Question

in the given fug abc is a right angled triangle at b .d and e are points on it prove that 8ae^2 =3ac^2+5ad^2

Answer 1

Given a right triangle ABC, in which BC is trisected on points D and E

Given a right triangle ABC, in which BC is trisected on points D and E. Prove that 8AE2 = 3AC2 + 5AD2

Class-X Maths

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Asked by Harshit

Sep 9

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Agam , SubjectMatterExpert

Member since Sep 21 2013


Given right DABC, right angled at B.
D and E are points of trisection of the side BC
Let BD = DE = EC = k
Hence we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem, we get
AD2�= AB2�+ BD2
AD2�= AB2�+ k2
Similarly, in ΔABE we get
AE2�= AB2�+ BE2
Hence AE2�= AB2�+ (2k)2
= AB2�+ 4k2�and
AC2�= AB2�+ BC2
�= AB + (3k)2
AC2= AB2�+ 9k2
Consider, 3AC2�+ 5AD2�= 3(AB2�+ 9k2) + 5(AB2�+ 4k2)
= 8AB2�+ 32k2
= 8(AB2�+ 4k2)
∴ 3AC2�+ 5AD2�= 8AE2

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Kishore , Student

Member since Dec 10 2008


Given right DABC, right angled at B.
D and E are points of trisection of the side BC
Let BD = DE = EC = k
Hence we get BE = 2k and BC = 3k
In ΔABD, by Pythagoras theorem, we get
AD2 = AB2 + BD2
AD2 = AB2 + k2
Similarly, in ΔABE we get
AE2 = AB2 + BE2
Hence AE2 = AB2 + (2k)2
= AB2 + 4k2 and
AC2 = AB2 + BC2
 = AB + (3k)2
AC2= AB2 + 9k2
Consider, 3AC2 + 5AD2 = 3(AB2 + 9k2) + 5(AB2 + 4k2)
= 8AB2 + 32k2
= 8(AB2 + 4k2)
∴ 3AC2 + 5AD2 = 8AE2