In triangle ABC, side AC = 6, C = 135 degrees and height BD = 2. Find the area of the triangle ABD.

The problem can be solved in different ways, let’s consider them.
Option 1.
Height BD is lowered to the extension of the AC side. Consider a right-angled triangle BDC, in it:
∠С = 180 ° – 135 = 45 °, ∠В = 90 ° – ∠С = 90 ° – 45 ° = 45 °. Therefore, the triangle is isosceles, BD = CD = 2.
By the Pythagorean theorem, we find the hypotenuse of BC:
ВС = √ (BD² + CD²) = √8 = 2√2.
Consider a triangle ABC, in which we know two sides and the angle between them.
We find the area by the formula:
S ABC = 1/2 * AC * BC * sin 135 ° = 1/2 * 6 * 2√2 * √2 / 2 = 6.

Option 2.
Find the area of ​​a right-angled isosceles triangle BDC ^
S BDC = 1/2 * CD * BD = 1/2 * 2 * 2 = 2
Find the area of ​​a right-angled triangle ADB. AD = AC + CD = 6 + 2 = 8, BD = 2
S ADB = 1/2 * AD * BD = 1/2 * 8 * 2 = 8.
Find the area of ​​the triangle ABC:
S ABC = S ADB – S BDC = 8 – 2 = 6.
Answer: the area of ​​the triangle ABC 6.