Through the vertex B of the obtuse angle of the parallelogram ABCD, a perpendicular MB

| May 16, 2021 | education

Through the vertex B of the obtuse angle of the parallelogram ABCD, a perpendicular MB equal to 12 cm is drawn to its plane. The area of the parallelogram is 144 cm2. AB = 12 cm, BC = 18 cm. Find the distances from point M to lines AD and CD.

Let us draw from the top B of the parallelogram of the heights of VK and BH to the sides AD and CD.

Since the lengths of the opposite sides of the parallelogram are equal, then AD = BC = 18 cm, CD = AB = 12 cm.

Let’s apply the formula for the area of ​​a parallelogram.

S = AD * BK and S = CD * BH.

S = 18 * BK = 144.

BK = 144/18 = 8 cm.

From the right-angled triangle MBK, by the Pythagorean theorem, we determine the length of the hypotenuse of MK.

MK ^ 2 = BK ^ 2 + MB ^ 2 = 8 ^ 2 + 12 ^ 2 = 64 + 144 = 208.

MK = 4 * √13 cm.

S = CD * BH.

S = 12 * BH = 144.

BK = 144/12 = 12 cm.

From the right-angled triangle МBН, according to the Pythagorean theorem, we determine the length of the hypotenuse МН.

MH ^ 2 = BH ^ 2 + MB ^ 2 = 12 ^ 2 + 12 ^ 2 = 144 + 144 = 228.

MH = 2 * √12 cm.

Answer: The distance from point M to line AD is 4 * √13 cm, to line CD is 2 * √12 cm.



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