# In an isosceles trapezoid, the bases are 50 cm and 14 cm, the diagonal is 40 cm. Find the perimeter of the trapezium.

Let us draw the height of the trapezoid from the top of the obtuse angle C, which divides the base of the blood pressure into two segments, the length of the larger of which is equal to the half-sum of the lengths of the bases, and the smaller, their half-difference.
AH = (AD + BC) / 2 = (50 + 14) / 2 = 64/2 = 32 cm.
DN = (AD – BC) / 2 = (50 – 14) / 2 = 36/2 = 18 cm.
From the right-angled triangle ASN, we determine the length of the leg CH, and from the triangle CHD, the hypotenuse of the SD.
CH2 = AC2 – AH2 = 1600 – 1024 = 576.
CH = 24 cm.
CD2 = CH2 + DH2 = 576 + 324 = 900.
CD = 30 cm.
Since the trapezoid is isosceles, then AB = CD = 30 cm.
Let’s define the perimeter of the trapezoid.
Ravsd = AB + BC + CD + AD = 30 + 14 + 30 + 50 = 124 cm.
Answer: The perimeter of the trapezoid is 124 cm.

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