Answer:
1. The applicable theorem is the midsegment theorem
y = 5 cm
Step-by-step explanation:
The given quadrilateral MATH is an isosceles trapezoid,
The bases of the isosceles trapezoid are
b₁ = [tex]\overline{MA}[/tex] and b₂ = [tex]\overline{HT}[/tex]
The median of the isosceles trapezoid = [tex]\overline{LV}[/tex]
MA = 3·y - 2
HT = 2·y + 4
LV = 8.5 cm
The midsegment theorem states that the median of a trapezoid is parallel to the bases of the trapezoid and is equal to half the length of the sum of the bases of the trapezoid
The median of a trapezoid = (1/2)×(b₁ + b₂)
Therefore, we have;
The median of the isosceles trapezoid, LV, given as follows;
LV = 8.5 = (1/2)×((3·y - 2) + (2·y + 4))
LV = 8.5 = (1/2)×(5·y + 2) =
2 × 8.5 = (5·y + 2)
17 = 5·y + 2
∴ 5·y = 17 - 2 = 15
y = 15/3 = 5
y = 5 cm
