Question

ii) In A ABC, ZABC = 135°, AB = 18, BC = 12 then find area of A ABC​

Answer 1

Answer:

76.3 cm square

Step-by-step explanation:

Given:

∠ABC = 135°

Construction:

Extend AB to a point D such that CD⊥AD

Solving:

∠CBD = 180 - 135 = 45°

In right triangle CDB, right angeled at D

$sin(B) = \frac{CD}{BC}\\sin(45) = \frac{CD}{12}\\\frac{1}{\sqrt2} =\frac{CD}{12}\\CD = \frac{1}{\sqrt2}*12 = 6\sqrt2$

Area = 0.5 * CD * AB

= 0.5 * $6\sqrt2$ * 18 ≈ 76.3