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 √{

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9.Straight Line

hard

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

$2$

$9$

$4$

$5$

(IIT-2009)

$\text { Using cosine rule } \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{a^2+16-8}{2 \times a \times 4}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+8}{8 a}$

$\Rightarrow a^2-4 \sqrt{3} a+8=0$

$\Rightarrow a_1+a_2=4 \sqrt{3}, a_1 a_2=8$

$\Rightarrow\left|a_1-a_2\right|^2=\left(a_1+a_2\right)^2-4 a_1 a_2=48-32=16$

$\Rightarrow\left|a_1-a_2\right|=4$

$\therefore\left|\Delta_1-\Delta_2\right|=\frac{1}{2}\left|a_1-a_2\right| \cdot c \sin$ $B=\frac{1}{2} \times 4 \sin 30^{\circ} \times 4=4$

Standard 11

Mathematics

normal

easy

hard

(JEE MAIN-2019)

normal

normal