The base angle of an isosceles trapezium is 45°. If the shorter side and both the equal sides are 10 cm each, what is the area of the trapezium?

In $$\triangle$$ ADE, $$cos 45 = \frac{AE}{AD}$$

=> $$\frac{1}{\sqrt{2}} = \frac{AE}{10}$$

=> $$AE = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$

Similarly, $$BF = 5\sqrt{2}$$

Also, in $$\triangle$$ ADE, $$sin 45 = \frac{DE}{AD}$$

=> $$\frac{1}{\sqrt{2}} = \frac{DE}{10}$$

=> $$DE = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$

=> AB = AE + EF + BF = $$5\sqrt{2} + 10 + 5\sqrt{2} = 10 + 10\sqrt{2}$$

$$\therefore$$ Area of trapezium = $$\frac{1}{2} \times$$ sum of parallel sides $$\times$$ height

= $$\frac{1}{2} \times (CD + AB) \times DE$$

= $$\frac{1}{2} \times (10 + 10 + 10\sqrt{2}) \times (5\sqrt{2})$$

= $$(10 + 5\sqrt{2}) \times 5\sqrt{2}$$

= $$50\sqrt{2} + 50$$ sq. cm