Question

Length of two sides of a triangle are 20cm and 16cm and the angle between them is 135°.

a. Draw a rough figure and mark the measurements.

b. Find the perpendicular distance of the vertices to the side of length 20cm.

c. Find the area of the triangle.

Answer 1

b. Now AADB is a right triangle with angles 45°, 45°, 90°. Since the side opposite to 90° angle is 16cm, the other two sides are 8√2 each.

= \(8\sqrt{2}\) ie 20 cm

c. Area of ∆ABC = \(\frac{1}{2}\) bh

= \(\frac{1}{2}\) BC x AD = \(\frac{1}{2}\) x 20 x \(8\sqrt{2}\) = \(80\sqrt{2}\) cm²