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$\triangle _{A B_1 C_1}$ and $\triangle{A B_2 C_2}$ are not congruent. If $AB_1 = AB_2 =8$ and $AC_1 = AC_2 = 6$. Area of the $\triangle{AB_1C_1}$ is equal to the area of the $\triangle{AB_2C_2}=12\sqrt3$, then find $\left|(B_1 C_1)^{2} - (B_2 C_2)^{2}\right|$.

I tried this question by herons formula but its getting very complicated.

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    $\begingroup$ Heron's formula is a good idea. Begging for help may actually discourage potential answerers. Instead, show the work you've done so far. $\endgroup$ – Matthew Leingang Nov 28 '17 at 14:57
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Hint : put $BC_2$ =2x . Now use herons formula on both . Answer comes as 2.

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  • $\begingroup$ I got a different answer. Using your method, we get $s=\frac{6+8+2x}{2}=7+x$ and applying Heron's formula: $12\sqrt 3=\sqrt {(7+x)(7-x)(1+x)(-1+x)}\Rightarrow 432=(49-x^2)(x^2-1)$. This only has two positive solutions of $\sqrt{13}$ and $\sqrt{37}$, meaning these must be $\frac12 B_1C_1$ and $\frac12 B_2C_2$, so $\lvert B_1C_1-B_2C_2\rvert=2\sqrt{37}-2\sqrt{13}$ $\endgroup$ – Riley Nov 28 '17 at 18:58
  • $\begingroup$ Riley substitute $x^2$ =a $\endgroup$ – Atharva Shetty Nov 28 '17 at 19:05
  • $\begingroup$ Substituting $x^2=a$ into $432=(49-x^2)(x^2-1)$ yields $432 = (49-a)(a-1)\Rightarrow -a^2+50a-49=432\Rightarrow a^2-50a+481=0$ Quadratic formula gives us $a=\frac{50\pm \sqrt{2500-1924}}{2}=25\pm 12=13,37$ Therefore, $x=\sqrt{37},\sqrt{13}$ if we only take the positive solutions. This is what I have already said. $\endgroup$ – Riley Nov 28 '17 at 19:11
  • $\begingroup$ Could you edit your answer directly? Even though you told the OP your correction, everybody else who sees your answer will see the wrong answer. $\endgroup$ – Riley Nov 29 '17 at 15:42
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One formula for the area of a triangle is $\frac12 ab\sin \theta$ where $a$ and $b$ are the lengths of two sides and $\theta$ is the angle between those two sides. We can apply this formula to the first triangle by setting $a=AB_1$, $b=AC_1$, and once again for the second triangle by setting $a=AB_2$, $b=AC_2$. Since the areas are equal, we find that $\sin(\angle B_1AC_1)=\sin(\angle B_2AC_2)$. But the triangles are not congruent, so the angles must be different. The only way for this to be possible is if $\angle B_1AC_1 = 180^\circ -\angle B_2AC_2$. Do you think you can proceed from here?

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  • $\begingroup$ Actually I told him the correction in global $\endgroup$ – user506171 Nov 29 '17 at 11:14

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