Question

An isosceles triangle of equal sides is 6 cm each is inscribed in a circle of radius 5 cm. The length of the third side is​

Answer 1

Let O be the centre of the circle and let P be the mid-point of BC. Then, OP⊥BC.

Since △ABC is isosceles and P is the mid-point of BC. Therefore, AP⊥BC as median from the vertex in an isosceles triangle is perpendicular to the base.

Let AP=x and PB=CP=y.

Applying Pythagoras theorem in △s APB and OPB, we have

AB

2

=BP

2

+AP

2

and OB

2

=OP

2

+BP

2

⇒36=y

2

+x

2

. . . (i) and, 81=(9−x)

2

+y

2

. . . (ii)

⇒81−36=(9−x)

2

+y

2

−y

2

+x

2

[Subtracting (i) from (ii)]

⇒45=81−18x

⇒x=2 cm

Putting x=2 in (i), we get

36=y

2

+4⇒y

2

=32⇒y=4

2

cm

∴BC=2BP=2y=8

2

cm

Hence, Area of △ABC=

2

1

(BC×AP)

=

2

1

×8

2

×2 cm

2

=8

2

cm

2