Bimal has a flexible string which is 132 cm long. He bent it into… - Vidyadip

Question

Bimal has a flexible string which is 132 cm long. He bent it into different shapes one by one.
Q.1. Bimal bent the string to form an equilateral triangle. What will be the length of each side of the triangle formed?
(a) 11 cm$\quad$(b) 22 cm$\quad$(c) 44 cm$\quad$(d) 48 cm
Q.2. Bimal bent the string to form a square. Find the area of the square.
(a) $1,936 cm^2$$\quad$(b) $1,729 cm^2$$\quad$(c) $1,464 cm^2$$\quad$(d) $1,089 cm^2$
Q.3. He bent it into a circle. Find the radius of the circle formed. [Use $\pi=\frac{22}{7}$ ]
(a) 7.5 cm$\quad$(b) 10.5 cm$\quad$(c) 15 cm$\quad$(d) 21 cm
Q.4. The area of the circle formed by bending the string will be [Use $\pi=\frac{22}{7}$ ]
(a) $1,323 cm^2$$\quad$(b) $1,386 cm^2$$\quad$(c) $1,545 cm^2$$\quad$(d) $1,594 cm^2$

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Answer

1. (c): Perimeter of equilateral triangle = 132 cm.
$\Rightarrow 3 \times$ length of each side $=132 cm \Rightarrow$ length of each side $=\frac{132}{3} cm=44 cm$.
2. (d): Perimeter of square = 132cm.
$\Rightarrow 4 \times$ side $=132 cm \Rightarrow$ side $=\frac{132}{4} cm=33 cm$.
$\Rightarrow$ area of square $=(\text { side })^2=(33)^2 cm^2=1.089 cm^2$.
3. (d): Circumference of the circle = 132 cm.
$\Rightarrow 2 \pi r=132 cm \Rightarrow 2 \times \frac{22}{7} \times r=132 cm \Rightarrow r=\left(\frac{132 \times 7}{2 \times 22}\right) cm =21 cm$
$\therefore$ radius of the circle $=21 cm$.
4. (b): Area of the circle
$=\pi r^2=\left(\frac{22}{7} \times 21 \times 21\right) cm ^2=1.386 cm^2$.

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