any quadrilateral triangle ABC D is a point on side BC such that BD equal to 1/3 BC prove that 980 square is equal to disable a b square
Answers
Answered by
4
In a ΔABC and ΔACE
AB = AC ( Given)
AE = AE ( common)
∠AEB = ∠AEC = 90°
∴ ΔABC ≅ ΔACE ( For RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
In a right angled triangle ADE
AD2 = AE2 + DE2 ---------(1)
In a right angled triangle ABE
AB2 = AE2 + BE2 ---------(2)
From equ (1) and (2) we obtain
⇒ AD2 - AB2 = DE2 - BE2 .
⇒ AD2 - AB2 = (BE – BD)2 - BE2 .
⇒ AD2 - AB2 = (BC / 2 – BC/3)2 – (BC/2)2
⇒ AD2 - AB2 = ((3BC – 2BC)/6)2 – (BC/2)2
⇒ AD2 - AB2 = BC2 / 36 – BC2 / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)
⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4
⇒ AD2 = (36AB2 + AB2– 9AB2) / 36
⇒ AD2 = (28AB2) / 36
⇒ AD2 = (7AB2) / 9
9AD2 = 7AB2 .
AB = AC ( Given)
AE = AE ( common)
∠AEB = ∠AEC = 90°
∴ ΔABC ≅ ΔACE ( For RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
In a right angled triangle ADE
AD2 = AE2 + DE2 ---------(1)
In a right angled triangle ABE
AB2 = AE2 + BE2 ---------(2)
From equ (1) and (2) we obtain
⇒ AD2 - AB2 = DE2 - BE2 .
⇒ AD2 - AB2 = (BE – BD)2 - BE2 .
⇒ AD2 - AB2 = (BC / 2 – BC/3)2 – (BC/2)2
⇒ AD2 - AB2 = ((3BC – 2BC)/6)2 – (BC/2)2
⇒ AD2 - AB2 = BC2 / 36 – BC2 / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)
⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4
⇒ AD2 = (36AB2 + AB2– 9AB2) / 36
⇒ AD2 = (28AB2) / 36
⇒ AD2 = (7AB2) / 9
9AD2 = 7AB2 .
Similar questions
India Languages,
7 months ago
History,
7 months ago
English,
1 year ago
Chemistry,
1 year ago
Hindi,
1 year ago