Question

Solve please
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Answer 1

In Given Quadrilateral ABCD.

Consider Triangle BCD.

➡️In which MC is Altitude Of Triangle.

➡️ BD is base of Triangle.

$\boxed{\mathtt{Ar\:of\: Triangle=\frac{1}{2}\times\:Base\:\times\:Altitude}}$

$\boxed{\mathtt{Ar\:of\: \triangle(BCD)=\frac{1}{2}\times\:BD\:\times\:MC}}$

Consider Triangle ABD

➡️In which LA is Altitude of Triangle.

➡️BD is base of Triangle.

$\boxed{\mathtt{Ar\:of\: \triangle(ABD)=\frac{1}{2}\times\:LA\:\times\:BD}}$

Area of Given Figure.

Area of Triangle BCD+Area of Triangle ABD

$( \frac{1}{2} \times mc \times bd )+( \frac{1}{2} \times la \times bd)$

Common can be Done

$\frac{1}{2} \times bd$

Now,

$\frac{1}{2} \times bd(mc + la)$

Hence Its Proven.

$\boxed{\mathtt{Ar.\:of\:ABCD=\frac{1}{2}\times\:BD\times(CM+LA)}}$

Answer 2

Given:

A quadrilateral ' ABCD '

In which ' BD ' is the diagonal.

And 'BCD' and 'ABD' are two triangles with same base 'BD'

'AL' is perpendicular to ' BD' and. 'CM' is perpendicular to 'BD'

To Prove:

ar of quad ABCD = 1/2 × BD × ( AL + CM)

Proof:

We know that: area of a triangle = 1/2 × base × height

So, ar of ∆ BCD = 1/2 × BD × CM

ar of ∆ BAD = 1/2 × BD × AL

Since, ar of quad ( ABCD ). = ( ar ∆ BCD ) + ( ar ∆ BAD )

ar of quad ( ABCD ) = (1/2 × BD × CM ) + (1/2 × BD × AL )

ar of quad ( ABCD )= 1/2 × BD ( CM + AL).

{ By taking " 1/2 × BD " as Common }

Ar of quad ( ABCD ) = 1/2 × BD ( CM + AL )

Hence proved.

(Note: Perpendicular of any triangle is always considered as height of the triangle)