In the picture below solve for question no. 8
Answers
Answer:
DE = 38/2 = 19.
Step-by-step explanation:
Draw a line BG onto CE. BG || AD. ( see the attached diagram)
In ∆CBG and ∆CAD,
∠CAD = ∠CBG, ∠CDA = ∠CGB and ∠ACD = ∠BCG.
Hence from A-A-A analogy, we can say ∆CBG similar to ∆CAD.
So CB/CA = CG/CD
We know that CB = CA/2, so CG / CD = ½
Hence we can say, CG = GD. (or G is mid point of CD).
CG = GD ---------------------------------------E1.
In ∆FED and ∆BEG,
∠EBG = ∠EFD, ∠BGE = ∠FDE and ∠BEG = ∠FED
From A-A-A analogy, we can say that ∆ FED is similar to ∆BEG
So EF/EB = ED/GD
Since EF/EB = ½, ED/DG = ½
D is mid point of GE.
GD = DE ----------------------------------E2.
CD = CG + GD = 2DE.
CD = 2DE.
DE = DE/2.
It is given that CD = 38.
DE = 38/2 = 19.
Answer:
19 cm , Option D
Step-by-step explanation:
In the picture below solve for question no. 8
in Δ DEF
DE/ Sin∠DFE = FE/Sin∠FDE
=> DE Sin∠FDE = FE Sin∠DFE - eq 1
in Δ ABF
AB/Sin∠AFB = BF/Sin∠A
=> AB Sin∠A = BF Sin∠AFB
∠AFB = ∠DFE & BF = FE
=> AB Sin∠A = FE Sin∠DFE - eq2
Equating both equation eq1 & eq2
DE Sin∠FDE = AB Sin∠A - eq3
in ΔACD
AC/Sin∠ADC = CD/Sin∠A
AC = AB + BC = AB + AB = 2AB
∠ADC = 180° - ∠FDE
Sinθ = Sin(180° - θ)
Sin∠ADC = Sin(180° - ∠FDE) = Sin∠FDE
2AB/Sin∠FDE = CD/Sin∠A
=> ABSin∠A = CDSin∠FDE/2 - eq 4
Equating eq 3 & Eq 4
DE Sin∠FDE = CDSin∠FDE/2
DE = CD/2
DE = 38/2
DE = 19 cm
option D is correct answer