Question

Q. A rectangular piece of paper ABCD measuring 4cm x 16 cm
is folded along the line MN so that vestes c concedes with
vortex A, as shown in the picture. What is the area of the
pentagon ABNMD' ?​

Answer 1

Answer

__________________

• Given Dimension of the piece of paper is 4 × 16cm

That means AB=4cm and BC=16cm

Let us find the area of rectangular piece of

___________________________________________

paper :-

_______

Area =

$= > 4 \times 16 \\ \\ = > \:64cm^{2}$

MN divided the Area of the rectangular piece so

$area \: of \: abnm = \frac{1}{2}abcd$

Area of ABNM= ½×64

Area of ABNM= 32cm²✓

Now:

Given when CD is folded the vertice C coincides with vertice A

so AD= CD= 4 cm

And DM= ½ × BC

=> ½ × 16=8cm

So area of ∆ ADM = ½ × AD × DM...( base ×height )/2

ar(∆ADM ) = ½×8×4=16cm²

Now area of the pantagon

________________________________________ BADMN is :-

________

Sum of areas of (ABNM+ADM)

$area \: of \: the \: \: pantagone = 16 + 32 \\ \\ = > 48cm ^{2}$