Question

In circle B, given isosceles triangle BTX, BA ⟂ WX, and WX = 10, what is the radius of circle B rounded to the nearest whole number?

Answer 1

Solution :

We observe that :

$\overline {BA} \perp \overline {WX}$

But BA is the perpendicular.

From the center B and WX is a chord.

Therefore, TW = TX (perpendicular from the centre of a circle to a chord bisects it)

Consider Δ BTX,

∠BTX = 90° (BA ⊥ WX)

BT = XT (Δ BTX is isosceles)

Since the angles opposite to equal sides are equal of a triangle arc are equal.

∠BTX = ∠BXT

But in the triangle,

∠TBX + ∠TXB + ∠BTX = 180°

∠TBX + ∠TBX + 90° = 180°

2 ∠TBX = 90°

∠TBX = 45°

From trigonometry, we get

$\sin \theta =\frac{TX}{BX}$    ...............(1)

WX = 10

i.e., TX + TW = 10

But TX  = TW

     2 TX = 10

         Tx = 5

BX = radius of circle.

∴ $\sin 45^\circ = \frac{5}{BX}$

  $\frac{1}{\sqrt2} = \frac{5}{BX}$

 $BX = 5\sqrt2$

        $=5 \times 1.41$

        = 7

Therefore, the radius of the circle is 7 units.