Question

⭐the lengths(in cm) of parallel sides of a trapezium are 2x and 3x-1,and the distance between the parallel sides is x+1.if the area of the trapezium be 28cm^2,find the smaller of the two parallel sides⭐

CLASS 10 CHAPTER :QUADRATIC EQUATIONS

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Answer 1

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$<B><I>HERE IS YOUR ANSWER$


$\underline{GIVEN}$


For a trapezium


parallel sides => 2x & 3x - 1


Distance between parallel sides i. e. height = x + 1


Area of trapezium = 28 cm²


$\underline{\:TO\:FIND}$


Smaller side of parallel sides = ?


$\underline{SOLUTION}$


We know that


area of trapezium = 1/2 × sum of parallel sides × height


28 = 1/2 × ( 2x + 3x - 1 ) × ( x + 1 )


=> 56 = ( 5x - 1 )( x + 1 )


=> 56 = 5x² - x + 5x - 1


=> 5x² + 4x - 57 = 0


=> 5x² - 15x + 19x - 57 = 0


=> 5x( x - 3 ) + 19( x - 3 ) = 0


=> ( x - 3 )( 5x + 19 ) = 0


=> x - 3 = 0. or. 5x + 19 = 0


=> x = 3. or. x = -19/5


x = - 19/5 is not acceptable since side can't be negative


so


x = 3 .


2x = 2 × 3 = 6 cm


3x - 1 = 3× 3 - 1 = 8 cm


So smaller side among the parallel sides is of 6cm .

Answer 2

Formula to be applied : -

$\boxed{\mathsf{ Area\:of\:trapezium= \dfrac{sum\:of\:lengths\:of\:parallel\:side}{2}\times height}}$

           Solution : -  

Given,

Length of 1st parallel side = 2x cm

Length of 2nd parallel side = 3x - 1 cm

Distance( height ) between ║ sides = x + 1 cm

Area of trapezium = 28 cm^2

On the basis of the formula given above : -

⇒ 28  = [ 2x + 3x - 1 ] / 2 × ( x + 1 )

⇒ 28 = [ ( 5x - 1 )( x + 1 ) ] / 2  

⇒ 56 = ( 5x - 1 )( x + 1 )

⇒ 56 = 5x^2 + 5x - x - 1

⇒ 56 = 5x^2 + 4x - 1

⇒ 5x^2 + 4x - 1 - 56 = 0

⇒ 5x^2 + 4x - 57 = 0

⇒ 5x^2 + ( 19 - 15 ) x - 57 = 0

⇒ 5x^2 + 19x - 15x - 57 = 0

⇒ 5x^2 - 15x + 19x - 57 = 0

⇒ 5x( x - 3 ) + 19( x - 3 ) = 0

⇒ ( x - 3 )( 5x + 19 ) = 0

∴ x = 3  or x - 19 / 5

Substitute the value of x in 2x and 3x - 1.

Length of 1st ║ side = 2x cm = 2( 3 ) cm = 6 cm

Length of 2nd ║ side = 3x - 1 cm = 3( 3 ) - 1 cm = 9 - 1 cm = 8 cm

Lengths of parallel sides are 6 cm and 8 cm.

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