The length of a certain rectangle is 4 inches longer than its width.

Let the width of the rectangle be x,then the length will be x+4.
The area is 221.
so,
x(x+4)=221
or \(x^2+4x-221=0\)
or \((x-13)(x+17)=0\)
or x=13

width is 13 so the lenght will 13+4= 17
So, The diagonal of the rectangle will be
\(\sqrt{13^2 + 17^2}\) = \(\sqrt{458}\)

Clearly, the sum lies between \(21^2\) i.e 441 and \(22^2\) i.e 484

Answer:- C

L = 17, B = 13
169+289 = 458 is between 441 and 484 hence between 21 and 22. answer C

Width=x, Length=x+4
x*x+4=221
Solving for x, x=13,x+4=17
Therefore using pythagoras theorem, Diagonal^2=169+289=458
Diagonal=More than 21 but less than 22
Answer C

MANHATTAN GMAT OFFICIAL SOLUTION:

First, set up an equation for the area of the rectangle. If x is the width, then we have

x(x + 4) = 221

Note that if we put this into standard quadratic form and then try to factor, we wind up back where we started, in some sense: we are looking for two numbers that multiply to 221 and that differ by 4.

x^2 + 4x – 221 = 0

Beyond pure trial and error, we can look for nearby squares. 221 is nearly 225, which equals 15^2. So we might try numbers near 15. As it turns out, 221 = 13 × 17. We might even get there by noticing a difference of squares:

221 = 225 – 4 = 15^2 – 2^2 = (15 – 2)(15 + 2) = 13 × 17.

As a last resort, we could always use the quadratic formula, which gets us the roots of the equation as well.

Now, the diagonal of the rectangle will be given by the Pythagorean Theorem:

d^2 = 13^2 + 17^2
= 169 + 289
= 458.

The square root of 458 is definitely larger than 20, since 20^2 = 400. Going up, we can compute 21^2 = 441 < 458, whereas 22^2 = 484 > 458. So the length of the diagonal must be between 21 and 22.

The correct answer is C.

SHORTER METHOD

Let the dimensions be L & B
Given: L - B= 4 & L.B= 221
Length of diagonal = (L^2 + B^2)^(1/2)

L^2 + B^2= √{(L-B)^2 + 2L.B}=√{(4)^2 + 2.(221)}= √{16 + 442}= √(458)

Now 21^2=441 & 22^2= 484.

Hence 21< Diagonal length< 22. Hence C

W* (w+4)= 221

221 has four factors 1, 13, 17,and liself

W= 13
length= 13 +4 = 17

According to , Pythagorean theoram
C^2= a^2 + b^2
C^2= 13^2+ 17^2

thus Diagonal could be \sqrt{458}
and could be Between 21 and 22.

Easier method:

Let width be x then length = x+4.
Given: x(x+4) = 221 => x^2+4x = 221 (don't solve, just keep)...........(a)

We have to find diagonal. Using Pythagoras, x^2+(x+4)^2 = diagonal^2
=> x^2 + x^2 + 16 + 8x = diagonal^2
=> 2 (x^2+ 4x) +16 = diagonal^2
input value from (a)
=> 2(221) + 16 = 458.. so diagonal should be b/w 21 (441) and 22 (484)!
No need to find roots of (a)!
Cheers.

Area of a rectangle = (diagonal^2)/2
221 = x^2/2
442 = x^2

441 = (21)^2
therefore diagonal is slightly greater than 21, or C.