Exam

JEE Advanced

Year

2009

Paper

Paper 2

Subject

Mathematics

Chapter

Properties of Triangles

JEE Advanced 2009 · Paper 2 · Mathematics · Properties of Triangles

Let A B C and A B C^ be two non-congruent triangles with sides A...

Let $A B C$ and $A B C^{\prime}$ be two non-congruent triangles with sides $A B=4$, $A C=A C^{\prime}=2 \sqrt{2}$ and angle $B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

Step-by-step solution

In $\triangle A B C$, by sine rule, $$\begin{aligned}& \frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{{}^{\circ}}}=\frac{4}{\sin C}\\ \Rightarrow \quad & C=45^{{}^{\circ}}, C^{\prime}=135^{{}^{\circ}} \\ \Rightarrow & A=180^{{}^{\circ}}-\left(45^{{}^{\circ}}+30^{{}^{\circ}}\right)=105^{{}^{\circ}} \end{aligned}$$$\\ When$\quad C^{\prime}=135^{{}^{\circ}}$, then$$\begin{aligned}& \quad A=180^{{}^{\circ}}-\left(135^{{}^{\circ}}+30^{{}^{\circ}}\right)=15^{{}^{\circ}} \\ & \text{ Area of } \triangleA B C=\frac{1}{2}A B \timesA C \sin A \\ & =\frac{1}{2} \times4 \times2 \sqrt{2} \sin \left(105^{{}^{\circ}}\right) \\ & =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \\ & \text{ Area of } \triangleA B C^{\prime}=\frac{1}{2}A B \timesA C \sin A \\ & =\frac{1}{2} \times4 \times2 \sqrt{2} \sin \left(15^{{}^{\circ}}\right)=2(\sqrt{3}-1) \end{aligned}$$Difference of areas of triangles =|2($\sqrt{3}+$1)-2($\sqrt{3}-$1)|=4$$\\ A D=2, D C=2 Difference of areas of$\triangle A B C and $\triangle A B C^{\prime}$$\begin{aligned}& =\text{ Area of } \triangleA C C^{\prime} \\ & =\frac{1}{2}A D \timesC C^{\prime}=\frac{1}{2} \times2 \times4=4 \end{aligned}$$