In an equilateral triangle ABC with a side equal to 10 cm, points K and M are the midpoints of sides AB and BC

| January 15, 2021 | education

In an equilateral triangle ABC with a side equal to 10 cm, points K and M are the midpoints of sides AB and BC, respectively. a) Prove that AKMS is a trapezoid; b) Find the perimeter of the AKMС.

An equilateral triangle is a triangle in which all sides are equal, such a triangle is also called regular:
AB = BC = AC = 10 cm.
The segment KM, which connects the midpoints of the sides AB and BC, is the middle line of this triangle, which is parallel to the side of AC and is half its length:
KM = AC / 2;
KM = 10/2 = 5 cm.
a) Since a trapezoid is a quadrilateral in which only two sides are parallel, then the AKMD quadrilateral is indeed a trapezoid in which the CM side is parallel to the ABP side.
b) The perimeter of a trapezoid is the sum of the lengths of its sides:
R = AK + KM + MС + AC.
Since the middle line divides the sides in half, then:
AK = AB / 2;
AK = 10/2 = 5 cm;
MС = BC / 2;
MС = 10/2 = 5 cm.
P = 5 + 5 + 5 + 10 = 25 cm.
Answer: the perimeter of the AKMС trapezoid is 25 cm.



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