Find the area of a trapezium whose parallel sides are 25 cm,13 cm and other sides are 15 cm each.

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Solution

In trapezium ABCD,parallel sides are AB and DC.

$∴ A B = 25 c m, C D = 13 c m a n d o t h e r s i d e s a r e 15 c m e a c h i . e, A D = C B = 15 c m F r o m C, d r a w E | | D A a n d C L ⊥ A B ∴ A E = D C = 13 c m a n d E B = A B - A E = 25 - 13 = 12 c m ∴ P e r p e n d i c u l a r C L b i s e c t s t h e b a s e E B o f t h e i s o s c e l e s t r i a n g l e C E B t h e r e f o r e E L = L B = \frac{12}{2} = 6 c m N o w i n r i g h t △ C L B, C B^{2} = C L^{2} + L B^{2} (P y t h a g o r a s t h e o r e m) \Rightarrow (15)^{2} = C L^{2} + (6)^{2} \Rightarrow 225 = C L^{2} + 36 \Rightarrow C L^{2} = 225 - 36 = 189 = 9 \times 21 ∴ C L = \sqrt{9 \times 21} = 3 \sqrt{21} c m N o w, a r e a o f t r a p e z i u m = \frac{1}{2} (S u m o f p a r a l l e l s i d e s) \times h = \frac{1}{2} (25 + 13) \times 3 \sqrt{21} c m^{2} = \frac{1}{2} \times (38) \times 3 \sqrt{21} c m^{2} = 19 \times 3 \sqrt{21} c m^{2} = 57 \sqrt{21} c m^{2}$

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