:- A triangle ABC is drawn to a circumscribe a circle of radius 4 cm, such that the segment 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 AC and AB.
Please help me
Answers
Answer:
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
BC=BD+DC
⟹BC=8+6=14=a
CA=CF+FA
⟹CA=6+x=b
Now
Semi-perimeter, s=
2
(AB+BC+CA)
s=
2
(x+8+14+6+x)
s=
2
(2x+28)
⟹s=x+14
Area 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=
2
1
×OD×BC
=
2
1
×4×14=28
Area of △OBC=
2
1
×OF×AC
=
2
1
×4×(6+x)
=12+2x
Area of ×OAB=
2
1
×OE×AB
=
2
1
×4×(8+x)
=16+2x
Now, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB
4
3x(14+x)
=28+12+2x+16+2x
4
3x(14+x)
=56+4x=4(14+x)
3x(14+x)
=14+x
On squaring both sides, we get
3x(14+x)=(14+x)
2
3x=14+x ------------- (14+x=0⟹x=−14 is not possible)
3x−x=14
2x=14
x=
2
14
x=7
Hence
AB=x+8
AB=7+8
AB=15
AC=6+x
AC=6+7
AC=13
So, the value of AB is 15 cm and that of AC is 13 cm.
Answer:
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