CUET Preparation Today
A wire of length 30 m is cut into two pieces. One of the pieces is folded into a square and the other into a circle. If 'x' m is the length of one of the pieces, the combined area of the square and the circle is given by: |
$\frac{x^2}{4 \pi}+\frac{1}{4}(30-x)^2$ $\frac{x^2}{4 \pi}+\frac{1}{16}(30-x)^2$ $\frac{(30-x)^2}{4 \pi}+\frac{x^2}{8}$ $\frac{(30-x)^2}{16 \pi}+\frac{x^2}{4}$ |
$\frac{x^2}{4 \pi}+\frac{1}{16}(30-x)^2$ |
The correct answer is Option (2) → $\frac{x^2}{4 \pi}+\frac{1}{16}(30-x)^2$ |
