IIT-JEE 2009 · Paper 2 · Mathematics Q33

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MathematicsInteger+4 / -1Question 33 of 57

Let $ABC$ and $ABC'$ be two non-congruent triangles with sides $AB=4$ , $AC=AC'=2\sqrt2$ and angle $B=30°$ . The absolute value of the difference between the areas of these triangles is

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Answer and worked solution

Official answer: 4

Side BC. By the cosine rule with $c=4$ , $b=2\sqrt2$ , $B=30°$ :
$8=a^2+16-4\sqrt3\,a\implies a^2-4\sqrt3a+8=0\implies a=2\sqrt3\pm2.$

Areas. $\frac12ac\sin B=a$ , so the two areas are $2\sqrt3\pm2$ and their difference is 4.

Answer: 4

Topics:Trigonometric ratios and identities

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IIT-JEE 2009 · Paper 2, Mathematics question 33