From point M, a perpendicular MD is drawn, equal to 6 cm, to the plane of the square ABCD. The inclined MB

From point M, a perpendicular MD is drawn, equal to 6 cm, to the plane of the square ABCD. The inclined MB makes an angle of 60º with the plane of the square. a) Prove that triangles MAB and MCB are rectangular; b) Find the side of the square; c) Find the area of the triangle ABD.

Proof:

MD ⟂ (ABC); AD⟂ AB hence AM⟂ AB by TPP

Hence, ∠MAB = 90 ° hence the triangle MAB is right-angled

MD⟂ (ABC); BC ⟂ DC hence MC⟂ BC by TTP

Hence, ∠ MCB = 90 ° hence the triangle MBC is right-angled

Solution :

In a triangle MBD tg∠60 ° = MD / BD, hence BD = 6 / √3 = 2√3

BD = AD√2 hence AD = BD / √2 = 2√6 / 2 = √6

Area of a triangle ABD = √6 * √6 / 2 = 3

ANSWER: 3