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(D) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the identity for the exradii $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$,the condition $rr_3 = r_1 r_2$ becomes $\frac{\Delta}{s} \cdot \frac{\Delta}{s-c} = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b}$.This simplifies to $(s-a)(s-b) = s(s-c)$.Expanding this,$s^2 - s(a+b) + ab = s^2 - sc$.Since $a+b+c = 2s$,we have $a+b = 2s-c$,so $s^2 - s(2s-c) + ab = s^2 - sc$,which simplifies to $ab = s^2 - sc - s^2 + 2s^2 - sc = s^2 - 2sc + s^2 = s^2 - sc$ is incorrect; let us re-evaluate: $s^2 - s(2s-c) + ab = s^2 - sc \implies s^2 - 2s^2 + sc + ab = s^2 - sc \implies ab = 2s^2 - 2sc = 2s(s-c)$.Using $s = \frac{a+b+c}{2}$,$s-c = \frac{a+b-c}{2}$,we get $ab = 2 \cdot \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} = \frac{(a+b)^2 - c^2}{2}$.$2ab = a^2 + b^2 + 2ab - c^2 \implies a^2 + b^2 = c^2$.Thus,the triangle is a right-angled triangle with $\angle C = 90^{\circ}$.However,we know $B = 60^{\circ}$,so $A = 30^{\circ}$.In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the sides are in the ratio $a:b:c = 1:\sqrt{3}:2$.Given $c = 10$,we have $2k = 10 \implies k = 5$.So $a = 5$ and $b = 5\sqrt{3}$.Then $a^2 + b^2 + c^2 = 5^2 + (5\sqrt{3})^2 + 10^2 = 25 + 75 + 100 = 200$.

In a triangle $ABC$,if $A, B$ and $C$ are in arithmetic progression,$rr_3 = r_1 r_2$ and $c = 10$,then $a^2 + b^2 + c^2 =$

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