Properties of Triangles 2 Question 17
17. 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
(2009)
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Solution:
- In $\triangle A B C$, by sine rule, $\frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{\circ}}=\frac{4}{\sin C}$
$$ \Rightarrow \quad C=45^{\circ}, C^{\prime}=135^{\circ} $$
When, $C=45^{\circ} \Rightarrow A=180^{\circ}-\left(45^{\circ}+30^{\circ}\right)=105^{\circ}$
When, $C^{\prime}=135^{\circ} \Rightarrow A=180^{\circ}-\left(135^{\circ}+30^{\circ}\right)=15^{\circ}$
Area of $\triangle A B C=\frac{1}{2} A B \times A C \sin A$
$$ =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(105^{\circ}\right) $$
$$ =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}} $$
$=2(\sqrt{3}+1)$ sq. units
Area of $\triangle A B C^{\prime}=\frac{1}{2} A B \times A C \sin A$
$$ \begin{aligned} & =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(15^{\circ}\right) \\ & =2(\sqrt{3}-1) \text { sq. units } \end{aligned} $$
Difference of areas of triangle
$$ =|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4 \text { sq units } $$
Alternate Solution
Here, $A D=2, D C=2$
Difference of areas of $\triangle A B C$ and $\triangle A B C^{\prime}$
$=$ Area of $\triangle A C C^{\prime}$
$=\frac{1}{2} A D \times C C^{\prime}=\frac{1}{2} \times 2 \times 4=4 sq$ units
