A rectangle ABCD is inscribed in a circle who | vedclass

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(B) The length of the side $AB$ is the distance between $(-6, 7)$ and $(4, 7)$,which is $|4 - (-6)| = 10$. Let the center of the circle be $O(h, k)$.

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(B) The length of the side $AB$ is the distance between $(-6, 7)$ and $(4, 7)$,which is $|4 - (-6)| = 10$. Let the center of the circle be $O(h, k)$. Since the center lies on $3y = x + 10$,we have $3k = h + 10$,or $h = 3k - 10$. The perpendicular bisector of $AB$ passes through the center $O$. The midpoint of $AB$ is $(\frac{-6+4}{2}, \frac{7+7}{2}) = (-1, 7)$. Since $AB$ is a horizontal line $(y=7)$,its perpendicular bisector is the vertical line $x = -1$. Thus,the $x$-coordinate of the center is $h = -1$. Substituting $h = -1$ into $h = 3k - 10$,we get $-1 = 3k - 10$,so $3k = 9$,which means $k = 3$. The center is $O(-1, 3)$. The distance from the center $O(-1, 3)$ to $A(-6, 7)$ is the radius $R$. $R^2 = (-1 - (-6))^2 + (3 - 7)^2 = 5^2 + (-4)^2 = 25 + 16 = 41$. Let the height of the rectangle be $2d$. The distance from the center $O(-1, 3)$ to the line $AB$ $(y=7)$ is $d = |7 - 3| = 4$. In the right triangle formed by the center,the midpoint of $AB$,and vertex $B$,we have $R^2 = d^2 + (AB/2)^2$. Here,$R^2 = 4^2 + 5^2 = 16 + 25 = 41$. This matches. The distance from the center to the side $CD$ is also $d = 4$. The total height of the rectangle is $2d = 2 	imes 4 = 8$. The area of the rectangle is $	ext{length} 	imes 	ext{height} = 10 	imes 8 = 80$.

$A$ rectangle $ABCD$ is inscribed in a circle whose center lies on the line $3y = x + 10$. If $A$ and $B$ are the points $(-6, 7)$ and $(4, 7)$ respectively,find the area of the rectangle.

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