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(B) Let the vertices of the rectangle be $A(-8, 5)$ and $B(6, 5)$. Since $AB$ is a side of the rectangle,the length of $AB = |6 - (-8)| = 14$.Let the

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$84$

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(B) Let the vertices of the rectangle be $A(-8, 5)$ and $B(6, 5)$. Since $AB$ is a side of the rectangle,the length of $AB = |6 - (-8)| = 14$.Let the other two vertices be $D(-8, k)$ and $C(6, k)$.The center of the circle is the midpoint of the diagonal $AC$ (or $BD$).Midpoint of $AC = \left( \frac{-8 + 6}{2}, \frac{5 + k}{2} ight) = \left( -1, \frac{5 + k}{2} ight)$.Since the center lies on the diameter $3y = x + 7$,we substitute the coordinates into the equation:$3\left( \frac{5 + k}{2} ight) = -1 + 7$$3\left( \frac{5 + k}{2} ight) = 6$$\frac{5 + k}{2} = 2$$5 + k = 4$$k = -1$.The length of the side $BC = |5 - (-1)| = 6$.Area of the rectangle = $AB 	imes BC = 14 	imes 6 = 84 	ext{ sq. units}$.

$A$ rectangle is inscribed in a circle with a diameter lying along the line $3y = x + 7$. If the two adjacent vertices of the rectangle are $(-8, 5)$ and $(6, 5)$,then the area of the rectangle (in $sq. units$) is

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