Question

Prove that the area of a trapezium is equal to 1/2×b×h​

Answer 1

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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 !