Answer:
(C) 18(1 + √ 3)
Step-by-step explanation:
Given,
In triangle ABC,
AC = 12 unit,
m∠A = 30°,
m∠B = 45°,
Let D∈AB such that CD ⊥ AB,
In triangle ADC,
[tex]sin A = \frac{CD}{AC}[/tex]
[tex]sin 30^{\circ}=\frac{CD}{12}[/tex]
[tex]\frac{1}{2}=\frac{CD}{12}[/tex]
[tex]\implies CD=6\text{ unit}[/tex],
By the pythagorean theorem,
[tex]AC^2=CD^2+AD^2\implies 144 = 36 + AD^2\implies 108=AD^2\implies AD=6\sqrt{3}\text{ unit}[/tex]
Now, in triangle CDB,
[tex]tan 45^{\circ}=\frac{CD}{DB}[/tex]
[tex]1=\frac{6}{DB}\implies DB=6\text{ unit}[/tex]
[tex]AB=AD+DB=6\sqrt{3}+6=6(\sqrt{3}+1)\text{ unit}[/tex]
Hence, the area of the triangle ABC = [tex]\frac{1}{2}\times AB\times CD[/tex]
[tex]=\frac{1}{2}\times 6(\sqrt{3}+1)\times 6[/tex]
[tex]=18(\sqrt{3}+1)\text{ square unit}[/tex]
Option 'C' is correct.
