Answer:
JK = 83 , m∠A = 70° , m∠ALM = 110°
Explanation:
* Lets explain how to solve the problem
∵ ABCD is a trapezoid
∴ DC // AB
∴ m∠D + m∠A = 180° ⇒ interior supplementary angles
∵ m∠D = 110°
∴ 110° + m∠A = 180° ⇒ subtract 110° from both sides
∴ m∠A = 70°
∵ L is the midpoint of AD, and M is the midpoint of BC
∴ LM is the median of trapezoid ABCD
∴ LM // AB and DC
∴ m∠D = m∠ALM ⇒ corresponding angles
∵ m∠D = 110°
∴ m∠ ALM = 110°
- The length of the median is half the sum of the lengths of the two
parallel bases
∴ LM = 1/2 (AB + DC)
∵ AB = 96 units and DC = 44 units
∴ LM = 1/2 (96 + 44) = 1/2 (140) = 70 units
- In the quadrilateral ABML
∵ AB // LM
∵ AL ≠ BM
∴ ABML is a trapezoid
∵ JK is its median
∴ JK = 1/2 (AB + LM)
∵ AB = 96 units ⇒ given
∵ LM = 70 units ⇒ proved
∴ JK = 1/2 (96 + 70) = 1/2 (166) = 83
∴ JK = 83 units