Question

If an isosceles triangle ABC with sides AB = AC = 6 cm is drawn inside a circle of radius 9 cm, find area of the triangle.

Answer 1

FIGURE IS IN THE ATTACHMENT

GIVEN:
AB = AC = 6 cm
Radius of circle(OB = OC ) = 9 cm

Let O is the Centre of the circle & let P be the midpoint of BC. Then OP ⟂ BC.

∆ABC is Isosceles & P is the midpoint of BC. AP ⟂ BC as median from the vertex in an Isosceles ∆ is ⟂ to the base.

Let AP = x  & PB = CP = y

In ∆APB ,
AB² = BP² + AP²
6² = y² + x²
36 = y² + x²…………..(1)

In ∆OPB,
OB² = OP² + BP²
9² = (9-x)² + y²
81 =  (9-x)² + y²…………..(2)

On subtracting eq 1 from 2

81 - 36 = {(9-x)² + y²)} - {y² + x²}
45 = 81 + x² - 18x + y² - y² - x²
18x = 81 - 45
18x = 36
x = 36/18= 2 cm
AP = 2 cm.

Putting x = 2 in eq 1,
36 = y² + x²
36 = y² + 2²
36 = y² + 4
36 - 4 = y²
32 = y²
y² = 32
y =√32 = √16×2 = 4√2
y = 4√2 cm

BC = 2BP
BC = 2 × y
BC= 2 ×4√2 = 8√2 cm
BC = 8√2 cm

Area of ∆ABC = ½(base × height)
Area of ∆ABC = ½(BC × AP)
Area of ∆ABC = ½( 8√2 × 2) = 8√2 cm

Hence, the Area of ∆ABC = 8√2 cm.

HOPE THIS WILL HELP YOU...

Answer 2

▶ Given :- In isosceles ∆ABC, AB = AC = 6 cm and radius BO = OC = 9 cm.

▶ To Find :- Area of ∆ABC

▶ Solution :- AM perpendicular to BC.
let AM = x cm
OM = ( 9 - x ) cm

In ∆AMC, angle AMC = 90
AM^2 + MC^2 = AC^2
AM^2 = AC^2 - MC^2
AM^2 = (6)^2 - PC^2
AM = √( 36 - PC^2 )

and

MC^2 = AC^2 - AM^2
MC^2 = 36 - x^2

Now, In ∆ OMC
OC^2 = OM^2 + MC^2
(9)^2 = ( 9 - x )^2 + 36 - x^2
81 = 81 - 18x + x^2 + 36 - x^2
18x = 36
x = 36/18
x = 2
AM = x = 2 cm.

MC^2 = 36 - x^2
MC^2 = 36 - 2^2
MC^2 = 36 - 4
MC = √32
MC = 4√2

BC = 2 ( MC )
BC = 2( 4√2 )
BC = 8√2 cm

Area of ∆ABC = 1/2 × AM × BC
= 1/2 × 2 × 8√2
= 8√2 cm^2

HOPE IT HELPS YOU...

THANKS^-^