A geometry problem by Aaryan Maheshwari

Let the length of the rectangle be denoted by $l$ and its breadth by $b$ . Then, ATQ, $2(l+b)=46 \Rightarrow l+b = 23$ and $lb=120$ .

Now, we know that:

$(l+b)^2= l^2+b^2+2lb \Rightarrow l^2+b^2=(l+b)^2-2lb$ .

Putting values, we get:

$l^2+b^2= (23)^2-(2*120) = 289$ .

$\therefore$ Length of the Diagonal: $\sqrt{l^2+b^2}=\sqrt{289}=17$

let $L$ be the length and $W$ be the width and $d$ be the diagonal of the rectangle.

$P = 2(W + L) = 46$

After simplifying the above equation, we get

$W + L = 23$ $\implies$ (equation 1)

$A = LW = 120$

After simplifying the above equation, we get

$L = \frac{120}{W}$ $\implies$ (equation 2)

Substitute equation 2 in equation 1, then simplify and solve for $W$ and $L$

$W + \frac{120}{W} = 23$

After simplifying the above equation, we get

$W^2 - 23W + 120 = 0$

Solving the above equation by using the quadratic formula, we get

$W = 15$

$W = 8$

Based from the above values, we can conclude that $L$ can be $15$ or $8$ .

Now by Pythagorean Theorem, we get

$d = \sqrt{15^2 + 8^2} = 17$ $cm$