Sphere P has a radius of 5 inches. Find the area of the cross section formed by the intersection of a plane 3 cm above the center of sphere P and parallel to its equator.
O None of the other answers are correct
O 16pi in
O4pi in
O 7pi in
O 9pi in

pi*16 in^2

Step-by-step explanation:

In the image, we can see a triangle rectangle.

Where one cathetus is equal to 3 inches, the hypotenuse is equal to the radius of the sphere, 5 inches, and the other cathetus is the radius of the cross-section of which we can find the area.

Using the Pythagorean theorem, we will have that:

x^2 + (3in)^2 = (5in)^2

x = √( (5in)^2 - (3in)^) = 4 in

Then the radius of the cross-section is 4 inches.

And we know that the area of a circle of radius R is:

A = pi*R^2

Then the area of the cross-section will be:

A = pi*(4in)^2 = pi*16 in^2

Then the correct option is the first one.

the answer is d on edge

step-by-step explanation:

$75.00

step-by-step explanation:

i = prt

i = 1,5000 * 0.025 * 2

i = 75