pls clear my Doubt...
Q.) 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 Fig. 10.14). Find the sides AB
and AC
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Verified answer
Answer:
[tex]AB=15cm,AC=13cm[/tex]
Step-by-step explanation:
Let there is a circle having the 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 ΔABC,
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,
[tex]AB=AE+EB[/tex]
[tex]⇒AB=x+8=c[/tex]
[tex]BC=BD+DC[/tex]
[tex]⇒BC=8+6=16=a[/tex]
[tex]CA=CF+FA[/tex]
[tex]⇒CA=6+x=6[/tex]
Now,
Semi-perimeter, [tex]s=\frac{(AB+BC+CA)}{2}[/tex]
[tex]s=\frac{(x+8+14+6+x)}{2}[/tex]
[tex]s=\frac{(2x+28)}{2}[/tex]
[tex]⇒s=x+14[/tex]
Area of the [tex]ΔABC=\sqrt{s(s-a)(s-b)(s-c)}[/tex]
[tex]⇒\sqrt{(14+x)((14+x)-14)((14+x)-(6+x))((14+x)-(8+x))}[/tex]
[tex]⇒\sqrt{(14+x)(x)(8)(6)}[/tex]
[tex]⇒\sqrt{(14+x)(x)(2×4)(2×3)}[/tex]
Area of the [tex]ΔABC=\sqrt[4]{3x(14+x)}[/tex]
Area of [tex]ΔOBC=\frac{1}{2}×OD×BC[/tex]
[tex]→\frac{1}{2}×4×14=28[/tex]
Area of [tex]ΔOBC=\frac{1}{2}×OF×AC[/tex]
[tex]→\frac{1}{2}×4×(6+x)[/tex]
[tex]→12+2x[/tex]
Area of [tex]ΔOAB=\frac{1}{2}×OE×AB[/tex]
[tex]→\frac{1}{2}×4×(8+x)[/tex]
[tex]→16=2x[/tex]
Now, Area of the ΔABC=Area of ΔOBC+Area of ΔOBC+Area ofΔOAB
[tex]→\sqrt[4]{3x(14+x)}=28+12+2x+16+2x[/tex]
[tex]→\sqrt[4]{3×(14+x)}=56+4x=4(14+x)[/tex]
[tex]→\sqrt{3x(14+x)}=14+x[/tex]
Squaring on both sides,
We get,
[tex]→3x(14+x)=(14+x)²[/tex]
[tex]→3x=14+x[/tex]
[tex]→3x-x=14[/tex]
[tex]→2x=14[/tex]
[tex]→\frac{14}{2}[/tex]
[tex]→x=7[/tex]
[tex]Hence,[/tex]
[tex]AB=x+8[/tex]
[tex]AB=7+8[/tex]
[tex]\boxed{AB=15}[/tex]
[tex]AC=6+x[/tex]
[tex]AC=6+7[/tex]
[tex]\boxed{AC=13}[/tex]
So, the value of AB is 15cm
an the valuse of AC is 13cm
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