Question

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​

Answer 1

Answer:

$AB=15cm,AC=13cm$

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,

$AB=AE+EB$

$⇒AB=x+8=c$

$BC=BD+DC$

$⇒BC=8+6=16=a$

$CA=CF+FA$

$⇒CA=6+x=6$

Now,

Semi-perimeter, $s=\frac{(AB+BC+CA)}{2}$

$s=\frac{(x+8+14+6+x)}{2}$

$s=\frac{(2x+28)}{2}$

$⇒s=x+14$

Area of the $ΔABC=\sqrt{s(s-a)(s-b)(s-c)}$

$⇒\sqrt{(14+x)((14+x)-14)((14+x)-(6+x))((14+x)-(8+x))}$

$⇒\sqrt{(14+x)(x)(8)(6)}$

$⇒\sqrt{(14+x)(x)(2×4)(2×3)}$

Area of the $ΔABC=\sqrt[4]{3x(14+x)}$

Area of $ΔOBC=\frac{1}{2}×OD×BC$

$→\frac{1}{2}×4×14=28$

Area of $ΔOBC=\frac{1}{2}×OF×AC$

$→\frac{1}{2}×4×(6+x)$

$→12+2x$

Area of $ΔOAB=\frac{1}{2}×OE×AB$

$→\frac{1}{2}×4×(8+x)$

$→16=2x$

Now, Area of the ΔABC=Area of ΔOBC+Area of ΔOBC+Area ofΔOAB

$→\sqrt[4]{3x(14+x)}=28+12+2x+16+2x$

$→\sqrt[4]{3×(14+x)}=56+4x=4(14+x)$

$→\sqrt{3x(14+x)}=14+x$

Squaring on both sides,

We get,

$→3x(14+x)=(14+x)²$

$→3x=14+x$

$→3x-x=14$

$→2x=14$

$→\frac{14}{2}$

$→x=7$

$Hence,$

$AB=x+8$

$AB=7+8$

$\boxed{AB=15}$

$AC=6+x$

$AC=6+7$

$\boxed{AC=13}$

So, the value of AB is 15cm

an the valuse of AC is 13cm

Hope it helps :-)

MARK ME BRAINLIST.....