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
_9th_April_Evening_Slot_en_6_2.png)
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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