Question

a square has two of its vertices on a circle and the other two on the tangent to the circle if the diameter of the circle is 10 determine the side of the square​

Answer 1

Given ABCD is a square .

Diameter of the circle is 10 .

$Radius = PB = PT = \frac{diameter}{2} \\= \frac{10}{2} \\= 5$

1) Let QP= x .

In ∆PQB , <Q = 90° .

QB² = PB² - QP² [ By Pythagoras Theorem ]

= 5² - x²

= 25 - x² ----(1)

2) BC = QT

= QP + PT

= x + 5 ----(2)

3) AB² = BC² [ Area of the square ]

=> ( 2QB )² = BC²

=> 4QB² = BC ²

=> 4(25-x²) = (x+5)² [ From (1) and (2) ]

=> 100 - 4x² = x² + 10x + 25

=> 0 = 5x² + 10x -75

/* Divide each term by 5, we get */

=> 0 = x² + 2x - 15

/* Splitting the middle term,we get */

=> x² + 5x - 3x - 15 = 0

=> x( x + 5)- 3( x + 5) = 0

=> (x+5)(x-3) = 0

=> x + 5 = 0 Or x - 3 = 0

=> x = -5 Or x = 3

/* x should not be negative */

Therefore.,

x = 3

Side of the square = QT = x + 5

= 3 + 5 = 8 Units

$\red { Side \:of \:the \:square }\green { = 8 }$

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