The side edge of a regular quadrangular pyramid is 18, and the side of the base is 8.

| May 31, 2021 | education

The side edge of a regular quadrangular pyramid is 18, and the side of the base is 8. Find the area of the section drawn through the side of the base and the middle of opposite sides.

The section of the pyramid is the isosceles trapezoid СDКН, since DК = СН as sides of equal triangles АDК and ВСН. Points K and H are the middle of the edges AM and BM, then HK is the middle line of the triangle ABM, then НK = AB / 2 = 8/2 = 4 cm.

Let’s draw the height ME of the triangle ADM, then AE = AD / 2 = 8/2 = 4 cm.

CosMAE = AE / MA = 4/1 = 2/9.

In triangle АDК, by the theorem of xines we define the length of the side DК.

DK ^ 2 = AD ^ 2 + AK ^ 2 – 2 * AD * AK * CosMAE = 64 + 81 – 2 * 8 * 9 * 2/9 = 145 – 32 = 113.

DК = √113 cm.

In the trapezium СDКН we will draw the height of the KR. Segment DP = (CD – НK) / 2 = (8 – 4) / 2 = 2 cm.

Then in a right-angled triangle KPD, KP ^ 2 = DK ^ 2 – DP ^ 2 = 113 – 4 = 109.

КР = √109 cm.

Let’s define the cross-sectional area. Ssection = (CD + KН) * KР / 2 = 12 * √109 / 2 = 6 * √109 cm2.

Answer: The cross-sectional area is 6 * √109 cm2



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.