Question

Prove that area of trapezium is 1\2*(l+b)*h

Answer 1

Hey friend, Harish here.

Here is your answer:

To prove:

Area of trapezium formula.

Proof:

We know that;

$Area\ of\ trapezium = Area\ of\ 2\ triangles + Area\ of\ rectangle$

Here,
We know that,

Area of rectangle = Length × Height =  DC × h.

Now,

In ΔADX a right angled triangle . 

 area of the triangle = 1/2 × AX × h

In ΔCBY a right angled triangle.

The area = 1/2 × YB × h.

Now adding all the three area we get area of the trapezium.


$(DC\times h) + ( \frac{1}{2}\times AX \times h) + ( \frac{1}{2}\times YB \times h)$

⇒  $( \frac{1}{2}\times h) (AX +YB + 2DC )$

(Here we took half times height as common).

We DC = XY.

Then,

$( \frac{1}{2}\times h) (AX +YB + 2DC ) = ( \frac{1}{2}\times h) (AX +YB + 2XY )$

⇒  $( \frac{1}{2}\times h) ((AX +YB + XY) + XY )$

We know that,

AX + YB + XY = AB

Then,

⇒ $( \frac{1}{2}\times h) ((AX +YB + XY) +XY )=( \frac{1}{2}\times h) (AB + DC )$

( Here we made One XY into DC as they both are 3 equal.)

Therefore Area of trapezium = 1/2 × height × (Sum of Parallel sides).
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Hope my answer is helpful to you.