Question

In ΔABC, AB = 10 cm. AC = 8 cm, ∠A = 45° 

a. Find the perpendicular distance from C to AB. 

b. Find the area of the triangle.

Answer 1

a. Draw CD perpendiculer to AB.

Angles of △ADC are 45°,45°,90°

CD = \(4\sqrt{2}\) cm.

b. Area = \(\frac{1}{2}\times10\times4\sqrt{2}\)

= \(20\sqrt{2}\) sq. cm