The sides of the rectangular trapezoid are 15 cm and 17 cm the middle line is 6 cm find the base of the trapezoid

Given: rectangular trapezoid LEDK, EL = 15 cm, DK = 17 cm, MN = 6 cm.
Find: LK, ED.
Draw the height DQ to the base of the LK.
DQ = EL = 15 cm, since the side that connects the right angles of the trapezoid is also its height (property of a rectangular trapezoid).
When the height DQ crosses the midline MN at the point U, the result is a rectangular △ DUN, ∠U = 90 °.
DN = 1/2 * DK = 1/2 * 17 = 8.5 (cm).
DU = 1/2 * DQ = 1/2 * 15 = 7.5 (cm).
Behind the Pythagorean theorem:
DN ^ 2 = DU ^ 2 + UN ^ 2.
Hence:
UN ^ 2 = DN ^ 2 – DU ^ 2 = 8.5 ^ 2 – 7.5 ^ 2 = 72.25 – 56.25 = 16 (cm).
UN = Sqrt16 = 4 (cm).
If MN = 6 cm, and UN = 4 cm, then:
MU = 6 – 4 = 2 (cm).
So the upper base of the trapezoid is equal to:
MU = ED = LQ = 2 (cm).
△ DQK – rectangular (∠Q = 90 °). Behind the Pythagorean theorem:
DK ^ 2 = DQ ^ 2 + QK ^ 2.
Hence:
QK ^ 2 = DK ^ 2 – DQ ^ 2 = 17 ^ 2 – 15 ^ 2 = 289 – 225 = 64 (cm).
QK = Sqrt64 = 8 (cm).
Now we find the bottom base of the trapezoid:
LK = LQ + QK = 2 + 8 = 10 (cm).
Answer: LK = 10 cm, ED = 2 cm.




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