12. 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 as shown in the figure
Find the sides AB and AC.
Answers
Answer:
Follow me and ill tell u the answer
Step-by-step explanation:
Follow me and ill tell u the answer
Step-by-step explanation:
let there is a circle having center o touches the side AB and AC of the triangle at point E and F respectively.
let the length of the segment AE is x.
Now in ∆ ABC,
CF=CD=6 ( tangent from c )
BE=BD=6 ( tangent from B )
AE=AF=x ( tangent from 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= ab+ bc+ca/2
s= x+8+14+6+x/2
s=2x+28/2
s=x+14
area of∆ abc = √ s(s-a)(s-b)(s-c)
= √ (14+x)((14+x)-14(14+x)-(6+x))
= √(14+x)(x)(8)(6)
=√(14+x)(8)(2×4)(2×3)
area of ∆ ABC = 4√3x(14+x)...................(1)
Area of∆ OBC = 1/2×od×bc
=1/2×4×14 =28
Area of∆ OBC = 1/2× of×ac
= 1/2×4×(6+x)
= 12+2x
Area of∆ OAB = 1/2×oe×ab
=1/2×4×(8+x)
= 16+2 x
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
Squaring both side, We get,
3x(14+x) = (14+x)^
3x=14+x ---------(14+x)=0 = x=-14 is not possible
3x-x=14
2x= 14
x=14/2
x=7
Hence,
AB = x+8
= 7+8
=15
AC= 6+x
=6+7
=13
Hope...it will help you