Question

prove that area of Trapezium is equal to
$1 \div2 \times base \times hieght$
​

Answer 1

area of trapezium is 1/2* sum of parallel sides *height
proving is in this way

Answer 2

$\underline{\Huge\mathfrak{Question-}}$

To prove that area of Trapezium ;-

$= \frac{1}{2} \times base \: \times height$

$\underline{\Huge\mathfrak{Question-}}$

To prove ;-

•Let assume a ||gm ABCD.
•In which , BD be the diagonal forming two triangles ; ∆BCD & ∆ABC.
•Let DE be the height of the trapezium.

Now ,
• Area of trapezium ABCD = Area of ∆ABD + Area of ∆BCD.

So ,

• The area of ∆ABD = 1/2 × b × h i.e. => 1/2 × AB × DE ..............(i)

• In the same way , area of ∆BCD = 1/2 × b × h i.e. => 1/2 × DC × DE.......................(ii)

Note ; Both the ∆'s will have the height as they lie between the same parallel lines.

Than ,
The area of Trapezium = Sum of (i) & (ii).

So ,
$= > (\frac{1}{2} \times ab \times de) + ( \frac{1}{2} \times dc \times de)$

$= > \frac{1}{2} \times \: de \: (ab \: + \: dc)$

Therefore , Proved !