the parallel side of a trapezium are 65m and 40m.its non parallel sides are 39m and 56m.find the area of the trapezium
Answers
--> That's an interesting case out of trapezium ...
--> However, a construction of altitudes would work just fine...
--> Suppose AB = 40m
BD = 39m
CD = 65m
AC = 56m
Let us assume, AE = k || CE = x ...
Pythagoras on ΔACE gives :->
-> k² + x² = 56² ---> ( i )
Further, AB = FE { since, AE, BF are ||s }
=> EF = 40m => DF = ( 65 - 40 ) - CE = 25 - x
However, BF = AE = k --> { ||gm }
--> Pythagoras on ΔBFD gives :->
--> k² + ( 25 - x )² = 39² ---> (ii)
--> ( ii ) - ( i ) :->
-50x + 625 = 39² - 56² = -1615
=> 50x = 2240 => x = 44.8 m
Putting this in ( i ) and solving for 'k' , we have :->
--> Height of trapezium = 33.6m
Hence, Area of Trapezium = 1/2 * [ 40 + 65 ] * k = 1/2 * 105 * 33.6 m²
And therefore, the req.d Area is 1764 m²....
Given : The parallel sides of a Trapezium are 65 metre and 40 metre
its non parallel sides are 39 and 56
To find : the area of the trapezium
Solution:
Ref attached figure :
DE and CF ⊥ AB
DE = CF = h cm
BF = x cm
=> AE = 65 + x - 40 - x = 25 + x cm
h² = 39² - x²
h² = 56² - (25 + x)²
=> 56² - (25 + x)² = 39² - x²
=> 3136 - 625 - x² - 50x = 1521 - x²
=> 50x = 990
=> x = 99/5
h² = 39² - x²
=> h² = 39² -(99/5)²
=> h² = 39² -(99/5)²
=> h = 168/5 cm
Area of trapezium = (1/2)(65 + 40) 168/5
= 1764 cm²
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