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anonymous

  • one year ago

A triangle is placed in a semicircle with a radius of 5 ft, as shown below. Find the area of the shaded region. Use the value 3.14 for pi and do not round your answer.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    area of triangle\[=\frac{ 1 }{ 2 }\times base \times altitude=?\] area of semi circle\[=\frac{ \pi \times r^2 }{ 2 }=?\] then subtract

  3. anonymous
    • one year ago
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    Do I find the answer for both

  4. anonymous
    • one year ago
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    Do you know that area of circle is\[Area = πr^2\] and the area of triangle is \[Area of \triangle = \frac{ 1 }{ 2 } * base * height\] Then now what you have the two areas with your hands , to get the shaded region you just simply subtract the smaller shape from the larger one , which in this question Larger : the semicircle and the smaller one is the triangle ! Now to calculate just subs. with given values \[Area of semicircle = \frac{ 1 }{ 2 } * π * 5^2 = 12.5 \] And \[Area of \triangle = \frac{ 1 }{ 2 } * 10 * 5 = 25\] Now the shaded region is :\[Area of shaded region : 12.5π - 25\] Let me know if you got it :)

  5. anonymous
    • one year ago
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    I completely understand :) so I subtract 12.5 pi - 25 and the answer is 14.26

  6. anonymous
    • one year ago
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    @Omar_Elboredy

  7. anonymous
    • one year ago
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    area of triangle\[=\frac{ 1 }{2 }\times10\times 5=25\] area of semi circle\[=\frac{ \pi r^2 }{ 2 }=\frac{ 3.14\times25 }{ 2 }=\frac{ 78.50 }{ 2 }=39.25\] reqd. area=39.25-25=?

  8. anonymous
    • one year ago
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    Exactly ! :D

  9. anonymous
    • one year ago
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    I have no idea who is right lol...

  10. anonymous
    • one year ago
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    both are correct

  11. anonymous
    • one year ago
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    @surjithayer would the answer be ft, ft^2 or ft^3?

  12. anonymous
    • one year ago
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    area is always in square units

  13. anonymous
    • one year ago
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    Thank you the both of ya'll!

  14. anonymous
    • one year ago
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    if I could give you both metals I would!

  15. anonymous
    • one year ago
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    Welcomes ! :) Feel free to ask anytime :D

  16. anonymous
    • one year ago
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    Will you help me with one more? @Omar_Elboredy

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