Question

in the given figure ABCD is a Quadrilateral. Prove that ar(quad. ABCD) =1/2*BD*(AM+CN), where AM and CN are perpendiculars on diagonal BD.

Answer 1

In∆ABD,
AM is height and BD is base
Area of ∆ADB=1÷2×base×height
=1÷2×AM×BD___(1)
In ∆CBD,
CN is height and BD is base
Area of ∆CBD=1÷2×base×height
=1÷2×CN×BD___(2)
AREA of ABCD=Area of ∆ADB+Area of CBD
=1÷2×AM×BD+1÷2×CN×BD
=1÷2×BD×(AM+CN)

Answer 2

Area(quad. ABCD) = $\frac{1}{2}\times \text{BD} \times (\text{AM + CN)}$.

Step-by-step explanation:

We are given the following figure in which ABCD is a Quadrilateral.

To prove: ar(quad. ABCD) =1/2*BD*(AM+CN), where AM and CN are perpendiculars on diagonal BD.

As we can see that in the figure; there are two triangles ABD and BCD.

Firstly, considering the triangle ABD;

The area of the triangle ABD = $\frac{1}{2}\times \text{Base} \times \text{Height}$

                                                 = $\frac{1}{2}\times \text{BD} \times \text{AM}$

Similarly, considering the triangle BCD;

The area of the triangle BCD = $\frac{1}{2}\times \text{Base} \times \text{Height}$

                                                = $\frac{1}{2}\times \text{BD} \times \text{CN}$

Now, adding both the areas of triangles, we get;

Area(ABD) + Area(BCD) = $\frac{1}{2}\times \text{BD} \times \text{AM}$ + $\frac{1}{2}\times \text{BD} \times \text{CN}$

Area(quad. ABCD) = $\frac{1}{2}\times \text{BD} \times (\text{AM + CN)}$

Hence, proved.