Question

A piece of wire 132 dm long is bent successively in the shape of an equilateral triangle, a square, a regular hexagon and a circle. Then the area included is largest when the shape is
(a) Triangle
(b) square
(c) Hexagon
(d) Circle

Answer 1

Hint: First, we will equate the perimeter of each shape given in question with total length of wire i.e. 132dm. From this we will find the side length of each shape. Then we will put that value in finding areas of each individual shape. At the end we will compare all the four areas of each shape which is largest and that will be the answer.

Complete step-by-step answer:
Here, we are given that 132 dm long wire is bent into different shapes. So, first we will equate the perimeter of each shape with 132 and then find the area of that shape.
Here, taking an equilateral triangle having 3 sides equal. Assuming each side as a, so the perimeter will be 3a.

$perimeter=3a=132$
On solving, we get
$a=\dfrac{132}{3}=44dm$
Now, we will find the area of the equilateral triangle using the formula $\dfrac{\sqrt{3}}{4}{{a}^{2}}$ . So, we get
$area=\dfrac{\sqrt{3}}{4}{{\left( 44 \right)}^{2}}$
On solving, we will get as
$area=484\sqrt{3}d{{m}^{2}}$ ………………………..(1)
Similarly, we will find a side of square.

We will get as
$perimeter=4a=132$
On dividing it by 4 on both sides, we get
$a=\dfrac{132}{4}=33dm$
So, now finding the area of the square given as ${{a}^{2}}$ .
$area={{a}^{2}}={{\left( 33 \right)}^{2}}$
On solving we get
$area=1089d{{m}^{2}}$ ……………………………(2)
Now, finding the side of regular hexagon by equation perimeter with 132.

So, we will get
$perimeter=6a=132$
On dividing both sides by 6, we get
$a=\dfrac{132}{6}=22dm$
To find the area of a regular hexagon, we will use the formula $\dfrac{3\sqrt{3}}{2}{{a}^{2}}$ . On putting values, we get
$area=\dfrac{3\sqrt{3}}{2}{{\left( 22 \right)}^{2}}$
$area=726\sqrt{3}d{{m}^{2}}$ ………………………………..(3)
Now, taking the circle and finding radius of circle using perimeter of circle.

So, we get
$perimeter=2\pi r=132$
$r=\dfrac{132\times 7}{2\times 22}=21dm$ (using $\pi =\dfrac{22}{7}$ )
Now, finding area of circle, we get
$area=\pi {{r}^{2}}=\dfrac{22}{7}\times {{\left( 21 \right)}^{2}}$
$area=1386d{{m}^{2}}$ ………………………………..(4)
Now, we will compare all the four equations i.e. (1), (2), (3), (4) we can see that the area of the circle is highest compared to all.
Thus, the area included is largest when the shape is a circle.
Hence, option (d) is the correct answer.

Note: Remember that we have to first find the side length of each and every shape and then area we have to find. Do not directly equate it with area formula and find side length. For example: area of equilateral triangle $\dfrac{\sqrt{3}}{4}{{a}^{2}}=132$ . On solving this we will get $a=13.26dm$ . So, by doing this in all the four-given shapes, we will get the side of the equilateral triangle that is largest as compared to all other sides. So, the answer will be (a) which is wrong. So, do not make this mistake.