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Consider a square sheet of side $1$ unit. The sheet is first folded along the main diagonal. This is followed by a fold along its line of symmetry. The resulting folded shape is again folded along its line of symmetry. The area of each face of the final folded shape, in square units, equal to _________

  1. $\frac{1}{4}$
  2. $\frac{1}{8}$
  3. $\frac{1}{16}$
  4. $\frac{1}{32}$
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Migrated from GO Mechanical 3 years ago by gatecse

2 Answers

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6 votes
Best answer

Given that a square sheet has a side length of $1$ unit.

Now, the sheet is first folded along the main diagonal. 

Again, this is followed by a fold along its line of symmetry. And this operation is performed two times.

We get the final folded sheet.

Let the height of the triangle be $h$ units.

Using the Pythagorean theorem, we get.

$\left(\frac{1}{\sqrt{2}}\right)^{2} = h^{2} + \left(\frac{1}{2}\right)^{2}$

$\implies \frac{1}{2} = h^{2} + \frac{1}{4}$

$\implies h^{2} = \frac{1}{2} – \frac{1}{4}$

$\implies h^{2} = \frac{2 – 1 }{4} = \frac{1}{4}$

$\implies h = \frac{1}{2}$

Now, the area of triangle $\Delta = \frac{1}{2} \times \text{Base} \times \text{Height}$

$\implies \Delta = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\;\text{unit}^{2}.$

$\textbf{Short Method:}$ Area follows the symmetry.

  • Initially the area of square $ = 1$
  • Folded first time, then the area of square $ = \frac{1}{2}$
  • Folded second time, then the area of square $ = \frac{1}{4}$
  • Folded third time, then the area of square $ = \frac{1}{8}$

So, the correct answer is $(B).$

edited by

2 Comments

Question :-

The area of each face of the final folded shape, in square units, equal to-

I got it how we are getting 1/8 square units but the question is asking about each face. when we have folded the last time then their have to be two faces - front or back side. Just like cube has 6 faces.

So, isn’t area has to be 2* 1/8 = ¼ square units.??

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edited by

@ASNR1010 question is asking about area of EACH face . which means one face . 

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Ans- B ) 1/8

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