Inside a 9cm radius circle an isosceles triangle ABC exists with side AB = 6cm and AB = AC - Maths - Circles

Question

Inside a 9cm radius circle an isosceles
triangle ABC exists with side AB = 6cm and AB = AC
Find the area of triangle

Priyanka Kedia · Accepted Answer

Let AP = x cm

In quadrilateral OBAC,

OB = OC = 9 cm ( Radius of the circle )

And, AB = AC = 6 cm ( Given )

Thus, OBAC forms a kite and we know that, the diagonals of a kite are perpendicular to each other.

Therefore,

$O A ⊥ B C$

$I n r i g h t a n g l e d Δ A P C, U \sin g P y t h a g o r a s T h e o r e m, w e h a v e, {A P}^{2} + {P C}^{2} = {A C}^{2} \Rightarrow {P C}^{2} = {A C}^{2} - {A P}^{2} \Rightarrow {P C}^{2} = 6^{2} - x^{2} ...... (1)$

And,

$I n r i g h t a n g l e d Δ O P C, U \sin g P y t h a g o r a s T h e o r e m, w e h a v e, {O P}^{2} + {P C}^{2} = {O C}^{2} \Rightarrow {(9 - x)}^{2} + (6^{2} - x^{2}) = 9^{2} \Rightarrow 81 + x^{2} - 18 x + 36 - x^{2} = 81 \Rightarrow 18 x = 36 \Rightarrow x = 2 ...... (2)$

Substitute value of x in equation (1), we get,

${P C}^{2} = 36 - 2^{2} = 36 - 4 = 32 \Rightarrow P C = 4 \sqrt{2} c m S i n c e O A ⊥ B C, S o, B C = 2 P C = 2 \times 4 \sqrt{2} = 8 \sqrt{2} c m$

Therefore,

$A r e a o f Δ A B C = \frac{1}{2} \times B C \times A P = \frac{1}{2} \times 8 \sqrt{2} \times 2 = 8 \sqrt{2} {c m}^{2}$

Hence, area of triangle is $8 \sqrt{2} {c m}^{2}$ .

Inside a 9cm radius circle an isosceles triangle ABC exists with side AB = 6cm and AB = AC . Find the area of triangle .