The bisector ∟A is drawn in the APMC rectangle, which intersects the PM side at the point E, and PE: EM = 2: 3.

univerkov | May 6, 2021 | education

The bisector ∟A is drawn in the APMC rectangle, which intersects the PM side at the point E, and PE: EM = 2: 3. Find PM if the perimeter of the APMC is 56.

Let the length of the segment PE be equal to 2 * X cm, then EM = 3 * X cm.

Since AE is the bisector of the angle PAK, then the triangle APE is isosceles, PA = PE = 2 * X cm.

The length of the segment is PM = PE + EM = 2 * X + 3 * X = 5 * X cm.

Since APMK is a rectangle, then AK = PM = 5 * X, MK = AP = 2 * X cm.

Then the perimeter of the rectangle is: P = 2 * (2 * X + 5 * X) = 14 * X = 56 cm.

X = 56/14 = 4 cm.

Then PM = 5 * 4 = 20 cm.

Answer: The length of the PM side is 20 cm.

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