Ex.6.5 (Q.5)
Q. ABC is an isosceles triangle with AC = BC. If AB² = 2AC²,
prove that ABC is a right triangle.
C
A
B
Proof :
AB² = 2AC2
... (i)
AB2 =
AC2 +
AC2
∴
AC = BC
AB2 =
AC2
+ BC2
∴
... From (i) and (ii)
∴
Δ ABC is a right angled triangle
...[by converse of
Pythagoras theorem]
∴
... (ii)
Ex.6.5 (Q.5)
In a quadrilateral ABCD. ∠B = 90o, AD2 = AB2 + BC2 + CD2.
Prove : ∠ACD = 90o
Proof :
In right triangle ABC,
∠B
=
90o
=
AC2
+
CD2
AD2
∠ACD
=
90o
[By Converse of Pythagoras
theorem]
∴
…(i)
…(ii)
[Pythagoras theorem]
[Given]
[From (i) and (ii)]
∴
C
D
A
B
AC2
=
AB2
+
BC2
=
AB2
+
BC2
AD2
+
CD2
…(iii)
∴
In ΔADC,
=
AC2
+
CD2
AD2
[From (iii)]
OC2 = O
OC2 = OQ2 + CQ2 …(3) OC2 = OQ2
Ex.6.5 (Q.17)
B
C
A
Sol:
AB2
=
108
AC2
=
144
BC2
=
36
AB2
+
BC2
=
AC2
∠B
=
90º
∴
Hence, the correct answer is (C).
In Δ ABC,
(A) 120º (B) 60º (C) 90º (D) 45º
The angle B is :
Tick the correct answer and justify :
6 cm
12 cm
AB2
+
BC2
=
108 + 36
=
144
?
The given triangle is a right angled triangle at B.
[By Converse of Pythagoras Theorem]
90º
∴