A circle is inside a square. The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are decreasing at a rate of 5 meters per hour. When the radius is 4 meters, and the sides are 23 meters, then how fast is the AREA outside the circle but inside the square changing? The rate of change of the area enclosed between the circle and the square is square meters per hour.

Step-by-step explanation:

Let r = radius of circle

     s = length of a side of the square

dr/dt = -3m/hour

  ds/dt = 5m/hour

we know that

area of square= s^2

area of circle=πr^2

Let A be the area of outside

A = (area of square) - (area of circle)

          = s^2 - πr^2

dA/dt=?

when r = 4 and s = 23

 substituting we have

dA/dt = 2s(ds/dt) - 2πr(dr/dt)

             =2(23)(5) - 2π(4)(-3)

            = (230 + 24π) m^2/hour

            = (230 + 75.408) m^2/hour

            = 305.408m^2/hour

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.