Question

The size of a square computer screen is measured by the length of its diagonal. How much bigger is the visible area of a square 24-inch screen than the area of a square 20-inch screen ?

• A. $46$ sq. inches
• B. $72$ sq. inches
• C. $60$ sq. inches
• D. $88$ sq. inches

Answer 1

Answer: (D)

$88$ sq. inches

88 in${}^2$: If the diagonal of the larger screen is 24 inches, and it is always true for a square that $d=s\sqrt{2}$ , then:

$s=\frac{d}{\sqrt{2}}=\frac{24}{\sqrt{2}}=\frac{24\left(\sqrt{2}\right)}{\left(\sqrt{2}\right)\left(\sqrt{2}\right)}=12\sqrt{2}$

By the same reasoning, the side length of the smaller screen is $\frac{20}{\sqrt{2}}=\frac{20\sqrt{2}}{\left(\sqrt{2}\sqrt{2}\right)}=10\sqrt{2}$.

The areas of the two screens are:

Large screen: $A=12\sqrt{2}\times x12\sqrt{2}=288$

Small screen: $A=10\sqrt{2}\times 10\sqrt{2}=200$

The visible area of the larger screen is $88$ square inches bigger than the visible area of the smaller screen.