In the given figure if ∠GHE=∠DFE=90∘, DH = 8 cm, DF = 12 cm, DG = 3x - 1 and DE = 4x + 2, then DG and DE are respectively
DG = 20 cm and DE = 30 cm
In ΔDFE&ΔDHG
∠D=∠D (common)
∠DFE=∠DHG (given 90∘)
So ΔDFE∼ΔDHG (AA similarity)
DEDF=DGDH (corrosponding side are proportional)
⇒ 4x+212=3x−18
⇒ 8(4x + 2) = 12 (3x - 1)
⇒ 2(4x + 2) = 3 (3x - 1)
⇒ 8x + 4 = 9x - 3
⇒ x = 7 cm
So DG = 3x - 1 = 3 (7) - 1 = 20 cm
DE = 4x + 2 = 4(7) + 2 = 30 cm