Isosceles trapezoid ABCD is inscribed in ⊙O with radius 5. AD=6 and the median of ABCD has length 7. Find the distance from AD to BC. this was the only info given!

The distance from AD to BC is 7

Step-by-step explanation:

The information given are;

The type of inscribed quadrilateral ABCD = Isosceles trapezoid

The radius of the circle = 5

Segment AD of ABCD = 6

The median of the trapezoid ABCD = 7

Given the trapezoid theorem, the median is equal to half the length of the two bases added together, we have;

(AD + BC)/2 = 7

Which gives;

(6 + BC)/2 = 7

BC = 7×2 - 6 = 8

Therefore the distance from AD to BC is given by the distance from BC to the median line added to the distance from AD to the median line given as follows;

The distance from BC to the median = √(Radius² - (BC/2)²) = √(5² - (8/2)²) = 3

The distance from BC to the median = 3

The distance from AD to the median = √(Radius² - (AD/2)²) = √(5² - (6/2)²) = 4

Which gives;

The distance from AD to BC = 3 + 4 = 7

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)