A sphere with a diameter of 10 cm is cut by a plane 3 cm above its center. what is the area of the circle formed? (Leave answer in units of (pi))

Question

Area of a circle with a diameter of cm that is cut by a plane 3 cm above its center.

Answer 1

The radius of the sphere is 10/2=5cm. The way to picture the problem is to represent the plane as a line parallel to the horizontal diameter of the sphere, represented by a circle, cutting through the circle. If you draw a perpendicular radius at right angles to this line from the centre of the circle, and another radius from where the line meets the circumference, you will have a right-angled triangle where the sloping radius, length 5, is the hypotenuse, the vertical side, length 3, and a horizontal segment of the line, length sqrt(5^-3^)=sqrt(16)=4. This line segment forms the radius of the circular cross-section cut by the plane, so the area of the cross-section is (pi)*4^2=16(pi).

Answer 2

This be sekter av serkel...triangel from senter (isosoleez trinagel) & ark abuv

triangel hite=3 & 2 side leng=radius=5

Kut it in half with vertikal line & yu get 2 identikal rite triangels...hite=3 & diagonal=5

Yu get triagel1...angel=106.26 deg

chord leng=8, so radius av  top serkel=4  (3, 4, 5 triangel)

sekter area=23.18238, triangel area=12, segment area=11.18238

Comments

The chord length of 8cm is the diameter of the circular cross-section cut by the plane, so you can find its area (pi)*16, which is what the question is asking for. So you were nearly there! But you over-complicated the answer by finding the area of a segment, and you didn't show the answer in terms of pi!