Question

A square piece of paper is folded three times along its diagonal to get an isosceles triangle whose equal sides are $10$ cm. What is the area of the unfolded original piece of paper?

• A. $400\ \text{sq.\ cm}$
• B. $800\ \text{sq.\ cm}$
• C. $800\sqrt{2}\ \text{sq.\ cm}$
• D. $1600\ \text{sq.\ cm}$
• E. Insufficient data to answer

Hint:

Repeated folds along the {same} line keep the outline aligned with that line; after two such folds the visible legs along adjacent sides become half the original side length. So with diagonal-folds on a square, the final isosceles right triangle’s equal sides are $s/2$.

Answer 1

The Correct Option is A

Step 1: Track what folding along the {same diagonal does.}
Let the square be $ABCD$ with side $s$ and diagonal $AC$.
- After the first fold along $AC$, the outline becomes an isosceles right triangle with legs along $AB$ and $AD$ of length $s$.
- Folding again along the {same} diagonal $AC$ superposes the previous triangle onto itself; the right-angle corner now lies at the midpoints of $AB$ and $AD$. Hence the visible legs become {half} of $s$, i.e., $s/2$.
- A third fold along $AC$ only increases thickness; the outline (and leg lengths) remain $s/2$. Step 2: Use the given equal side of the final triangle.
The equal sides of the final isosceles right triangle are $10$ cm, so \[ \frac{s}{2}=10 \;\Rightarrow\; s=20\ \text{cm}. \] Step 3: Area of the original square.
\[ \text{Area}=s^{2}=20^{2}=\boxed{400\ \text{sq.\ cm}}. \]