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12.2 question 3

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12.2 question 3
  • 8 answers

Sahil Tewatia 4 years, 4 months ago

Thx

Yashi .... 4 years, 5 months ago

Mera bhi dekh lena..bht neend aayi h

Abhinav ? 4 years, 5 months ago

Sorry ..neend me dekha ..?

Yashi .... 4 years, 5 months ago

Here, r= 14cm.. area swept= thetha/360°×πr^2....30/360×22/7×14×14....30/360×44×14....154/3cm^2

Yashi .... 4 years, 5 months ago

Answer will be 154/3 cm^2.

Yashi .... 4 years, 5 months ago

Abhinav.. its from exercise 12.3 not 12.2

Abhinav ? 4 years, 5 months ago

◆area of square ABCD - (area of semi circle APD + area of semicircle BPC) ◆14×14 - [1/2 × 22/7(14/2)^2 + 22/7(14/2)^2】 ..solve this then u can get answer..

Abhinav ? 4 years, 5 months ago

42cm^2
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