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

Always make a sketch when it concerns geometry.

First, I'll need to ask you if you know the cosine rule?

See here some some tips:

Isosceles Triangle Equations Formulas Calculator - Inscribed Circle Radius Geometry (http://www.ajdesigner.com/phptriangle/isosceles_triangle_inscribed_circle_radius_r.php)

You can do this without the cosine rule, using only Heron's formula (http://en.wikipedia.org/wiki/Heron's_formula).

If we simply knew the length of side BC, we'd be able to use Heron's formula to compute the area, and we'd be done. But unfortunately, we aren't given that information.

The circumcenter of a triangle (the center of a circle which circumscribes the triangle) is at the intersection of the perpendicular bisectors of the triangle's sides, as shown here (http://www.mathopenref.com/trianglecircumcircle.html).

Given your triangle, ABC, we can call the circumcenter O. (It will help if you draw a diagram similar to the one on the web link I just provided).

Since the circumcircle has a radius of 9 cm, we know that the distance from the circumcenter to any of the triangle's vertices is 9 cm:

OA = OB= OC = 9

Now using Heron's formula from the link above, we can calculate the area of triangle AOC with sides of 9, 9, and 6. (I'll leave it up to you to do that).

Meanwhile, we also know the area of that same triangle can be computed as

A=\frac{1}{2}b \cdot h,

where b is the base and h is the height. If you consider side AO to be the base, then the height, h, will be half of the length of side BC (we know it's exactly half because we're told that triangle ABC is isosceles, therefore it's symmetric).

h = \frac{BC}{2}

So we can say

\text{Area by Heron's formula} = \frac{1}{2}9h = \frac{9 \cdot BC}{4}

From that equation, you can compute the length of side BC.

Once you know the length BC, you can now use Heron's formula once again to compute the area of triangle ABC.

Tell us what you get for an answer!

Hmm... nice to know there is yet another formula to compute the area of a triangle :)

Sorry ,I don't know the cosine rule.

Please tell me.

Hm... Did you take a look at the other answers? Those don't include the cosine rule, but if you understand them, it's better.

You have a great explanation of the rule here on wikipedia: Law of cosines - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Law_of_cosines)

Basically, you draw lines, each starting from the centre of the circle to the vertex of the isosceles triangle. You now have 3 triangles.

You can get the angle AOB using the cosine rule, and using symmetry to get the angle AOC. Then, finding the angle BOC, you can find the area of each triangle using the formula:

Area = \frac12 ab \sin\angle{c}

where a and b are two sides of a triangle and c is the angle between those two sides.

The given information can be represented using a figure as:

http://www.meritnation.com/app/webroot/img/userimages/mn_images/image/04Feb2011-1-i.png
Let AM = x cm
In quadrilateral OBAC, OB = OC = 9 cm and AB = AC = 6 cm
Thus, OBAC is a kite. It is known that the diagonals of a kite are perpendicular to each other.
∴ OA ⊥ BC
In right angled ∆AMC,
AM2 + MC2 = AC2 [Pythagoras theorem]
⇒ MC2 = (62 – x 2) cm2

Using Pythagoras theorem for ∆OMC, it is obtained
OM2 + MC2 = OC2
⇒ (9 – x)2 + (62 – x 2) = 92
⇒ 81 + x 2 – 18x + 36 – x 2 = 81
⇒18x = 36
⇒x = 2

http://www.meritnation.com/app/webroot/img/userimages/mn_images/image/04Feb2011-1-iI.png