A triangle ABC is drawn to circumscribe a cir | vedclass

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(D) Let the circle touch the sides $AB$ and $AC$ of the triangle at points $E$ and $F$ respectively. Let the length of the tangent $AF$ be $x \, cm$.S

Vedclass · Accepted Answer

$15, 13$

Vedclass · Answer

(D) Let the circle touch the sides $AB$ and $AC$ of the triangle at points $E$ and $F$ respectively. Let the length of the tangent $AF$ be $x \, cm$.Since tangents drawn from an external point to a circle are equal in length:$CF = CD = 6 \, cm$$BE = BD = 8 \, cm$$AE = AF = x \, cm$Thus,the sides of the triangle are:$AB = AE + EB = x + 8$$BC = BD + DC = 8 + 6 = 14 \, cm$$AC = AF + FC = x + 6$Semi-perimeter $s = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14$.Using Heron's formula,Area of $	riangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$$= \sqrt{(x+14)(x+14-14)(x+14-(x+6))(x+14-(x+8))}$$= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} = 4\sqrt{3x(x+14)}$.Also,Area of $	riangle ABC = 	ext{Area}(	riangle OBC) + 	ext{Area}(	riangle OCA) + 	ext{Area}(	riangle OAB)$$= \frac{1}{2} 	imes 4 	imes 14 + \frac{1}{2} 	imes 4 	imes (x+6) + \frac{1}{2} 	imes 4 	imes (x+8)$$= 28 + 2(x+6) + 2(x+8) = 28 + 2x + 12 + 2x + 16 = 56 + 4x = 4(14+x)$.Equating the areas: $4\sqrt{3x(x+14)} = 4(14+x)$$\sqrt{3x(x+14)} = 14+x$Squaring both sides: $3x(x+14) = (14+x)^2$$3x^2 + 42x = 196 + x^2 + 28x$$2x^2 + 14x - 196 = 0 \Rightarrow x^2 + 7x - 98 = 0$$(x+14)(x-7) = 0$. Since $x > 0$,$x = 7$.Therefore,$AB = 7+8 = 15 \, cm$ and $AC = 7+6 = 13 \, cm$.

$A$ 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 figure). Find the sides $AB$ and $AC$ (in $cm$).

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