A rectangle has a length L and a width W, where L > W. If the width, W, is increased by 10%, which one of the following statements is correct for all values of L and W?

A.

Perimeter increases by 10%.

B.

Length of the diagonals increases by 10%.

C.

Area increases by 10%.

D.

The rectangle becomes a square.

Solution:

We are given:

  • A rectangle with length L and width W, where L > W.

  • The width W is increased by 10% → new width = 1.1W.

  • We are to determine which of the options is always correct for all values of L and W.

Let's evaluate each option:

(A) Perimeter increases by 10%

Original perimeter:

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

New perimeter:

P′=2(L+1.1W)=2L+2.2WP' = 2(L + 1.1W) = 2L + 2.2W

Percent increase in perimeter:

P′−PP=(2L+2.2W)−(2L+2W)2L+2W=0.2W2L+2W\frac{P' - P}{P} = \frac{(2L + 2.2W) - (2L + 2W)}{2L + 2W} = \frac{0.2W}{2L + 2W}

This is not always 10% — it depends on the ratio W/L. So ❌ Not always true.

(B) The Length of the diagonals increases by 10%

Original diagonal:

d=L2+W2d = \sqrt{L^2 + W^2}

New diagonal:

d′=L2+(1.1W)2=L2+1.21W2d' = \sqrt{L^2 + (1.1W)^2} = \sqrt{L^2 + 1.21W^2}

This is not 10% more than d, because square roots are nonlinear.
Try an example:

  • L = 10, W = 5

  • d = √(100 + 25) = √125 ≈ 11.18

  • new W = 5.5 → d' = √(100 + 30.25) = √130.25 ≈ 11.41

  • % increase ≈ (11.41 - 11.18) / 11.18 ≈ 2%

So ❌ Not always 10% — false.

(C) Area increases by 10%

Original area:

A=L×WA = L \times W

New area:

A′=L×1.1W=1.1LWA' = L \times 1.1W = 1.1LW

→ Area increases by exactly 10%, regardless of L and W.

✅ Always true

(D) The rectangle becomes a square

Only happens if new width = length
→ L = 1.1W → L/W = 1.1
But question says L > W, not necessarily L = 1.1W

So this is only true for one specific case, not all values → ❌ False

✅ Final Answer: (C) Area increases by 10%.