The bases of an isosceles trapezoid are 12 cm and 16 cm, and its diagonals are mutually perpendicular.
The bases of an isosceles trapezoid are 12 cm and 16 cm, and its diagonals are mutually perpendicular. Find the area of the trapezoid.
Consider a right-angled triangle AOD, which, by the property of the diagonals of an isosceles trapezoid, AO = OD, that is, an isosceles trapezoid.
Then, by the Pythagorean theorem, AD ^ 2 = AO ^ 2 + DO ^ 2 = 2 * AO ^ 2.
AO ^ 2 = AD ^ 2/2 = 256/2 = 128.
Then from the triangle AOH, OH ^ 2 = AO ^ 2 – AH ^ 2 = 128 – 82 = 64.
OH = 8 cm.
Likewise in the BOC triangle.
ВO ^ 2 = BC ^ 2/2 = 144 / = 72.
KO ^ 2 = ВO ^ 2 = ВK ^ 2 = 72 – 62 = 72 – 36 = 36.
KO = 6 cm.
Then the height KH = OH + KO = 8 + 6 = 14 cm.
Let’s define the middle line of the trapezoid.
PE = (AD + BC) / 2 = (16 + 12) / 2 = 28/2 = 14 cm.
Then the area of the trapezoid is: S = PE * KH = 14 * 14 = 196 cm2.
Answer: The area of the trapezoid is 196 cm2.