Question

parallel sides of trapezium are 65 and 40 its non parallel sides are 39 and 56 metre find the area of trapezium​

Answer 1

Given:

The parallel sides of trapezium = 65 and 40 units and

The non parallel sides of trapezium = 39 and 56 units

We have to find, the area of trapezium.

Solution:

∴ $x^2+h^2= 56^2$

⇒ $(25-x)^2+h^2 = 39^2$

⇒ $x^2-56^2 = (25-x)^2 - 39^2$

⇒ x = $\dfrac{224}{5}$ units

∴ $h^2 = 56^2-x^2$

⇒ $h^2 = 56^2-(\dfrac{224}{5} )^2$

⇒ $h^2 = \dfrac{28224}{25}$

⇒ $h^2 = (\dfrac{168}{5})^2$

⇒ $h = \dfrac{168}{5}$units

We know that:

The area of trapezium​ = $\dfrac{1}{2} \times$ (Sum of parallel sides) × (Distance between them)

= $\dfrac{1}{2} \times$ (40 + 65) × $\dfrac{168}{5}$

=  (105) × $\dfrac{84}{5}$

= 21 × 84 square units

= 1764 square units

Thus, the area of trapezium​ = 1764 square units

Answer 2

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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