Isosceles triangle has , and a circle with radius is tangent to line at and to line at . What is the area of the circle that passes through vertices , , and

The circle that passes through the vertices of triangle ΔABC (A, B, C) is the

circumscribing circle of triangle ΔABC.

The area of the circle that passes through vertices A, B, and C, is (C) 26·π

Reasons:

The given parameters are;

Side length of isosceles triangle ΔABC; [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex] = 3·√6

Radius of circle tangent to [tex]\overline{AB}[/tex] at B and [tex]\overline{AC}[/tex] at C = 5·√2

Required:

Area of the circle that passes through vertices A, B, and C

Solution:

Angle ∠BAO is given as follows;

[tex]\angle BAO = arctan\left(\dfrac{5 \cdot \sqrt{2} }{3 \cdot \sqrt{6}} \right) = \mathbf{arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)}[/tex]

Therefore;

[tex]\angle BOA = 90^{\circ} - arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)[/tex]

[tex]\overline{BC} = 2 \times 5 \cdot \sqrt{2} \times sin\left(90^{\circ} - arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right) = 15\cdot \sqrt{\dfrac{6}{13} }[/tex]

∠ABO' = ∠BAO' (Base angles of isosceles triangle ΔABO')

[tex]\angle BAO' = \angle BAO = \mathbf{arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)}[/tex]

Therefore;

[tex]\angle BO'A = 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)[/tex]

From sine rule, we have;

[tex]\dfrac{\overline{AB}}{sin \left(\angle BO'A \right)} = \mathbf{\dfrac{\overline{BO'}}{sin \left(\angle BAO' \right) \right)}}[/tex]

Which gives;

[tex]\mathbf{\dfrac{3 \cdot \sqrt{6} }{sin \left( 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right)}} = \dfrac{\overline{BO'}}{sin \left(arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right) \right)}[/tex]

Using a graphing calculator, we get;

[tex]\overline{BO'} = \dfrac{3 \cdot \sqrt{6} }{sin \left( 180^{\circ} - 2 \times arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right)\right)} \times sin \left(arctan\left(\dfrac{5 \cdot \sqrt{3} }{9} \right) \right) = \sqrt{26}[/tex]

The radius of the circumscribing circle [tex]\overline{BO'}[/tex] = √(26)

Therefore, area of the circumscribing circle, [tex]A_{O'}[/tex] = π·(√(26))² = 26·π

The area of the circle that passes through vertices A, B, and C, is (C) 26·π

Learn more here:

https://brainly.com/question/17147358

The possible question options obtained from a similar question online are;

(A) 24·π (B) 25·π (C) 26·π (D) 27·π (E) 28·π

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