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 (see figure). Find the
sides AB and AC.​

Answer 1

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Let there is a circle having center O touches the sides AB and AC of the triangle at point E and F respectively.

Let 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.

Let 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,

Let 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)

Let 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)

Let 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)

Let 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

Let 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=c

Let 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

Let 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=a

Let 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

Let 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=b

Let 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=bNow

Semi-perimeter, s= 2

Semi-perimeter, s= 2(AB+BC+CA)

Semi-perimeter, s= 2(AB+BC+CA)

Semi-perimeter, s= 2(AB+BC+CA) s= 2(x+8+14+6+x)

Semi-perimeter, s= 2(AB+BC+CA) s= 2(x+8+14+6+x)s= 2(2x+28)

⟹s=x+14

Area of ×OAB= 21×OE×AB= 21 ×4×(8+x)=16+2x

Area of ×OAB= 21×OE×AB= 21 ×4×(8+x)=16+2xNow, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB 4 3x(14+x)

4 3x(14+x)

4 3x(14+x) =28+12+2x+16+2x4 3x(14+x)

4 3x(14+x) =28+12+2x+16+2x4 3x(14+x)=56+4x=4(14+x)3x(14+x) =14+x

3x(14+x) =14+xOn squaring both sides, we get

3x(14+x) =14+xOn squaring both sides, we get3x(14+x)=(14+x) 2

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(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= 214x=7

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= 214x=7Hence AB=x+8AB=7+8AB=15AC=6+xAC=6+7AC=13

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= 214x=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.