Question

A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8cm and 6cm respectively. find the sides AB and AC .​

Answer 1

Answer:

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. Option B.

Step-by-step explanation:

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Answer 2

Answer:

$\large \green{Solution}$☟︎︎︎

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF = x

We observed that,

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

Now semi perimeter of circle s,

⇒ 2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒s = 14 + x

Area of ΔABC = √s (s - a)(s - b)(s - c)

= √(14 + x) (14 + x - 14)(14 + x - x - 6)(14 + x - x - 8)

= √(14 + x) (x)(8)(6)

= √(14 + x) 48 x .

(i)

Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

= 2×1/2 (4x + 24 + 32) = 56 + 4x .

(ii)

Equating equation (i) and (ii) we get,

√(14 + x) 48 x = 56 + 4x

Squaring both sides,

48x (14 + x) = (56 + 4x)2

⇒ 48x = [4(14 + x)]2/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7 cm

$\large \green{Hence}$,

AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm.

$\large \green{Answer}$➪ 13Cm.

$\large \pink{Mabelrose♡︎}$