Given that, \overline{AD}=45 and \overline{DG}=60, find the lengths of other line segments in this right triangle \triangle ABC given that the DEFG is a square?

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2 Answers
Feb 5, 2018

DE = EF = FG = 60

AE = 75, EC = 36, BG = 80, BF = 100, CF = 48

Explanation:

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DE = EF = FG = 60 as DEFG is a square

tan A = 60 / 45 = (4/3)^c

hat A = tan ^(-1) (4/3) = 0.9273

hat B = pi - pi/2 - 0.9273 = 0.6435

BF = sqrt((BG)^2 + (GF)^2) = sqrt(80^2 + 60*3) = 100

hat A = hat E = 0.9273 (corresponding angles)

In triangle CEF,

CE = EF cos hat E = 60 cos 0.9273 = 36

CF = EF sin hatE = 60 sin 0.9273 = 48

In triangle BFG,

BG = (FG) / tan hatB = 60 / tan (0.6435) = 80#

In triangle AEF,

AE = sqrt((AD)^2 + (DE)^2) = sqrt(45^2 + 60^2) = 75

Verification :

In right triangle ABC,

AB = sqrt((AC)^2 + (BC)^2)

AB = = 45 + 60 + 80 = color(green)(185

sqrt((AC)^2 + (BC)^2) = sqrt((75+36)^2 + (100 + 48)^2) = color(green)(185

Hence Proved.

Feb 5, 2018

Please see below.

Explanation:

.

Because DEFG is a square, it has all equal sides:

DE=EF=FG=DG=60

Because EF and |AB are parallel,

/_CFE=/_CBA and /_CEF=/_CAB

This means that by virtue of Angle Angle theorem:

DeltaADE~=DeltaECF

DeltaFGB~=DeltaECF

DeltaACB~=DeltaCEF

Therefore, all four right angle triangles are similar which mean the ratios of their corresponding sides are equal.

(AD)/(FG)=(ED)/(GB)

45/60=60/(GB)

GB=3600/45=80

AB=AD+DG+GB=45+60+80=185

(AE)^2=(AD)^2+(DE)^2

(AE)^2=(45)^2+(60)^2=2025+3600=5625

AE=75

(DE)/(CF)=(AE)/(EF)

60/(CF)=75/60

CF=3600/75=48

(AD)/(CE)=(AE)/(EF)

45/(CE)=75/60

CE=2700/75=36

AC=AE+EC=75+36=111

(ED)/(GB)=(AE)/(FB)

60/80=75/(FB)

FB=6000/60=100

CB=CF+FB=48+100=148