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

Perimeter is the sum of length of the boundaries. In a square, the sides act as the boundaries. Since a square has $4$ sides,
so perimeter of a square is the sum of the lengths of all the 4 sides.

Distance covered by $1$ student $= 1.7$
$\therefore$ Distance covered by 80 students $= (1.75 × 80)m = 140m$
$\therefore$ Perimeter of the garden $= 140m$

The radius of the circle $= 28\ cm$
So, the circumference of the circle $ = {2}\pi\text{r} = {2}\pi \times {28} = {176}\text{cm}$
the perimeter of the square is equal to the circumference of the circle
$4 \times $ side $= 176$ side $= 44\ cm$

Perimeter of square$= 4a$ Diagonals of square $ = \text{D} = \sqrt{2}\text{a}$
$\therefore\text{a} = \frac{16\sqrt{2}}{\sqrt{2}} = {16}$
$\therefore$ Perimeter $= 16 \times 4 = 64\ cm​$

Diagonal of square $= \sqrt{2}\text{a}$
${7}\sqrt{2} = \sqrt{2}\text{a}$
So a $= 7a$ Perimeter $= 4a$
$= 4 \times 7$
$28\ cm.$

Let the side of the square be $xm$. We have, perimeter of the square $= 124m$ i.e. 4x side of the square $= 124$
$\Rightarrow {4}\text{x} = {124}$
$\Rightarrow\text{x} = \frac{124}{4} = {31}\text{m}$
the length of the side of the square is $31m$

Perimeter of the square park $= 4 \times s = 4$
times $5 = 20$
Perimeter of the square $= 20m$

We know that Area $= l \times bl \times b = 630$
$l \times 15 = 630$
${1} = \frac{630}{15} = {42}\text{cm}$

$l = 6m ; b = 3m$
Area of one plank $= 6 \times 3 = 18sq m$
Number of wooden planks $= 5$
Area of $5$ wooden planks $= 18 \times 5$
$= 90sq m$

Let the breadth $= x$
the length $= x + 15$
given perimeter $= 150$
we know Perimeter of rectangle $= 2(l + b)150 = 2( x + x + 15) 75 = 2x + 15 60 = 2 x 30= x$
$\therefore$ breadth $= 30\ cm$
length $= 30 + 15 = 45$
mcheck $150 = 2( (75) 150 = 150$
$LHS = RHS$ hence proved.

Length of top of the table $= 3m 15\ cm$
$= (300 + 15)cm = 315\ cm$
Breadth of top of the table $= 90\ cm$
Perimeter $= 2(315 + 90) = 2(405) = 810\ cm$
$= 8m 10\ cm$

The perimeter of a shape is the distance around it. In particular, the perimeter of a rectangle is given by the formula $P = 2W + 2L.$ Substitute the correct values of
the variables into this formula $(P = 30$ and $L = 6)$ and then solve for the width $W:$
$30 = 2W + 2(6)$
$30 = 2W + 12$
$18 = 2W$
$W = 9$
Therefore, the width of the garden is $9$ feet.

$l = 900, b = 700$
Perimeter $= 2 (900 + 700)$
$= (1600)$
$= 3200$
$3$ rounds of fence:$= 3(3200)$
$= 9600m$
$89600 \times 8$
$= Rs. 76800$

Length of the lane $= 150m$
Breadth of the lane$ = 9m$
Area of the lane $= (150 \times 9)m^2$
$= 1350m^2$
Area of the brick $= 22.5\ cm \times 7.5\ cm$
$= 168.75\ cm^2$
$=\frac{168.75}{10000}\text{m}^2$
$=0.016875\text{m}^2$
$\therefore$ Number of bricks required $=\frac{\text{Area of lane}}{\text{Area of brick}}$
$=\frac{1350}{0.016875}$
$=1350\times \frac{1000000}{46875}$
$=80000$

Cost of fencing the rectangular field $= Rs. 600$
Rate of fencing the field $= Rs. 7.50/m$
Therefore, perimeter of the field $=\frac{\text{Cost of fencing}}{\text{Rate of fencing}}$
$=\frac{600}{7.50}=80\text{m}$
Now, length of the field $= 24m$
Therefore, breadth of the field $=\frac{\text{Perimeter}}{2}-\text{Length}$
$=\frac{80}{2}-24=16\text{m}$

Length of side of square $= \frac{3\text{x}}{4} + {1}$
Perimeter of a square $= 4 \times $ side
$ = {4} \times \big(\frac{3\text{x}}{4} + {1}\big)$
$= 3x + 4$
So, perimeter of square with side $\big(\frac{3\text{x}}{4} + {1}\big) \text{ is } ({3}\text{x} + {4})$

Let breadth of rectangle be b.
Perimeter $= 170m$
$\Rightarrow 2(l + b) = 170$
$\Rightarrow 2(50 + b)=170$
$\Rightarrow 50 + b = 85$
$\Rightarrow b = 35m$

The perimeter of a figure is the total distance around the edge of the figure.
Example: A rectangle whose length and width are 2m and 3m has a perimeter of $2 + 3 + 3 + 2 = 10m.$

Length of rectangular park $= 30m$
Breadth of rectangular park $= 20m$
$\therefore$ Perimeter of park $= 2(30 20) = 100m$
So, distance covered by the athlete in $15$ rounds $= 15 \times 100 = 1500m$

Given, side of a square $= 10\ cm$
We know that, perimeter of a square $= 4 \times $ Side $= 4 \times 10$
$= 40\ cm$
$\therefore$ Perimeter of old square $= 40\ cm$
Now, according to the question, side of the square is doubled.
New side $= 2 \times 10 = 20\ cm$
Again, perimeter of new square$ = 4 \times $Side
$= 4 \times 20 = 80\ cm$
$\therefore$ New perimeter
$= 2 \times $(Old perimeter)
$= 2 \times 40 = 80\ cm$
Hence, the new perimeter is $2$ times of the old perimeter.

