Question

ABCD is a square of side x CM. PQRS is a rectangle of dimensions (x+3) CM by (x-3) CM. Show that ABCD And PQRS have equal perimeters. Find the area of PQRS. Which figure has the greater area? How much more is it's area in square centimeters?​

Answer 1

Step-by-step explanation:

Side of square ABCD = x cm

Therefore,

Perimeter of square ABCD = 4x cm.... (1)

Dimensions of rectangle PQRS = (x+3)cm by (x-3) cm

Therefore,

Perimeter of rectangle PQRS = 2[(x+3)+(x-3)]

=2[x+3+x-3]

=2*2x

Perimeter of rectangle = 4x..... (2)

From equations (1) & (2), we find:

Perimeter of square ABCD = Perimeter of rectangle PQRS

$Area \: of \: square \: ABCD = {x}^{2} \: square \: cm\\ Area \: of \: rectangle \: PQRS \\ = (x + 3)(x - 3) = {x}^{2} - 9 \: square \: cm\\ \therefore \: Area \: of \: rectangle \: PQRS \: = Area \: of \: square \: ABCD - 9$

Which means, area of square is greater than the area of rectangle by 9 square cm.

Hence proved.

Answer 2

Answer:

Side of square ABCD = x cm

Therefore,

Perimeter of square ABCD = 4x cm.... (1)

Dimensions of rectangle PQRS = (x+3)cm by (x-3) cm

Therefore,

Perimeter of rectangle PQRS = 2[(x+3)+(x-3)]

=2[x+3+x-3]

=2*2x

Perimeter of rectangle = 4x..... (2)

From equations (1) & (2), we find:

Perimeter of square ABCD = Perimeter of rectangle PQRS

Which means, area of square is greater than the area of rectangle by 9 square cm.

Hence proved.

Step-by-step explanation: