# In a regular quadrangular MAВСD pyramid, the side AB of the base is 6 roots of 2 cm, and the lateral edge of the MA is 12 cm

In a regular quadrangular MAВСD pyramid, the side AB of the base is 6 roots of 2 cm, and the lateral edge of the MA is 12 cm. Find the area of the lateral surface of the pyramid.

The base of a regular pyramid is a square, and its lateral faces are isosceles triangles. In the MCD triangle, we draw the height MH, which will also be the median of an isosceles triangle, then CH = DН = СD / 2 = 6 * √2 / 2 = 3 * √2 cm.
In a right-angled triangle MCH, according to the Pythagorean theorem, we determine the length of the leg MH.
MH ^ 2 = MC ^ 2 – CH ^ 2 = 144 – 18 = 126.
MH = √126 = 3 * √14.
Determine the area of the triangle MСD.
Smsd = СD * MН / 2 = 6 * √2 * 3 * √14 / 2 = 9 * √28 = 18 * √7 cm2.
Then S side = 4 * Smsd = 4 * 18 * √7 = 72 * √7 cm2.
Answer: The lateral surface area is 72 * √7 cm2.

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