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Question

# A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

A
13 and 15
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B
15 and 13
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C
16 and 12
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D
None of These
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Solution

## The correct option is B 15 and 13Let there is a circle having center O touches the sides AB and AC of the triangle at point E and F respectively.Let the length of the line segment AE is x.Now in △ABC,CF=CD=6 (tangents on the circle from point C)BE=BD=6 (tangents on the circle from point B)AE=AF=x (tangents on the circle from point A)Now AB=AE+EB⟹AB=x+8=cBC=BD+DC⟹BC=8+6=14=aCA=CF+FA⟹CA=6+x=bNowSemi-perimeter, s=(AB+BC+CA)2s=(x+8+14+6+x)2s=(2x+28)2⟹s=x+14Area of the △ABC=√s(s−a)(s−b)(s−c)=√(14+x)((14+x)−14)((14+x)−(6+x))((14+x)−(8+x))=√(14+x)(x)(8)(6)=√(14+x)(x)(2×4)(2×3)Area of the △ABC=4√3x(14+x) .............................(1)Area of △OBC=12×OD×BC=12×4×14=28Area of △OBC=12×OF×AC=12×4×(6+x)=12+2xArea of ×OAB=12×OE×AB=12×4×(8+x)=16+2xNow, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB4√3x(14+x)=28+12+2x+16+2x4√3x(14+x)=56+4x=4(14+x)√3x(14+x)=14+xOn squaring both sides, we get3x(14+x)=(14+x)23x=14+x ------------- (14+x=0⟹x=−14 is not possible)3x−x=142x=14x=142x=7Hence AB=x+8AB=7+8AB=15AC=6+xAC=6+7AC=13So, the value of AB is 15 cm and that of AC is 13 cm. Option B.

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