CBSE Board Exam 2023/24 | SQP | 1 Mark MCQ
There is a square board of side '$2a$' units circumscribing a red circle. Jayadev is asked to keep a dot on the above said board. The probability that he keeps the dot on the shaded region is:
(A) $\frac{\pi}{4}$ (B) $\frac{4-\pi}{4}$
(C) $\frac{\pi-4}{4}$ (D) $\frac{4}{\pi}$
(A) $\frac{\pi}{4}$ (B) $\frac{4-\pi}{4}$
(C) $\frac{\pi-4}{4}$ (D) $\frac{4}{\pi}$
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1
Geometric Probability: When throwing a dart or placing a random dot on a surface, the probability of hitting a specific zone is simply the ratio of areas: $P(E) = \frac{\text{Favourable Area}}{\text{Total Available Area}}$.
2
The Circumscribed Property: If a square "circumscribes" a circle, it wraps perfectly around it. This means the Diameter of the circle is exactly equal to the Side of the square. If the side is $2a$, the diameter is $2a$, making the radius strictly $a$.
3
The Shaded Area Setup: The shaded region consists of the four corners of the square outside the red circle. Therefore, $\text{Favourable Area} = \text{Area of Square} - \text{Area of Circle}$.
1
Calculate the Total Area (Area of the Square). The side is $2a$, so $\text{Area} = (2a) \times (2a) = 4a^2$. This is your denominator!
2
Find the radius of the inscribed circle. Since the diameter equals the square's side ($2a$), the radius $r = a$.
3
Calculate the Area of the Circle: $\pi(r)^2 = \pi(a)^2 = \pi a^2$.
4
Find the Favourable Area (Shaded corners): $\text{Square Area} - \text{Circle Area} = 4a^2 - \pi a^2$.
5
Set up the probability fraction: $\frac{4a^2 - \pi a^2}{4a^2}$. Factor out the $a^2$ from the top and cancel it with the bottom to get the final ratio.
!
The "Radius = Side" Trap: A very common and fatal mistake is misreading the setup and assuming the radius is $2a$. If you do that, the area of the circle becomes $4\pi a^2$, which breaks the subtraction completely. Another mistake is calculating the probability of hitting the RED CIRCLE ($\frac{\pi}{4}$) and picking Option A. The question asks for the SHADED REGION (the outside corners)!
🚀
The Dummy Variable Trick: Don't let algebra confuse you. The variable '$a$' is just a placeholder and will cancel out anyway. Let's just pretend $a = 1$!
• Side of square = $2$. Area of Square = $4$.
• Radius of circle = $1$. Area of Circle = $\pi$.
• Area of Shaded Corners = $4 - \pi$.
• Probability = $\frac{\text{Shaded}}{\text{Total}} = \frac{4 - \pi}{4}$.
Solved in 10 seconds mentally!
• Side of square = $2$. Area of Square = $4$.
• Radius of circle = $1$. Area of Circle = $\pi$.
• Area of Shaded Corners = $4 - \pi$.
• Probability = $\frac{\text{Shaded}}{\text{Total}} = \frac{4 - \pi}{4}$.
Solved in 10 seconds mentally!
Topper's Answer Sheet
Given: Side of the square board = $2a$. The circle is inscribed within the square.
Step 1: Calculate Total Area of the board. $\text{Total Area (Square)} = \text{side} \times \text{side}$ $= (2a) \times (2a) = 4a^2$ sq units. [½ Mark]
Step 2: Calculate Favourable Area (Shaded Region). Since the square circumscribes the circle, $\text{Diameter of circle} = \text{Side of square} = 2a$. $\therefore \text{Radius } (r) = a$. $\text{Area of red circle} = \pi r^2 = \pi(a)^2 = \pi a^2$. $\text{Shaded Area} = \text{Area of Square} - \text{Area of Circle}$ $\text{Shaded Area} = 4a^2 - \pi a^2 = a^2(4 - \pi)$ sq units. [½ Mark]
Step 3: Calculate the Probability. $P(\text{Shaded}) = \frac{\text{Favourable Area}}{\text{Total Area}}$ $P = \frac{a^2(4 - \pi)}{4a^2}$ Cancelling $a^2$ from numerator and denominator: $P = \frac{4 - \pi}{4}$.
Ans: (B) $\frac{4 - \pi}{4}$
Step 1: Calculate Total Area of the board. $\text{Total Area (Square)} = \text{side} \times \text{side}$ $= (2a) \times (2a) = 4a^2$ sq units. [½ Mark]
Step 2: Calculate Favourable Area (Shaded Region). Since the square circumscribes the circle, $\text{Diameter of circle} = \text{Side of square} = 2a$. $\therefore \text{Radius } (r) = a$. $\text{Area of red circle} = \pi r^2 = \pi(a)^2 = \pi a^2$. $\text{Shaded Area} = \text{Area of Square} - \text{Area of Circle}$ $\text{Shaded Area} = 4a^2 - \pi a^2 = a^2(4 - \pi)$ sq units. [½ Mark]
Step 3: Calculate the Probability. $P(\text{Shaded}) = \frac{\text{Favourable Area}}{\text{Total Area}}$ $P = \frac{a^2(4 - \pi)}{4a^2}$ Cancelling $a^2$ from numerator and denominator: $P = \frac{4 - \pi}{4}$.
Ans: (B) $\frac{4 - \pi}{4}$
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