Question

Square $ABCD$ has sides with length $20$. Each of the smaller figures is formed by connecting midpoints of the next larger figure.
What is the area of $EFGH$?

• A. $\frac{25}{64}$
• B. $\frac{25}{16}$
• C. $\frac{25}{4}$
• D. $25$
• E. $100$

Answer 1

The correct option is D $25$

Given side of $ABCD=a$

Let $P,Q,R,S$ be the mid points of $AB,BC,CD,AD$ respectively.

$RC=QC=a/2$

From Pythagoras theorem

$R{Q}^2=R{C}^2+Q{C}^{2}R{Q}^2=2R{C}^{2}RQ=\sqrt{2}RC=\sqrt{2}\frac{a}{2}=\frac{a}{\sqrt{2}}$

Side of square $PQRS=\frac{a}{\sqrt{2}}$

Similarly, side of $3^{rd}$ square $=\frac{a}{2}$
This forms a GP: $a,\frac{a}{\sqrt{2}},\frac{a}{2},\frac{a}{2\sqrt{2}},\frac{a}{4},.....$

Square $EFGH$ is the $5^{th}$ square,

Its side $=a_5=a{r}^4=a{\left(\frac{1}{\sqrt{2}}\right)}^4=\frac{a}{4}=\frac{20}{4}=5$

Area os square $EFGH=(side{)}^2=5^2=25$