If the radius of the circle is increased by 5 moles1, its area will be increased by 88 square decimeters. Request the area of the original circle

If the radius of the circle is increased by 5 moles1, its area will be increased by 88 square decimeters. Request the area of the original circle

The equation can be solved as: π (R + 0.2R) 2 = π R2 + 88
So π R2 = 200 square decimeters
So the original area is 200 square decimeters

If the radius of the circle is increased by 15, its area will be increased by 88 DM2

Let the radius of the original circle be x, then π (x + 15) 2 = π x2 + 88, and the solution is x = 200 π. Then π x2 = 200

If the radius of the circle is increased by 15, its area will be increased by 88 DM2

Let the radius of the original circle be x, then π (x + 15) 2 = π x2 + 88, and the solution is x = 200 π. Then π x2 = 200

If the radius of the circle is increased by one fifth, its area will be increased by 88 square decimeters, and the area of the original circle will be requested
It's due tomorrow
Write popular point, little brother mathematics is not very --

Make a table
Area radius
The original 1 times 1 times π '1'
Now six fifths of the square times π 1 times (one fifth of the original 1 + increase)
Six fifths of the square is thirty-six out of twenty-five - unit one = eleven out of twenty-five
That's 88. Divide 88 by 11 / 25 and that's the original area
Because eleven out of twenty-five is eleven out of twenty-five of unit one, which is the original area. If you don't understand, I can only suggest you do more