A10
B18
C14
D20
Answer:
C. 14
Read Explanation:
Understanding the Problem: Geometry and Algebra Combined
This problem requires us to use our knowledge of geometric formulas for squares and circles and then solve a system of algebraic equations.
We are given a relationship between the side of a square and the radius of a circle, and another relationship involving their perimeters/circumferences.
Key Geometric Formulas:
Perimeter of a Square: $P = 4s$, where $s$ is the length of a side.
Circumference of a Circle: $C = 2\pi r$, where $r$ is the radius.
For competitive exams, it's crucial to have these formulas memorized.
Setting up the Equations:
Let $s$ be the side of the square and $r$ be the radius of the circle.
Condition 1: "Side of square is 4 more than the radius of circle." This translates to the equation: $s = r + 4$.
Condition 2: "Sum of perimeter of square and circumference of circle is 160." This translates to: $4s + 2\pi r = 160$.
These are the two fundamental equations we will work with.
Solving the System of Equations:
Substitution Method: Since we have an expression for $s$ from the first equation ($s = r + 4$), we can substitute this into the second equation.
Substituting $s$ in the second equation: $4(r + 4) + 2\pi r = 160$.
Simplifying the Equation: Distribute the 4: $4r + 16 + 2\pi r = 160$.
Collect terms with $r$: $(4 + 2\pi)r + 16 = 160$.
Isolate the term with $r$: $(4 + 2\pi)r = 160 - 16$.
$(4 + 2\pi)r = 144$.
Using the Value of $\pi$: For many competitive exams, using $\pi \approx \frac{22}{7}$ is common and simplifies calculations.
Substitute $\pi = \frac{22}{7}$: $(4 + 2 \times \frac{22}{7})r = 144$.
$(4 + \frac{44}{7})r = 144$.
Find a common denominator for the terms inside the parenthesis: $(\frac{28}{7} + \frac{44}{7})r = 144$.
$(\frac{72}{7})r = 144$.
Solving for $r$: $r = 144 \times \frac{7}{72}$.
Simplify the fraction: $r = 2 \times 7$.
$r = 14$.