Question

Let \( S_1 \) be a square of side 5 cm. Another square \( S_2 \) is drawn by joining midpoints of the sides of \( S_1 \). Square \( S_3 \) is drawn similarly and so on. Then \( \text{Area}(S_1) + \cdots + \text{Area}(S_{10}) \) is

• A. \(25\left(1-\frac{1}{2^{10}}\right)\)
• B. \(50\left(1-\frac{1}{2^{10}}\right)\)
• C. \(2\left(1-\frac{1}{2^{10}}\right)\)
• D. \(1-\frac{1}{2^{10}}\)
• E. \(10\left(1-\frac{1}{2^{10}}\right)\)

Hint:

Joining midpoints of a square always halves the area — remember this key geometric fact.

Answer 1

The Correct Option is B

Understanding the Concept: Each new square formed by joining midpoints has half the area of the previous.

Step 1: First area
\[ A_1 = 5^2 = 25 \]

Step 2: Ratio of areas
\[ A_2 = \frac{1}{2} A_1 = \frac{25}{2} \] Thus G.P. with: \[ a=25,\quad r=\frac{1}{2} \]

Step 3: Sum formula
\[ S_n = a \frac{1-r^n}{1-r} \]

Step 4: Substitute values
\[ S_{10} = 25 \cdot \frac{1-(1/2)^{10}}{1-1/2} \] \[ = 25 \cdot \frac{1-(1/2)^{10}}{1/2} \] \[ = 50\left(1-\frac{1}{2^{10}}\right) \]

Step 5: Final Answer
\[ \boxed{50\left(1-\frac{1}{2^{10}}\right)} \]