Case Study Based-2
THREE FRIENDS
18. Three friends stand in such a way that they form a right angled triangleABC as shown.
Point D is mid-point of BC
(a) AC^2=
(i)AD^2+CD^2
(ii)AB^2+4BD^2
(iii)BD^2+CD^2
(iv)AD^2+BD^2
(b)AD^2=
(i)AB^2+BD^2
(ii)BD^2-AB^2
(iii)4AB^2
(iv)3CB^2
(c)AC^2=
(i)AD^2-3AB^2
(ii)4AD^2-AB^2
(iii) 4AD^2-3AB^2
(iv) 4AD^2+3AB^2
(d) BC =
(i)BD. (ii)CD. (iii)2AB. (iv)4BD
(e) CD^2=
(i) AB^2- AD^2
(ii) AD^2 - AB^2
(iii) AB^2 + AD^2
(iv) 2AD
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Answer:
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To prove:- AB
2
=4AD
2
−3AC
2
Proof:-
In △ACD
AC
2
+CD
2
=AD
2
(Pythagoras Theorem)\
AC
2
+(
2
BC
)
2
=AD
2
(Pythagoras Theorem)-------------(1)
In △ABC,
AC
2
+BC
2
=AB
2
⇒BC
2
=AB
2
−AC
2
------------(2)
Substituting Equation(2) in (1), we get
⇒AC
2
+
4
AB
2
−
4
AC
2
=AD
2
⇒
4
3AC
2
+
4
AB
2
=AD
2
⇒AB
2
=4AD
2
−3AC
2
solution
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