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In the figure shown above, each inside square is formed by joining the midpoints of the sides of the next larger square. The area of the smallest square (shaded) as shown, in $\text{cm}^{2}$ is:

1. $12.50$
2. $6.25$
3. $3.125$
4. $1.5625$
Migrated from GO Electrical 1 year ago by Arjun

It is given that the side of the outer square  $= 10\;\text{cm} = \frac{10}{(\sqrt{2})^{0}}\;\text{cm}.$

The side of the second outer square $= \sqrt{5^{2} + 5^{2}} = \sqrt{25+25} = \sqrt{50} = 5 \sqrt{2}\;\text{cm} = \frac{10}{\left(\sqrt{2}\right)^{1}}\;\text{cm}$

The side of the third outer square $= \sqrt{\left(\frac{5\sqrt{2}}{2}\right)^{2} + \left(\frac{5\sqrt{2}}{2}\right)^{2}} = \sqrt{\frac{50}{4}+\frac{50}{4}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5 \;\text{cm} = \frac{10}{\left(\sqrt{2}\right)^{2}}\;\text{cm}$

The side of the fourth outer square $= \sqrt{\left(\frac{5}{2}\right)^{2} + \left(\frac{5}{2}\right)^{2}} = \sqrt{\frac{25}{4}+\frac{25}{4}} = \sqrt{\frac{50}{4}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \;\text{cm} = \frac{10}{\left(\sqrt{2}\right)^{3}}\;\text{cm}$

The side of the fifth outer square $= \sqrt{\left(\frac{5}{2\;\sqrt{2}}\right)^{2} + \left(\frac{5}{2\;\sqrt{2}}\right)^{2}} = \sqrt{\frac{25}{8} + \frac{25}{8}} = \sqrt{\frac{50}{8}}= \sqrt{\frac{25}{4}} = \frac{5}{2} \;\text{cm} = \frac{10}{\left(\sqrt{2}\right)^{4}}\;\text{cm}$

The side of the inner (shaded) square $= \sqrt{\left(\frac{5}{4}\right)^{2} + \left(\frac{5}{4}\right)^{2}} = \sqrt{\frac{25}{16} + \frac{25}{16}} = \sqrt{\frac{50}{16}}= \sqrt{\frac{25}{8}} = \frac{5}{2\;\sqrt{2}} \;\text{cm} = \frac{10}{\left(\sqrt{2}\right)^{5}}\;\text{cm}$

$\therefore$ The area of the smallest square (shaded) $= \left(\frac{5}{2\;\sqrt{2}}\right)^{2} = \frac{25}{8} = 3.125\;\text{cm}^{2}.$

$\textbf{PS:}$  The Pythagorean theorem states that if a triangle has one right angle, then the square of the longest side, called the hypotenuse, is equal to the sum of the squares of the lengths of the two shorter sides, called the legs. So if $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse, then $a^{2}+b^{2}=c^{2}.$

$\textbf{Short Method:}$

It is given that the side of the outer square  $= 10\;\text{cm}.$

Now, we can create some patterns, and find the area of inner (shaded) square.

• The area of outer square $= 10^{2} = 100\;\text{cm}^{2}$
• The area of second outer square $= \frac{100}{2} = 50\;\text{cm}^{2}$
• The area of third outer square $= \frac{50}{2} = 25\;\text{cm}^{2}$
• The area of fourth outer square $= \frac{25}{2} = 12.50\;\text{cm}^{2}$
• The area of fifth outer square $= \frac{12.50}{2} = 6.25\;\text{cm}^{2}$
• The area of inner (shaded) square $= \frac{6.25}{2} = 3.125\;\text{cm}^{2}.$

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

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