Question

If an isosceles triangle ABC, in which AB=AC=6 cm. is inscribed in a circle of radius 9 cm. Find the area of the triangle ABC.

Answer 1

hope this will help you...........

Answer 2

Answer:

The area of triangle ABC is $2\sqrt{32}$ cm square.

Step-by-step explanation:

Given information: ABC is an isosceles triangle. AB=AC=6 cm and radius of the circle is 9 cm.

Using Pythagoras theorem in triangle ADC,

$DC=\sqrt{6^2-x^2}$              ...(1)

Using Pythagoras theorem in triangle ODC,

$DC=\sqrt{9^2-(9-x^2}$         ...(2)

Equating (1) and (2),

$\sqrt{6^2+x^2}=\sqrt{9^2+(9-x^2)}$

Squaring both sides,

$36-x^2=81-81-x^2+18x$

$36=18x$

$x=2$

The value of x is 2 cm.

The value of DC is

$DC=\sqrt{6^2-2^2}=\sqrt{32}$

Area of a triangle is

$A=\frac{1}{2}\times base \times height$

Area of triangle ADC,

$A=\frac{1}{2}\times\sqrt{32}\times 2=\sqrt{32}$

Area of triangle ABC is twice of area of triangle ADC.

$A=2\sqrt{32}$

Therefore the area of triangle ABC is $2\sqrt{32}$ cm square.