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The radius of the small round field is increased by 5 m to get the large round field, and the area of the field is increased by 3 times
The radius of the small site is 5m
Specific process: set the radius of small site as R
Then: small site area π R & sup2;
The field area is π (R + 5) & sup2;
For the quadratic equation of one variable, find the value of R 4 π R & sup2; = π (R + 5) & sup2;
4r²=r²+10r+25
3r²-10r-25=0
(3r+5)(r-5)=0
Then r = 5 or r = - 5 / 3, because it is radius, it cannot be negative. Then r = 5, that is 5m
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