 Hi and welcome to the session. Let us discuss the following question. Question says figure 12.3 depicts an archery target marked with its five scoring areas from the center outwards as gold, red, blue, black and white. The diameter of the region representing gold score is 21 cm and each of the other bands is 10.5 cm white. Find the area of each of the five scoring regions. This is the given figure 12.3. First of all let us understand that area of circle is equal to pi r square where r is the radius of the circle. Now if we are given two circles and we have to find area enclosed between them then area of the region lying between the two circles is equal to area of outer circle minus area of inner circle. So we can write area of region enclosed between two circles is equal to area of outer circle minus area of inner circle. Area of outer circle whose radius is capital R is pi r square and area of inner circle whose radius is small r is pi r square. So we get area of region enclosed between two circles is equal to pi r square minus pi r square. Now this can be further written as pi multiplied by r square minus r square. Now we will use these formulas as our key idea to solve the given question. Let us now start with the solution. Now we are given the diameter of the region representing the gold score is equal to 21 cm. So we can write diameter of inner circle representing gold score is equal to 21 cm. So the radius of the inner circle is equal to 21 upon 2 cm. So we can write radius equal to r is equal to 21 upon 2 cm. We know radius is equal to half of diameter. From key idea we know area of circle is equal to pi r square. So area of inner circle or we can say area of region representing the gold score is equal to 22 upon 7 multiplied by 21 upon 2 multiplied by 21 upon 2 cm square. We know value of pi is equal to 22 upon 7 and value of r is equal to 21 upon 2. So substituting corresponding values of pi and r in this formula we get this expression. Now simplifying we get 693 upon 2 cm square. Now this is further equal to 346.5 cm square. So required area of region representing gold score is equal to 346.5 cm square. Now we will find area of this region. This region represents the red score. Now from key idea we know area enclosed between the two circles is equal to area of outer circle minus area of inner circle. Now we know radius of the inner circle is equal to 21 upon 2 cm. And we also know that each of the other band is 10.5 cm white. This is given in the question. So radius of the outer circle is equal to 21 upon 2 plus 10.5 cm. Now let us consider inner two circles separately. Now we know radius of inner circle is equal to 21 upon 2 cm. And radius of outer circle is equal to 21 upon 2 plus 10.5 cm. And we also know that area enclosed between two circles is equal to or we can say area enclosed between two concentric circles is equal to pi multiplied by r square minus r square. Now here capital R is the radius of the outer circle and small r is the radius of the inner circle. Now here radius of the outer circle is equal to 21 upon 2 plus 10.5 cm. So we can say capital R is equal to 21 upon 2 plus 10.5 cm. Now simplifying we get capital R is equal to 21 cm. And small r is equal to 21 upon 2 cm. Or we can say it is equal to 10.5 cm. Now area of region representing red score is equal to pi multiplied by square of 21 minus square of 10.5 Substituting corresponding values of capital R and small r in this expression we get this expression. Now we know pi is equal to 22 upon 7. So we get 22 upon 7 multiplied by square of 21 minus square of 10.5. Now we know square of 21 is equal to 441 square of 10.5 is equal to 110.25. Now this expression is further equal to 22 upon 7 multiplied by 330.75. Subtracting these two terms we get 330.75. Now we will cancel common factor 7 from numerator and denominator both. And we get area of region representing red score is equal to 1039.50 cm. Now we will find out area of region representing blue score. Now we will consider these two circles this one and this one. If we subtract area of this circle from area of this circle we get area of region representing blue score. Now we know radius of this circle is equal to 21 cm. And radius of this circle is equal to 21 plus 10.5 cm. Now let us consider these two circles separately. We have to find area enclosed between these two circles. Now clearly we can see capital R that is the radius of the outer circle is equal to 21 plus 10.5 cm. Now this is further equal to 31.5 cm. We know width of each band is equal to 10.5 cm. And we know radius of inner circle is equal to 21 cm. So we will add 10.5 in radius of inner circle to find the radius of the outer circle. Now a radius of inner circle that is small r is equal to 21 cm. Now area of region representing the blue score is equal to pi multiplied by r square minus r square. Now substituting corresponding values of pi capital R and small r in this expression we get 22 upon 7 multiplied by square of 31.5 minus square of 21. Now we can apply formula of a square minus b square in this bracket. We know a square minus b square is equal to a minus b multiplied by a plus b. So we can write it as 22 upon 7 multiplied by 31.5 plus 21 multiplied by 31.5 minus 21. Now this is further equal to 22 upon 7 multiplied by 52.5 multiplied by 10.5 cm square. Now this expression can be further written as 22 upon 7 multiplied by 525 upon 10 multiplied by 105 upon 10 cm square. Now simplifying we get 3465 upon 2 cm square. Now this is further equal to 1732.5 cm square. So we get area of region representing blue score is equal to 1732.5 cm square. Now we will find area of region representing the black score. Now clearly we can see area of region representing the black score is the area enclosed between these two circles. So here we know area of the inner circle that is this circle is equal to 31.5 cm and area of this circle is equal to 31.5 plus 10.5 cm. Now let us consider these two circles separately. Now here clearly we can see value of small r is equal to 31.5 cm that is radius of the inner circle and value of capital R is equal to 31.5 plus 10.5 cm which is further equal to 42 cm. Now we will find area of the region representing the black score. It is equal to pi multiplied by r square minus r square where r is the radius of the outer circle and small r is the radius of the inner circle. Now substituting corresponding value of pi capital R and small r in this expression we get 22 upon 7 multiplied by square of 42 minus square of 31.5 cm square. Now this is equal to 22 upon 7 multiplied by 42 minus 31.5 multiplied by 42 plus 31.5 cm square. Here we have applied the formula of a square minus b square. We know a square minus b square is equal to a minus b multiplied by a plus b. Now simplifying this expression further we get 22 upon 7 multiplied by 10.5 multiplied by 73.5. Now we will cancel common factor 7 from numerator and denominator both. Now this is further equal to 22 multiplied by 110.25 cm square. Now multiplying these two terms we get 2425.50 cm square. So area of region representing the black score is equal to 2425.50 cm square. Now we will find the area of the region representing the white score. Now clearly we can see this is the area enclosed between the outer circle and this circle. Radius of this circle is equal to 42 cm and radius of this circle is equal to 42 plus 10.5 cm. Now let us consider these two circles separately. Now clearly we can see inner radius is equal to 42 cm and outer radius is equal to 42 plus 10.5 cm which is further equal to 52.5 cm. Now we know area of region representing the white score is equal to pi multiplied by r square minus r square. Now substituting corresponding values of pi capital R and small r in this expression we get 22 upon 7 multiplied by square of 52.5 minus square of 42 cm square. Now it can be further written as 22 upon 7 multiplied by 52.5 plus 42 multiplied by 52.5 minus 42 cm square. We know a square minus v square is equal to a plus v multiplied by a minus v. Now this is further equal to 22 upon 7 multiplied by 10.5 multiplied by 94.5 cm square. Now we will cancel common factor 7 from numerator and denominator both. And multiplying these three terms we get 3118.5 cm square. So we get area of region representing white score is equal to 3118.5 cm square. So this is our required answer. This completes the session. Hope you understood the solution. 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