JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 6)

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 :
72
84
56
98

Explanation


Distance beetween point A and B is = 14 unit

As y coordinate of point A(-8, 5) and B(6, 5) are same. So line AB is parallel to the x axis.

Point O is the center of the circle and OP is the perpendicular to the line AB. And P is the mis point of line AB.

$$ \therefore $$ Point P = $$\left( {{{ - 8 + 6} \over 2},{{5 + 5} \over 2}} \right)$$ = (-1, 5)

As OP is perpendicular to the line AB and AB is parallel to the x axis then OP is parallel to the y axis. So x coordinate of line OP is constant.

$$ \therefore $$ x coordinate of O = -1

And point O is lies on the line 3y = x + 7

$$ \therefore $$ 3y = -1 + 7

$$ \Rightarrow $$ y = 2

$$ \therefore $$ Center O = (-1, 2)

$$ \therefore $$ OP = 3

Then BC = 2OP = 6

$$ \therefore $$ Area of rectangle = (AB)(BC)

= 14 $$ \times $$ 6

= 84

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