Question

A square of sidelength a, a rectangle of length l and width w have the same exact area. Moreover, the diagonal of the rectangle is twice as long as the diagonal of the square. Then, what could be said about the perimeter P(r) of the rectangle compared to the perimeter P(s) of the square?

• A. $\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{2}$
• B. $\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{\frac{2}{3}}$
• C. $\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{\frac{3}{2}}$
• D. $\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{\frac{5}{2}}$

Answer 1

The correct option is D $\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{\frac{5}{2}}$

$Sidelengthofthesquare=aAreaofthesquare=a\times a=a^2$

$Lengthoftherectangle=lWidthoftherectangle=wAreaoftherectangle=l\times w$

As both the square and the rectangle have the same area, we have:

$a^2=lw$

Now, calculating the perimeter of the rectangle, we have:

$Perimeter\left(P\right(r\left)\right)=2\times (Length+Width)=2(l+w)$

Squaring the perimeter, we get:

$\left[P\right(r){]}^2=[2(l+w){]}^2=4(l+w{)}^2=4(l^2+2lw+w^2)=4(l^2+2a^2+w^2)$

We know that the diagonal of the rectangle is twice as long as that of the square. So, we find:


$D=2d⟹D^2=(2d{)}^2⟹D^2=4d^2$

Using the Pythagorean theorem in both the square and the rectangle, we find:

$D^2=4d^2⟹l^2+w^2=4(a^2+a^2)⟹l^2+w^2=8a^2$

Combining this with the square of the perimeter of the rectangle, we get:

$\left[P\right(r){]}^2=4(l^2+2a^2+w^2)=4(8a^2+2a^2)=40a^2$

Taking the square root of both sides of the equation, we find:

$P\left(r\right)=\sqrt{40}a=\sqrt{4\times 10}a=2\sqrt{10}a$

Now, the perimeter of a square of sidelength a is:

$P\left(s\right)=4a$

So, dividing the perimeter of the rectangle by the perimeter of the square, we get:

$\frac{P\left(r\right)}{P\left(s\right)}=\frac{2\sqrt{10}a}{4a}=\frac{\sqrt{10}a}{2a}=\frac{\sqrt{5\times 2}a}{\sqrt{2\times 2}a}=\sqrt{\frac{5}{2}}$

Hence, comparing the perimeters of the rectangle and the square, we get:

$\frac{P\left(r\right)}{P\left(s\right)}=\sqrt{\frac{5}{2}}$