\pi \left( {{{21}^2} - {{10.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {441 - 110.25} \right) = \dfrac{{22}}{7} \times 330.75 = 1039.5c{m^2}$
Hence, Area of red region=$1039.5c{m^2}$
Similarly as the red region, we can calculate the radius and area of the remaining three regions, i.e., blue, black and white.
Radius of blue region= Radius of red region+ width of band
$ = 21 + 10.5 = 31.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of blue region= \[\pi \left( {{{31.5}^2} - {{21}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {992.25 - 441} \right) = \dfrac{{22}}{7} \times 551.25 = 1732.5c{m^2}$
Hence, Area of blue region=$1732.5c{m^2}$
Radius of black region= Radius of blue region+ width of band
= $31.5 + 10.5 = 42cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of black region= \[\pi \left( {{{42}^2} - {{31.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {1764 - 992.25} \right) = \dfrac{{22}}{7} \times 771.75 = 2425.5c{m^2}$
Hence, Area of black region=$2425.5c{m^2}$
Radius of white region= Radius of black region+ width of band
$ = 42 + 10.5 = 52.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of white region= \[\pi \left( {{{52.5}^2} - {{42}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {2756.25 - 1764} \right) = \dfrac{{22}}{7} \times 992.25 = 3118.5c{m^2}$
Hence, Area of white region=$3118.5c{m^2}$
Note:While calculating the radius of red, blue, black or white region; it is necessary to add the width of the band in the radius of the previous region. This radius will act as the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.
Complete step-by-step answer:
Given, Diameter of gold region= $21$$cm$
Therefore, Radius of gold region =$\dfrac{{21}}{2}$$cm$ $ = 10.5$$cm$
Area of gold region= $\pi {r^2}$
$ = \pi {\left( {10.5} \right)^2} = \dfrac{{22}}{7} \times 110.25 = 346.5$$c{m^2}$
Hence, Area of gold region= $346.5c{m^2}$
Now, the radius of the red region is calculated by adding the width of the band in the radius of the previous region.
Radius of red region= Radius of gold region+ width of band
$ = 10.5 + 10.5 = 21cm$
Also, the area of red region is calculated by the formula of area of a ring, which is given by $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring .
The radius of a particular region is the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.
Area of red region= \[\pi \left( {{{21}^2} - {{10.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {441 - 110.25} \right) = \dfrac{{22}}{7} \times 330.75 = 1039.5c{m^2}$
Hence, Area of red region=$1039.5c{m^2}$
Similarly as the red region, we can calculate the radius and area of the remaining three regions, i.e., blue, black and white.
Radius of blue region= Radius of red region+ width of band
$ = 21 + 10.5 = 31.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of blue region= \[\pi \left( {{{31.5}^2} - {{21}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {992.25 - 441} \right) = \dfrac{{22}}{7} \times 551.25 = 1732.5c{m^2}$
Hence, Area of blue region=$1732.5c{m^2}$
Radius of black region= Radius of blue region+ width of band
= $31.5 + 10.5 = 42cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of black region= \[\pi \left( {{{42}^2} - {{31.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {1764 - 992.25} \right) = \dfrac{{22}}{7} \times 771.75 = 2425.5c{m^2}$
Hence, Area of black region=$2425.5c{m^2}$
Radius of white region= Radius of black region+ width of band
$ = 42 + 10.5 = 52.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of white region= \[\pi \left( {{{52.5}^2} - {{42}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {2756.25 - 1764} \right) = \dfrac{{22}}{7} \times 992.25 = 3118.5c{m^2}$
Hence, Area of white region=$3118.5c{m^2}$
Note:While calculating the radius of red, blue, black or white region; it is necessary to add the width of the band in the radius of the previous region. This radius will act as the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE