Find area of square abcd
area is 160 sq cmABCD is a square
Given <DEF = <EFD = 90
Given DE = 6, EF = 8, BF = 10
DB intersects EF at 7
DF = sqrt( DE^2 + EF^2) = sqrt(8^2 +6^2) = 10
DF = BF = 10
DEG and BFG are similar by AA
<DEG = <BFG = 90
<DGE =<FGD # vertically opposite
Thus DE/FB = EG/FG
EG/FG =6/10 = 3/5
EG = (3/5)FG
EG+FG = 8
Thus (8/5)FG = 8 , FG = 5 and EG = 3
In right angle triangle DEG EG = 3 DE=6 thus DG =
Sqrt(6^2+3^2) = sqrt(45) = 3sqrt(5)
In right angle triangle BFG
BG = sqrt(BG^2 + FG^2) = sqrt(10^2 + 5^2)
= sqrt(125) = 5sqrt(5)
Thus BD = BG + DG = 8sqrt(5)
Diagonal of ABCD is 8sqrt(5)
Thus area = (1/2)diagonal^2 = (1/2)( 8sqrt(5)) ^2 =160 sq cm
Given <DEF = <EFD = 90
Given DE = 6, EF = 8, BF = 10
DB intersects EF at 7
DF = sqrt( DE^2 + EF^2) = sqrt(8^2 +6^2) = 10
DF = BF = 10
DEG and BFG are similar by AA
<DEG = <BFG = 90
<DGE =<FGD # vertically opposite
Thus DE/FB = EG/FG
EG/FG =6/10 = 3/5
EG = (3/5)FG
EG+FG = 8
Thus (8/5)FG = 8 , FG = 5 and EG = 3
In right angle triangle DEG EG = 3 DE=6 thus DG =
Sqrt(6^2+3^2) = sqrt(45) = 3sqrt(5)
In right angle triangle BFG
BG = sqrt(BG^2 + FG^2) = sqrt(10^2 + 5^2)
= sqrt(125) = 5sqrt(5)
Thus BD = BG + DG = 8sqrt(5)
Diagonal of ABCD is 8sqrt(5)
Thus area = (1/2)diagonal^2 = (1/2)( 8sqrt(5)) ^2 =160 sq cm