The right-angled triangle with two lines (not the hypotenuse) a and b, is divided in two parts with the same area, by the segment MN which is perpendicular with the hypotenuse AB. Find the area of the circle which circumscribes the tetragon BCMN.
2007-02-05 08:24:19 · 5 answers · asked by Crystal 3 in Science & Mathematics ➔ Mathematics
This is a mess. Let b = BC, a = AC, so that AB = √(a² +b²). Let N lie on AB, and M lie on AC, so that the tetragon BCMN is formed therein. Let MN = x. Let AN = y. Because MN bisects the area of the triangle, we know:
2 x y = a b
Through proportions, we know:
b y = a x
With these, we find the following:
x = a / √2
y = b / √2
So that AM = √(a² +b²) / √2, and CM = a - √(a² +b²) / √2
Now, the area of the circle circumscribing BCMN has the area of
1/4 π BM²
But we know now that
BM² = a² + ( a - √(a² +b²) / √2)², so that the area of the circle is:
1/4 π ( a² + ( a - √(a² +b²) / √2)²)
It's not worth expanding this, nor will it simplify.
Maybe I should add the fact that because triangle BNM is also a right triangle, BCMN will fit perfectly inside of a circle? But proving this was not part of the problem.
2007-02-05 09:27:24 · answer #1 · answered by Scythian1950 7 · 0⤊ 1⤋
MN divided by 2 then squared. (mn/2)^2
Is MN equal to the diameter of the circle.
2007-02-05 16:29:07 · answer #2 · answered by Shmesh 3 · 0⤊ 0⤋
area for a circle is A=(pi)R2, R being the radius. 2 meaning squared.
2007-02-05 16:29:33 · answer #3 · answered by the bertis 2 · 0⤊ 0⤋
do ur own homework and go to extra help. i could solve this question better if there was a picture.
2007-02-05 16:27:08 · answer #4 · answered by Jeffrey W 3 · 0⤊ 1⤋
i have nooooooooooo idea. sorry :[
2007-02-05 16:27:12 · answer #5 · answered by sparkles 2 · 0⤊ 1⤋