Question

$\square \mathrm{ABCD}$ is a square.$\square \mathrm{PQRS}$ is a square formed by joining the mid-points of $\mathrm{AB},\mathrm{BC},\mathrm{CD},$ and $\mathrm{AD}$respectively $\square \mathrm{EFGH}$is a square formed by joining the mid-points of$\mathrm{PQ},\mathrm{QR},\mathrm{RS}$ and $\mathrm{PS}$ respectively. Find the sum of the perimeters of $\square \mathrm{ABCD},\square \mathrm{PQRS}$ and $\square \mathrm{EFGH}$ if the area of the square$\square \mathrm{EFGH}$ is $25c{m}^2$.

Answer 1

Step1: Finding the Sides of each of the squares:

Given area of square$\square \mathrm{EFGH}=25{\mathrm{cm}}^2$

$\mathrm{Side}\mathrm{of}\mathrm{square}=\sqrt{\mathrm{Area}\mathrm{of}\mathrm{Square}}$

Side of square $\square \mathrm{EFGH},\mathrm{a}=\sqrt{25}=5\mathrm{cm}$

The side of Square $\square \mathrm{PQRS}$ is equal to the diagonal of the $\square \mathrm{EFGH}$

$\mathrm{Diagonal}\mathrm{of}\mathrm{square}=\mathrm{side}\sqrt{2}$

Side of Square $\square \mathrm{PQRS}=\mathrm{a}\sqrt{2}=5\sqrt{2}\mathrm{cm}$

Side of square $\square \mathrm{ABCD}$ is equal to the diagonal of the $\square \mathrm{PQRS}$

Side of$\square \mathrm{ABCD}=5\surd 2\times \surd 2$

$=10\mathrm{cm}$

Step2: Finding the perimeter of Squares:

The perimeter of the Square $=4\times \mathrm{side}$

Perimeter of $\square \mathrm{ABCD}=4\times 10=40\mathrm{cm}$

Perimeter of $\square \mathrm{PQRS}=4\times 5\surd 2=20\surd 2\mathrm{cm}$

Perimeter of$\square \mathrm{EFGH}=4\times 5=20\mathrm{cm}$

The Sum of perimeters $=40+20\surd 2+20$

$=(60+20\surd 2)cm$

Hence, The Sum of perimeters is $(60+20\surd 2)cm$