Question

A square has two of its vertices on a circle and the other two are on the tangent to the circle. Find the area of the square if the radius of the circle is 5 units.

• A. 64 sq units
• B. 36 sq units
• C. 50 sq units
• D. 100 sq units

Answer 1

The correct option is C 50 sq units

Area of the square: (x + 5)${}^2$

If area is 50, then x = $\sqrt{50}$ - 5; So EF = $\sqrt{50}$, OE = $\sqrt{50}$ - 5 and EA = $\frac{\sqrt{50}}{2}$

If area is 36 then x = 1; So EF = 6, OE = 1, OA = 5 and EA = 3

$\angle$OEA = 90

If area is 64 then x = 3, EF = 8, EA = 4, OA = 5

OE, EA, and OA did not form Pythagoras triplet in first two cases.

Alternatively,
Clearly, from the figure, the side of the square is (5 + x) units.
Let the point of intersection of AD and the line parallel to the side of the square passing through the centre O of the circle be E.
Considering right-angled triangle ODE,
$O{D}^2=O{E}^2+E{D}^2$
$(5{)}^2=x^2+E{D}^2$
$E{D}^2=25-x^2$
$ED=\sqrt{(25-x^2)}$
$AD=2ED=2\sqrt{(25-x^2)}$ ------- 1)
But AD = 5 + x -------- 2)
Equating the above two, we get x = 3 , -5
x = 3 as -5 is not possible.
Side of the square = 5 + 3 = 8
Area = 64