# In a regular quadrangular pyramid SABCD with base ABCD, a section is drawn through the middle

**In a regular quadrangular pyramid SABCD with base ABCD, a section is drawn through the middle of edges AB and BC and vertex S. Find the cross-sectional area if the lateral edge = 5 and the side of the base = 4**

At the base of a regular quadrangular pyramid lies a square, and the side faces are equal isosceles triangles.

Let us denote the midpoints of the edges AB and BC as M and N, respectively. Then the section SMN, drawn through the midpoints of these edges and the apex of the base, is an isosceles triangle, in which the lateral sides are the equal heights SM and SN of the faces ASB and BSC.

Consider the triangle MBN: it is rectangular, since ABCD is a square, legs BM and BN are equal to half of sides AB and BC. Hence BM = BN = 4/2 = 2. The square of the hypotenuse is equal to the sum of the squares of the legs, we can find the base of the triangle SMN:

MN ^ 2 = BN ^ 2 + BM ^ 2 = 2 ^ 2 + 2 ^ 2 = 4 + 4 = 8;

MN = √8 = 2√2.

Consider a triangle SMB: the edge SB is the hypotenuse, SN and MB are legs, so SM ^ 2 = SB ^ 2-MB ^ 2 = 5 ^ 2-2 ^ 2 = 25-4 = 21; SM = √21.

Draw the height SK in triangle SMN from the vertex S to the base. Then from the triangle SKM:

SK ^ 2 = SM ^ 2-MK ^ 2 = SM ^ 2- (MN / 2) ^ 2 = 21- (2√2 / 2) ^ 2 = 21-2 = 19;

SK = √19.

The cross-sectional area SMN is found as half of the product of the base MN and the height SK.

Ssection = 0.5 * MN * SK = 0.5 * 2√2 * √19 = √2 * √19 = √38≈6.16.