Atoy has various shaped objects that a child is supposed to push through matching holes. the area of the square hole is 5 cm2. the area of the circular face of the round peg is 5 cm2. will the round peg fit through the square hole? use r = 3.14. round to the nearest hundredth as needed. explain

To answer this, let's first describe the two areas and obtain the pertinent dimensions from them.

The area of the square hole is 5 cm^2.  Since A = s^2, where s is the length of a side of the square, s = +√5 in this situation.  +√5 is approx. 2.24 cm.

The area of the round peg is 5 cm^2 also, but the area is calculated using a different formula:  A = πr^2, where r is the radius of the circle.  Solving for r^2, we get:

                 r^2 = A/π.   Here, r^2 = (5 cm^2)/π = 5π, so that:

r = +√(5π).  This is approx. 3.96 cm, and so the diameter is twice that, or 7.93 cm.

So there's plenty of room for the round peg to enter the square hole, but not the other way around!

25/6

step-by-step explanation:

you multiply 4 by the denominator (4*6=24) and add it to the numerator which makes it 25/6.