In the cube shown below, the distance between vertices 2 and 6 is 17 cm.

If the cube is divided into two equal parts by a plane parallel to the face defined by vertices 2, 3, 6, and 7, what will be the area of the cross-section?

Question

Esteban Fry

Mathematics

21 August, 17:55

In the cube shown below, the distance between vertices 2 and 6 is 17 cm.

If the cube is divided into two equal parts by a plane parallel to the face defined by vertices 2, 3, 6, and 7, what will be the area of the cross-section?

+2

Answers (1)

Know the Answer?

Chandler Gillespie · Answer 1

256 sq cm

Step-by-step explanation:

cube is a three-dimensional figure with six congruent square faces. Since each face is congruent, each of the edges have the same length.

The cross-section described is shown below.

The cross-section is also a square which is congruent to the other faces of the cubes. Therefore, the edges of the cross-section will each measure 16 cm. Find the area by using the formula for the area.

A=lw

= (16cm) (16cm)

=256 sq cm

So, the area of the cross-section is 256 sq cm.

In the cube shown below, the distance between vertices 2 and 6 is 17 cm.If the cube is divided into two equal parts by a plane parallel to the face defined by vertices 2, 3, 6, and 7, what will be the area of the cross-section?

In the cube shown below, the distance between vertices 2 and 6 is 17 cm.

If the cube is divided into two equal parts by a plane parallel to the face defined by vertices 2, 3, 6, and 7, what will be the area of the cross-section?