Question

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

Answer 1

Answer:

47 cm²

Step-by-step explanation:

In the given figure, ABCD is a rectangle.

$\angle ABN=90^\circ$   (∵ angle of a rectangle is right angle)

A rectangular piece of paper ABCD is folded along the line MN so that

vertex C coincides with vertex A.

Therefore, ANMD' is a parallelogram.

AD'M is a right angle triangle.

where, AD' = 4 cm and MD' = 8 cm

Area of triangle AD'M $=\dfrac{1}{2}\times AD'\times MD'$

                                    $=\dfrac{1}{2}\times 4\times 8$

                                    $=16\text{ cm}^2$

Area of parallelogram = 2 times ar(AD'M)

Area of parallelogram = 2 × 16

                                     = 32 cm²

BC = 16                               (side of rectangle)

BN + AN = 16------(1)           (∵ NC = AN )

In ΔABN, ∠B = 90°

$AB^2+BN^2=AN^2$                   (Pythagoreous theorem)

$AN^2-BN^2=16$---------(2)

Using (1) and (2)

$BN=\dfrac{15}{2}$

Area of triangle ABN $=\dfrac{1}{2}\times AB\times BN$

                                    $=\dfrac{1}{2}\times 4\times \dfrac{15}{2}$

                                    $=15\text{ cm}^2$

Area of pentagon ABNMD' = ar(ABMD') + ar(ABN)

                                            = 32 + 15

                                            = 47 cm²

Hence, the area of pentagon is 47 cm²