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exercise 12.1 Q no. 3 anyone …

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exercise 12.1 Q no. 3 anyone please i need only hint

    • Posted by Amanpreet Singh 7 years, 9 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Amanpreet Singh 7 years, 9 months ago

    thanks a lot

    0Thank You

    Keshav Bindal 7 years, 9 months ago

    Area of 1st region gold = πr^2 = 22/7 × 10.5 × 10.5 & Area of 2nd region red = π(R^2 - r^2) = 22/7 ×(21^2 - 10.5^2) & Area of 3rd region blue = π(R^2 - r^2) = 22/7 ×(31.5^2 - 21^2)

    1Thank You

    Jasmeen Kaur 7 years, 9 months ago

    Radius of gold =10.5 Area of gold = 22/7 × (10.5)^2 Radius of red =10.5+10.5 Area of red =22/7×(21)^2 - area of gold Similarly add 10.5 in radius for next and for its area 22/7×(21+10.5)^2 - 22/7×(21)^2

    3Thank You

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