In an isosceles triangle ABC in which AB=AC=6cm is inscribed in a circle of radius 9cm .find the area of triangle
To find the area of an isosceles triangle, we need to know the base and the height. In this case, since we are given the lengths of the sides AB and AC, we can conclude that BC is the base of the triangle.
In an isosceles triangle, the perpendicular bisector (a line that cuts a line segment into two equal halves at a right angle) of the base also passes through the center of the circle. This means that the height (h) of the triangle is equal to the radius of the circle (9 cm).
Now, to find the length of the base of the triangle, we can use the Pythagorean theorem. In a right triangle formed by the radius (9 cm), half of the base (BC/2), and the height (h), we have:
BC/2)^2 + h^2 = radius^2
(BC/2)^2 + 9^2 = 9^2
(BC/2)^2 = 0
BC/2 = 0
BC = 2 * 0
BC = 0
Since BC has a length of 0 cm, the triangle does not exist, and we cannot find its area.
Join A and B to O, the centre of the circle
So we have triangle ABO with all 3 sides known. We can use the cosine law to find angle Ø
6^2 = 9^2+9^2-2(9)(9)cosO
162cosØ = 126
cosØ = 126/162 = 7/9
Ø = 38.94...° (I stored it in calculator's memory)
so angleBAO = 70.528779...
angle BAC = 141.057...°
area of triangle ABC
= (1/2)(9)(9)sin 141.057..
= appr 25.5 cm^2
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we could also find it with "exact values"
Let angle OAB = angle OBA = x (isosceles)
then 2x + Ø = 180°
Ø = 180-2x
take cos of both sides:
cosØ = cos(180-2x)
7/9 = cos180cos(-2x) + sin180sin(2x)
7/9 = (-1)(cos(2x) + 0
cos(2x) = -7/9
we know 1 - 2sin^2 x = cos(2x)
1 + 7/9 = 2sin^2 x
sin^2 x = 8/9
sinx = √8/3
cosx = √( 1 - 8/9) = √(/9) = 1/3
sin(2x) = 2sinxcosx = 2(√8/3)(1/3)
area of triangle ABC = (1/2)(9)(9)(2(√8/3)(1/3))
= 9√8 cm^2
or appr 25.5 , the same as above!!!!
area of triangle