The area of a trapezium is $475 c m^{2}$ and the height is 19 cm, then the lengths of smaller side of two parallel sides if one side is 4 cm greater than the other is

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Solution

$L e t t h e l e n g t h o f o n e p a r a l l e l s i d e o f t r a p e z i u m i s a . S o l e n g t h o f o t h e r p a r a l l e l s i d e = a + 4 G i v e n t h e a r e a o f t r a p e z i u m = 475 & t h e h e i g h t = 19 N o w a r e a o f a t r a p e z i u m = \frac{1}{2} \times h e i g h t \times (s u m o f t h e p a r a l l e l s i d e s) \Rightarrow \frac{1}{2} \times 19 \times [a + (a + 4)] = 475 \Rightarrow 2 a + 4 = \frac{475 \times 2}{19} \Rightarrow 2 a + 4 = 25 \times 2 \Rightarrow 2 a + 4 = 50 \Rightarrow 2 a = 50 - 4 \Rightarrow 2 a = 46 \Rightarrow a = \frac{46}{2} = 23 ∴ L e n g t h o f s m a l l e r p a r a l l e l s i d e o f t r a p e z i u m = 23 & L e n g t h o f o t h e r p a r a l l e l s i d e o f t r a p e z i u m = 23 + 4 = 27$

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