More Questions from Area

In the given diagram, $ABCD$ is a square and semi-circular regions have been added to it by drawing two semi-circles with $AB$ and $CD$ as diameters. If the total area of the three regions is $350$ sq. cm, then the length of the side of the square is equal to abcd square semi circular regions added total area 350 sq cm

Aptitude Area Difficulty: Medium
Choose an option
  • A
    $5\sqrt{7}$ cm
  • B
    $7$ cm
  • C
    $13$ cm
  • D
    $14$ cm

Answer

Correct Answer: $14$ cm

Explanation

### Concept & Area of Composite Figures The figure is composed of a central square and two identical semi-circles extending outwards. Since the semi-circles share the side length of the square as their diameter, they can logically be combined into exactly one complete circle. Formulas: $$ \text{Area of square} = \text{side}^2 $$ $$ \text{Area of circle} = \pi r^2 = \frac{\pi d^2}{4} $$ ### Step-by-Step Solution 1. Let the side of the square $ABCD$ be $x$ cm. Consequently, the diameter of both semi-circles is also $x$ cm. 2. The area of the central square is $x^2$. 3. The two semi-circles combine to form a full circle with diameter $x$ (meaning its radius is $\frac{x}{2}$). Area of these two semi-circles combined $= \pi (\frac{x}{2})^2 = \frac{\pi x^2}{4}$. 4. Set up the equation for the total given area: $x^2 + \frac{\pi x^2}{4} = 350$ 5. Substitute $\pi = \frac{22}{7}$ to solve: $x^2 (1 + \frac{22}{28}) = 350$ $x^2 (1 + \frac{11}{14}) = 350$ $x^2 (\frac{25}{14}) = 350$ 6. Solve for $x$: $x^2 = 350 \times \frac{14}{25} = 14 \times 14 = 196$ $x = \sqrt{196} = 14$ cm. ### Exam Strategy & Shortcut Notice the total area equation: $x^2(1 + \pi/4) = x^2(25/14)$. The fractional factor $25/14$ tells us that $x^2$ must ideally be a multiple of $14$ to yield a clean integer total like $350$. Checking the options, $14$ is an extremely strong candidate. Testing it: $14^2 \times (25/14) = 14 \times 25 = 350$. It matches perfectly. ### Common Pitfall A common mistake is treating the two semi-circles as having a *radius* of $x$ instead of a *diameter* of $x$, which inflates the area of the circular components by a factor of 4. ### Final Answer Therefore, the correct answer is **14 cm**.
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