JEE Main 2025 — 2 April, Morning Shift — Question 25

$\operatorname{ar}(\triangle A B C)=5$

$\frac{1}{2} \times b c=5$

$\Rightarrow b c=10$

$\operatorname{ar}(\triangle A C D)=6$

$\frac{1}{2} \times c d=6$

$\Rightarrow c d=12$

$\operatorname{ar}(\triangle A B D)=7 \quad x^{\prime}(b, 0,0)$

$b d=14$ area $(\triangle B C D)=\frac{1}{2}|\overrightarrow{B C} \times \overrightarrow{B D}|$

$\overrightarrow{B C}=\langle-b, c, 0\rangle$

$\overrightarrow{B D}=\langle-b, 0, d\rangle$

$|\overrightarrow{B C} \times \overrightarrow{B D}|=\sqrt{c^{2} d^{2}+b^{2} d^{2}+b^{2} c^{2}}$

$=\sqrt{12^{2}+14^{2}+10^{2}}$

$=\sqrt{440}$

$\operatorname{ar}(\triangle B C D)=\sqrt{110}$