Question

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. Find the sides AB
and AC.​
Refer to the above attachment.

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Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies The \ radius \ of \ circle \ is \ 4cm. \\\sf\implies \triangle ABC \ circumscribes \ the \ circle. \\\sf\implies Value \ of \ CD \ is \ 6cm \ \& \ BD \ is \ 8cm.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf\implies Measure \ of \ sides \ AB \ and \ AC.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Let :- }}}$

$\sf\implies Value \ of \ AE = AF = x$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

$\underline{\purple{\sf Implemented \ Theorem :- }}$

• Tangents drawn from a external point to a circle are equal.

$\underline{\orange{\sf From\ the\ figure :- }}$

• AC = x + 6 cm
• AB = x + 8cm .
• BC = 8 cm + 6cm = 14 cm .
• ∠ ODC = ∠OFA = ∠OEA = 90°

$\rule{200}2$

We know that inradius ( r) of a triangle is k / s , where K is the area of the triangle and s is the semiperimeter .

$\qquad\boxed{\red{\bf Inradius = \dfrac{Area}{Semiperimeter}}}$

• Hence here ,

$\pink{\sf :\implies Semiperimeter = \dfrac{a+b+c}{2}}\\\\\sf:\implies S = \dfrac{ 14cm + x + 6cm + x + 8cm }{2}\\\\\sf:\implies S = \dfrac{28+2x}{2}\\\\\sf:\implies \boxed{\orange{\sf Semiperimeter = 14 + x}}$

$\rule{200}2$

$\underline{\blue{\boldsymbol {Using \ the \ stated \ Formula :- }}}$

$\sf:\implies Inradius = \dfrac{Area}{Semiperimeter}\\\\\sf:\implies 4 cm = \dfrac{\sqrt{ s(s-a)(s-b)(s-c) }}{14+x}\:\:\bigg\lgroup \red{\bf Area \ using \ Heron's Formula }\bigg\rgroup \\\\\sf:\implies 4cm = \dfrac{\sqrt{14+x(14+x-14)(14+x-8-x)(14+x-6-x )}}{14+x}\\\\\sf:\implies 4cm = \dfrac{\sqrt{14+x(x)(8)(6)}}{14+x}\\\\\sf:\implies 4(14+x) = \sqrt{ 48( 14+ x)x}\\\\\sf:\implies (56 + 4x )^2 = (14+x)48x \\\\\sf:\implies 4^2 ( 14+x)^2 = (14+x)48x \\\\\sf:\implies 16(14+x) = 48x \\\\\sf:\implies 14 + x = \dfrac{48x}{16}\\\\\sf:\implies 14cm + x = 3x \\\\\sf:\implies 3x - x = 14 cm \\\\\sf:\implies 2x = 14 cm \\\\\sf:\implies x =\dfrac{14cm}{2}\\\\\sf:\implies \boxed{\red{\sf x = 7cm }}$

$\underline{\blue{\sf Hence \ the \ value \ of \ x \ is \:\: \textsf{\textbf{\blue{7cm}}}}}$

Hence ,

$\large\purple{\sf Answer } \begin{cases} \dag \pink{\sf Value \ of \ AB \ = x + 8cm = \ 15cm }\\\dag \pink{\sf Value \ of \ AC \ = x + 6cm = 13cm . }\end{cases}$

Answer 2

$\Huge{\texttt{{{\color{Magenta}{⛄A}}{\red{N}}{\purple{S}}{\pink{W}}{\blue{E}}{\green{R}}{\red{♡}}{\purple{࿐⛄}}{\color{pink}{:}}}}}$

_______________________________________

▪BC is divided by the point of contact D as BD & CD with the length 8cm and 6cm respectively.

▪OD, OE and OF are the radii of the same circle with the length of 4cm each. So,

OD = OE = OF = 4cm.

▪ We Know that , the lengths of tangents from the external point to a circle are equal in length. So,

CD = CE = 6cm

BD = BF = 8cm

Let, the lenght of AE be y cm.

so, AE = AF = y.

==========

Step ---- I

==========

Area of Triangle BOC = 1/2 × base × height

= 1/2 × BC × OD

= 1/2 × 14 × 4

= 28

Thus ,

Area of Tri.. AOC = 1/2 × AC × OE

= 1/2 × [y+6] × 4

= 2y+12.

Similarly

Area Of Tri.. AOB = 1/2 × AB × OF

= 1/2 × [8+y] × 4

= 16+2y.

Now,

¤ Area Of Triangle ABC =

= ar [BOC ] + ar [AOC ] + ar [AOB ]

= 28 + 2y+12 + 16 + 2y

= 56 + 2y . . . . . . . . . . . . . . Eq. {1.}

============

Step ----- II

============

a = AB = [8+y]

b = BC = 14

c = AC = [y+6]

$\: \: \: \: s \: \: \: \: = \: \: \: \: \frac{a + b + c}{2}$

where, S is a semi-perimeter of Tri.. ABC.

$s = \frac{8 + y + 14 + y + 6}{2} \\ \\ s = \frac{2y + 28}{2} \\ \\ s = \frac{2(y + 14)}{2} \\ \\ s = y + 14$

Now, we find the Area of Triangle ABC by Heron's Formula ---------

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{(y + 14)(y + 14 - 8 - y)(y + 14 - 14)(y + 14 - y - 6)} \\ \\ = \sqrt{(y + 14) \times 6 \times y \times 8 }$

$\\ = \sqrt{48y(y + 14)} \\ \\ = \sqrt{16 \times 3y(y + 14)}$

$= 4 \sqrt{3y(y + 14)} .. . . . . . . . .eq.(2.)$

Both Equations (eq. 1 and eq.2) are equal because area of a same triangle ABC.

So,

▪ Eq. [1.] = Eq.[2.]

$56 + 4y = 4 \sqrt{3y(y + 14)}$

$4(y + 14) = 4 \sqrt{3y(y + 14)}$

$\\ squaring \: both \: sides$

${(y + 14)}^{2} = 3y(y + 14)$

$\\ 3y = \frac{ {(y + 14)}^{2} }{(y + 14)} \\ \\ 3y = y + 14 \\ \\ 3y - y = 14 \\ \\ 2y = 14 \\ \\ y = 7. . . .$

▪▪ Therefore,

¤¤ AB = [8+y] = [8+7] = 15 cm

and

¤¤ AC = [y+6] = [7+6] = 13 cm

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$\huge\red{\boxed{\orange{\mathcal{{{\fcolorbox{red}{i}{{\red{@Master}}}}}}}}}$