Question

Avneet buys$ 9$ square paving slabs, each with a side of $ \frac{1 }{2}$ m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement $ \text{[fig 10.7(i)]}$?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement $ \text{[(Fig 10.7 (ii)]}$?

(c) Which has a greater perimeter?

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this?

(The paving slabs must meet along complete edges i.e they cannot be broken.)

(i) (ii)

Answer 1

Determine the perimeter to the given points.

Given:

buying the square slabs $=9$

Side of each square slab $=\frac{1}{2}\mathrm{m}$

To find:

perimeter

Solution:

Calculate the perimeter for each of the given options:

For Option (a).

Side of each small square paving slab bought by Avneet $=\frac{1}{2}\mathrm{m}$

According to the arrangement shown in $\text{[fig 10.7(i)]}$,

Each Side of the whole square $=3\times$ side of the small square paving slab [Since the whole square is made up of 9 small square paving slabs]

$\Rightarrow 3\times \frac{1}{2}=\frac{3}{2}\mathrm{m}$

The perimeter of the whole square:

$$\begin{gathered} 4\times \mathrm{side}=4\times \frac{3}{2} \\ \Rightarrow =2\times 3 \\ \Rightarrow =6m \end{gathered}$$

Therefore, $6$ m is the perimeter of the square shown in $\text{[fig 10.7(i)]}$.

For Option (b).

The perimeter of the cross figure shown in $\text{[fig 10.7(ii)]}$ $=0.5+1+1+0.5+1+1+0.5+1+1+0.5+1+1$

$\Rightarrow =10m$

Therefore, the perimeter of Shari's arrangement is$10$ m.

For Option (c).

The second figure that is $\text{[fig 10.7(ii)]}$ has a greater perimeter.

For Option (d).

When all of the squares are lined up in a straight line as shown in the image given below

Total perimeter for the above arrangement:

$$\begin{gathered} AB+BC+CD+AD=\frac{9}{2}+\frac{9}{2}+\frac{1}{2}+\frac{1}{2} \\ \Rightarrow =\frac{20}{2} \\ \Rightarrow =10\mathrm{m} \end{gathered}$$

After such an arrangement, we will still have the greater perimeter as compared to $\text{[fig 10.7(i)]}$. that is,$10m>6m$