Problem solution
Whithin the circle ABCD , AC and BD are perpendicular. If the area of triangle ABD is 9, and the area of the circle is xπ, what is x.
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Re: Problem solution
Question : Within the circle ABCD , AC & BD are perpendicular. If the area of triangle ABD is 9 and the area of the circle is x$\pi$ ( E is the center of the triangle) , what is the value of x ?
Answer : The area of triangle ABCD = $\pi$x = $\pi\ r^2$ ....... ( r is the radius of triangle ABCD )
So, x = $\ r^2$
The area of triangle ABE = $\frac {1}{2}$ x AE x BE = $\frac {1}{2}$ x r x r = $\frac{r^2}{2}$
As , area of triangle ABD = 2 x triangle ABE
So , the area of triangle ABD = 2 x ABE = $r^2$ = 9
So , x = r = $\sqrt{9}$ = 3
The value of x is 3.
Answer : The area of triangle ABCD = $\pi$x = $\pi\ r^2$ ....... ( r is the radius of triangle ABCD )
So, x = $\ r^2$
The area of triangle ABE = $\frac {1}{2}$ x AE x BE = $\frac {1}{2}$ x r x r = $\frac{r^2}{2}$
As , area of triangle ABD = 2 x triangle ABE
So , the area of triangle ABD = 2 x ABE = $r^2$ = 9
So , x = r = $\sqrt{9}$ = 3
The value of x is 3.
Re: Problem solution
The question does not have enough information.
Re: Problem solution
triangle ADB = 9= x^(1/2) * x(1/2)=x
So, x=9
So, x=9