MATHS M.C.Q. [1 Marks Each]

$(b) 6r$, a regular hexagon comprises $6$ equilateral triangles, each of them having one of their vertices at the centre of the hexagon.
The sides of the equilateral triangle are equal to the radius of the smallest circle inscribing the hexagon.
each side of the hexagon is equal to the radius of the hexagon and the perimeter of the hexagon is $6r.$

Consider the given perimeter of the square is $= ( 4y + 12 )m$
We know, perimeter of the square $= 4 \times $side
$(4y + 12) = 4 \times $ side.
$\text{ side } \frac{4\text{y}+12}{4} = \text{y} + {3}$
Now, length of diagonal of the square
$ = \sqrt{(\text{side})^{2} + (\text{side}^{2})}$
$= \sqrt{(\text{y+3}^{2} + (\text{y + 3})^{2}}$
$ = \sqrt{2}.(\text{y} + {3})\text{m}$

A pentagonal prism has $15$ edges. Vertices $= 5 + 5 = 10$

Ratio in length and perimeter of a rectangle $= 1 : 3$
Let length $= x,$
then perimeter $= 3x$
$\therefore$ Breadth $=\frac{3\text{x}}{2}-\text{x}=\frac{\text{x}}{2}$
$\therefore$ Ratio in length and breadth $=\text{x}:\frac{\text{x}}{2}$
$=2:1$

Length of a rectangular field $(l) = 34m$
and breadth $(b) = 18m$
Circumference $= 2(l + b)$
$= 2(34 + 18)m$
$= 2 \times 52= 104m$
Rate of fencing $= Rs. 22.50$ per m
Total cost $= Rs. 22.50 \times 104$
$= Rs. 2340$

As we know that all the sides of a squqre are equal. So its perimeter will be $4$ times the length of the side.

Distance covered by the farmer = perimeter of the filed $= 2 \times $ (length + breadth) $= 2 \times (120 + 80) = 2 \times 200 = 400m$

$l = 900 b = 700$
Perimeter $= 2(900 + 700)$
$= 2(1600)$
$= 3200$
$3 rands fence:$
$= 3(3200)$
$= 9600m$
$= 9600 × 8 = Rs 76,800$

Area of rectangular floor $= 80sq.m.$
Length $\times $ Breadth $= 80sq.m$
$\Rightarrow 10 \times $ Breadth $= 80sq.m$
$\Rightarrow $ Breadth $= 8m$
$\therefore$ Perimeter $= 2$(Length $+$ Breadth)
$= 2(10 + 8) = 2 \times 18 = 36m.$

Let the width of the garden $= x$ meter
Then length $= (x + 4)$ meter
Half perimeter $= 36m$
So perimeter of garden $= (2 \times 36) = 72$ meters
According to the question
$\Rightarrow 2(l + b) = 72$
$\Rightarrow 2(x + x + 4) = 72$
$\Rightarrow 2x + 2x + 4 = 74$
$\Rightarrow 4x = 64$
$\Rightarrow x = 16$ meters
The length of the garden $= (16 + 4) = 20$ meters

We know, Perimeter $= 4 \times $ side $= 4 \times 5 = 20m$

Perimeter $= 444 = 4 \times s$
$\text{s} = \frac{444}{4}= {111}\text{m}$

Area of the poster $= 2.5 \times 2.5 = 6.25$
Area of the wall =$ 10.5 \times 8.5 = 89.25$
$= 89.25-6.25 = 83.00sq m$
$83 \times 12 = Rs. 996$

Perimeter of the square side $a = 4a$
hence, perimeter of this square of side $4m = 4 \times 4 = 16m$

Amount of wire needed = Perimeter of a rectangle $= 2 \times $ (length $+$ breadth) hence wire needed $= 2 \times (250 + 20) = 540m$

We have the perimeter of a square $= 4 \times a$
the perimeter of a square of length $25\ cm$ is $= 4 \times 25 = 100\ cm.$

The perimeter of a square is the sum of the lengths of its sides. Now, the sides of the square are all equal. say the side of a square is a units
thus, the perimeter of square $a + a + a + a = 4a$ Here perimeter is $4$ times length of sides. hence,

In regular polygon, all sides and angles are equal.
According to the question,
In figure $(i)$, all sides are not equal.
So, it is not a regular polygon.
In figure $(ii)$, it is a square and in square all sides are equal and all angles are of $90^\circ .$
So, it is a regular polygon.
In figure $(iii)$, it is a parallelogram and in parallelogram opposite sides are equal and
opposite angles are equal.
So, it is not a regular polygon.
In figure $(iv)$, all sides are not equal. So, it is not a regular polygon.

We know that,
Peri $= 4 \times $ side
$4 \times s = 728$
$\text{s} = \frac{728}{4} = {182}\text{m}$

Perimeter of a square is $= 4 \times s$ hence Perimeter of square $= 4 \times 4 = 16m$

Perimeter of hexagon $= x$ metres
6 (side)= $x$ metres
Side $= (x ÷ 6)$ metres
$\therefore$ Side $= (x÷6)$ metres $(c)$

For fencing the rectangular field, we need to find the perimeter of the rectangle.
Length of the rectangle $= 34m$
Breadth of the rectangle $= 18m$
Perimeter of the rectangle $= 2$(Length $+$ Breadth) $= 2(34 + 18)m$
$= 2 \times 52m$
$= 104m$
Cost of fencing the field at the rate of $Rs. 2.25$ per meter $= Rs. 104 \times 2.25$
$= Rs. 234$

Area of a rectangle $= 650\ cm^2$
and breadth $(b) = 13\ cm$
$\therefore$ Length (l) $=\frac{\text{Area}}{\text{Breadth}}$
$=\frac{650}{12}=50\text{cm}$
$\therefore$ Perimeter$ = 2(l + b)$
$= 2(50 + 13)cm$
$= 2 \times 63$
$= 126\ cm$

Since, Perimeter $= 2(l + b)$
Here length $=l$ and breadth $= b$
If the length and breadth of a rectangle are doubled.
then length $=2l$ and breadth $= 2b$
$\therefore$ perimeter of rectangle would be $2(2l + 2b) = 4(l + b) = 2.2(l + b)$
$\therefore$ if the length and breadth of a rectangle are doubled then its perimeter is also doubled.

Perimeter of square $={4}\times\text{s}$
$\therefore{4} \times\text{S} = {728}$
$\text{S} = \frac{728}{4} = {182}\text{m}$
thus the measure of the side of the square is $182\ cm.$

As per the given information, $\text{W} = \frac{\text{L}}{2} - {2}$
The Perimeter of the rectangle in terms of $\text{L} = {2}\big(\frac{\text{L}}{2} - {2} + \text{L}\big)$
$= L - 4 + 2L$
$= 3L - 4$

Each row contains $12$ plants.
there are $11$ gapes between the two corner trees $(11 \times 2)$ metres and $1$ metre on each side is left.
Length $= (22 + 2)m = 24m$