Question

In the figure ABCD is rectangle. BC-6 centimeters, CD = 4 centimeters. Triangle PCB is isosceles . If we put a dot in this figure, what is the probability of it being in the triangle PAB ?

• A. 1/4
• B. 1/3
• C. 1/2
• D. 1/5

Answer 1

Answer:

C. 1/2

To find the probability of a randomly placed dot landing inside triangle $PAB$, we calculate the ratio of the area of triangle $PAB$ to the total area of rectangle $ABCD$.

Step-by-Step Calculation

1. Total Area of Rectangle $ABCD$:

   
   

   $\text{Area of } ABCD = \text{length} \times \text{width} = BC \times CD = 6 \text{ cm} \times 4 \text{ cm} = 24\text{ cm}^2$

2. Area of Triangle $PAB$:

   
   

   • Base of $\triangle PAB$: The base is side $AB$. Since $ABCD$ is a rectangle, $AB = CD = 4\text{ cm}$.

   • Height of $\triangle PAB$: Point $P$ lies on the top side $AD$. The perpendicular distance from $P$ to the base $AB$ is equal to the length of side $AD$ (or $BC$), which is $6\text{ cm}$.

   • Area calculation:

     
     

     $\text{Area of } \triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \text{ cm} \times 6 \text{ cm} = 12\text{ cm}^2$

3. Probability:

   
   

   $\text{Probability} = \frac{\text{Area of } \triangle PAB}{\text{Area of Rectangle } ABCD} = \frac{12}{24} = \frac{1}{2} \text{ (or } 0.5 \text{ / } 50\%\text{)}$

Answer

The probability of the dot being in triangle $PAB$ is $\frac{1}{2}$ (or $0.5$).