10th Class Mathematics Areas Related to Circles Question Bank MCQs - Areas Related to Circles

  • question_answer
     
    In this section, each question has two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options [a], [b], [c] and [d] out of which one is correct.
    Figure depicts an archery target marked with its five scoring areas from the centres outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 14 cm and each of the other bands is 7 cm wide. Then match the following.
    List - I List - II
    P. Area of Gold score 1. \[462\,c{{m}^{2}}\]
    Q. Area of Red score 2. \[770\,c{{m}^{2}}\]
    R. Area of Blue score 3. \[154\,c{{m}^{2}}\]
    S. Area of Black score 4. \[1078\,c{{m}^{2}}\]

    A) Codes \[P\to 2,\,Q\to 1,\,R\to 4,\,S\to 3\]

    B) \[P\to 3,\,Q\to 1,\,R\to 2,\,S\to 4\]

    C) \[P\to 1,\,Q\to 4,\,R\to 2,\,S\to 3\]

    D) \[P\to 1,\,Q\to 3,\,R\to 4,\,S\to 2\]

    Correct Answer: B

    Solution :

    We have,
    r = Radius of the region representing gold score = 7 cm
    \[{{r}_{1}}\]= Radius of the region representing gold and red scoring areas
    =(7+7) cm =14 cm
    \[{{r}_{2}}\]= Radius of the region representing gold, red and blue scoring areas
    =(7+7+7)=21cm
    \[{{r}_{3}}\] = Radius of the region representing gold, red, blue and black scoring areas
    =(7+7+7+7) =28 cm
    (P) Now, \[{{A}_{1}}\]= Area of the region representing gold scoring area
    \[=\pi {{r}^{2}}=\frac{22}{7}\times 7\times 7=154\,c{{m}^{2}}\]
    (Q) \[{{A}_{2}}\]= Area of the region representing red scoring area
    \[=\pi r_{1}^{2}-\pi {{r}^{2}}=\frac{22}{7}\left( {{14}^{2}}-{{7}^{2}} \right)\]
    \[=\frac{22}{7}\left( 14+7 \right)\left( 14-7 \right)=462c{{m}^{2}}\]
    (R) \[{{A}_{3}}\] = Area of the region representing blue scoring area
    \[=\pi r_{2}^{2}-\pi r_{1}^{2}=\frac{22}{7}\left( {{21}^{2}}-{{14}^{2}} \right)\]
    \[=\frac{22}{7}\left( 21+4 \right)\left( 21-14 \right)=770c{{m}^{2}}\]
    (S) \[{{A}_{4}}\]= Area of the region representing black scoring area
    \[=\pi r_{3}^{2}-\pi r_{2}^{2}=\frac{22}{7}\left( {{28}^{2}}-{{21}^{2}} \right)\]
    \[=\frac{22}{7}\left( 28+21 \right)\left( 28-21 \right)=1078\,c{{m}^{2}}\]


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