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.

Answer 1

circumcentre of triangle=9
area =abc /4R
area=x
x=6*6*c/4*9
x=c=base
area of triangle=1/2*base*h----(1)

h^2+(b/2)^2=6^2
h^2=36-(b^2/4)
h^2=(144-b^2)/4
h=√(144-b^2)/2------(2)
(2)in (1)
c=c/4*√(144-b^2)
4=√(144-b^2)
16=144-b^2
b^2=128
b=8√2
area=b=8√2

Answer 2

The area of the triangle is half as big as the rectangle. So the area of triangle is half of the area of the rectangle.

Given:

$AB\quad =\quad AC\quad =\quad 6 \ cm$

Radius of circle = 9 cm

To find:

The “Area of the triangle”

Answer:

Let AP = x and OP = 9 - x

$OA\quad =\quad AP+OP$

$=\quad x+9-x$

$OA\quad =\quad 9$

$OB\quad =\quad 9$

$AB\quad =\quad 6$

$\Delta ABP$

${ BP }^{ 2 }={ AB }^{ 2 }-{ AP }^{ 2 }$

$In \ \Delta OBP,$

${ BP }^{ 2 }={ OB }^{ 2 }-{ OP }^{ 2 }$

Compare both sides

${ AB }^{ 2 }-{ AP }^{ 2 }={ OB }^{ 2 }-{ OP }^{ 2 }$

$6^{ 2 }-{ x }^{ 2 }=9^{ 2 }-\left( 9-{ x }^{ 2 } \right)$

$6^{2}-\mathrm{x}^{2}=9^{2}-\left(81+\mathrm{x}^{2}-18 \mathrm{x}\right)$

$6^{ 2 }-{ x }^{ 2 }=81-81-{ x }^{ 2 }+18x$

$36 = 18x$

$x = 2$

Area of triangle APC is

${ CP }=\sqrt { { AC }^{ 2 }-{ AP }^{ 2 } }$

$=\sqrt { 36-4 } =\sqrt { 32 }$

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

Area of triangle ABC is twice the area of triangle APC.

$A=2 \sqrt{32}$