# In DeltaABC,/_BAC=30^@;/_ACB=60^@ and BC = 6 cm. How will you find out the area of the triangle without using trigonometry?

Jun 18, 2016

$18 \sqrt{3} c {m}^{2}$

#### Explanation:

Given

• $\text{In } \Delta A B C$

$\to \angle A C B = {60}^{\circ}$

$\to \angle B A C = {30}^{\circ}$

$\to R e s t \angle A B C = {90}^{\circ}$

$\to \mathmr{and} B C = 6 c m$

Construction

$C D = B C \text{ is cut off from " CA " and B,D are joined}$

Analysis

$\text{In } \Delta B D C$

$\angle B C D = {60}^{\circ} , B C = C D \to \angle C D B = \angle C B D = {60}^{\circ}$

$\therefore \Delta B D C \text{ is equilateral } \to B C = C D = B D = 6 c m$

$\text{In } \Delta A B D$

$\angle A B D = \angle A B C - \angle D B C = \left(90 - 60\right) = {30}^{\circ}$

$\mathmr{and} \angle B A D = {30}^{\circ} \implies \Delta A B D \to \text{isosceles triangle}$

$\implies A D = B D = 6 c m$

$\text{Now } A C = A D + C D = \left(6 + 6\right) c m = 12 c m$

$\text{ Now applying Pythagoras theorem for } \Delta A B C$

$A B = \sqrt{A {C}^{2} - B {C}^{2}} = \sqrt{{12}^{2} - {6}^{2}} = 6 \sqrt{3} c m$

$\text{ Area of } \Delta A B C = \frac{1}{2} \times A B \times B C = \frac{1}{2} \cdot 6 \sqrt{3} \cdot 6 = 18 \sqrt{3} c {m}^{2}$