Let sides of rectangle are a and ba = length = width
We know perimeter $= 2(a + b)$
Area $= a \times b$
Here width $ = \frac{11}{4} = \text{b}{2}\big(\text{a} + \text{b}\big)$
$ = \text{a} \times \text{b} \Rightarrow{2}\Big(\text{a} + \frac{11}{4}\Big)$
$ = \text{a} \times \frac{11}{4} \Rightarrow\Big({2} - \frac{11}{4}\Big)$
$\text{a} = -\frac{11}{2} \Rightarrow - \frac{3}{4} \text{a} = - \frac{11}{2} $
$\Rightarrow \text{a} = \frac{22}{3}$

Let the sides of the rectangle be $5x$ and $4x$. (Since, they are in the ratio $5 : 4)$
Now, perimeter of rectangle = 2(Length + Breadth)
$72 = 2(5x + 4x)$
$72 = 2 \times 9x$
$72 = 18x$
$x = 4$
Thus, the length of the rectangle $= 5x$
$= 5 \times 4$
$= 20\ cm$

Total cost of fencing $= Rs. 2400$
Rate $= Rs. 30$ per m
Perimeter of the rectangular field $=\frac{2400}{30}$
$= 80m$
$\therefore$ Length + breadth $=\frac{80}{2}$
$= 40m$
Length of field $= 24m$
$\therefore$ Breadth $= 40 - 24$
$= 16m$

In order to find the length of the boundary of the remaining park, we add two flower beds each of length 10m and breadth $5m$, then remaining park is shown below:

Now, length of the boundary of the remaining park = Perimeter of remaining park $= (90 + 5 + 10 + 95 + 90 + 5 + 10 + 95)m = 400m$

The perimeter of any polygon is the sum of the lengths of all the sides.
Example: In a square whose side is given as $2m$, square has $4$ sides.
Perimeter $= 2 + 2 + 2 + 2 = 8m$

$\text{Area of wall} = \frac{\text{Total expenses}}{\text{ Rate}} = \frac{21000}{35}$
$= {600}\text{sq}.\text{m}$
$\text{Now}\text{ B} = \frac{2}{3} \text{L}$ $\text{and} \text{ L}\times\text{B} = {600}\text{m}^{2}$
$\Rightarrow\text{L}\times\frac{2}{3}\text{ L} = {600}$ $\text{L}^{2} = \frac{600\times}{2}{3} = {600}\text{m}^{2}$
$\Rightarrow \text{L} = {30} \text{m} \Rightarrow\text{B} = {20}\text{m}$
$\Rightarrow \text{perimeter} = {2}(\text{L+B})$
$\Rightarrow\text{perimeter} = {2} \big({30 + 20}\big)\text{m} = {100}\text{m}$

Length of bedsheet $= 10m$
Breadth of bedsheet $= 5m 60\ cm = 5.6m$
Perimeter of bedsheet $= 2(10 + 5.6)$
$2 \times (15.6) = 31.2m$
Cost of 1m border $= Rs 90$
$\therefore$ total cost $= Rs (90 \times 31.2) = Rs 2808$

Circumference $=\pi \text{d}$
$=\frac{22}{7}\times 7$
$=22\text{cm}$a

Ratio in the sides of a rectangle $= 7 : 5$
and perimeter $= 96\ cm$
$\therefore $ Length + Breadth $=\frac{96}{2}=48\text{cm}$
Let length $= 7x$
Then breadth $= 5x$
$\therefore $
$7x + 5x = 48$
$\Rightarrow 12x = 48$
$\Rightarrow \text{x}=\frac{48}{12}$
$= 4$
Length of the rectangle =$ 7x$
$= 7 \times 4$
$= 28\ cm$

Perimeter $= 80 \times 1.75 = 14000 = 140m$

let length of rectangle be lm, breadth be bm.
$\frac{1}{\text{p}} = \frac{1}{3}$
$\frac{1}{2}\big({1+\text{b}}\big) = \frac{1}{3}$
${31}={21} + {2}\text{b}$
${1} = {2}\text{b}$
$\therefore \frac{1}{\text{b}}=\frac{2}{1}$
$\therefore\text{required}\text{ ratio}: {2:1}$

Diameter $=\frac{\text{Circumference}}{\pi}$
$=\frac{88\times7}{22}$
$=28\text{cm}$

Let $AB$ be $x$
$\therefore AQ = QP = BP  = \frac{\text{x}}{3}$
Let $BC$ be $y$
$\therefore BR = RS = SC  = \frac{\text{y}}{3}$
Let $AC = z$
$AT = TU = UC = \frac{\text{z}}{3}$
Opposite sides of Hexagon are equal
$\therefore$ Perimeter of Hexagon $= PQ + QT + TU + US + RS + PR$
$ = \Big(\frac{\text{x}}{3} + \frac{\text{y}}{3} + \frac{\text{z}}{3}\Big) \times{2}$
$\therefore\frac{2}{3}$ Perimeter of hexagon is $\frac{2}{3}$ times the perimeter of $△ABC.$

Perimeter of square $a + a + a + a = 4a$ where a is side of square.Here side of square is $y$ hence perimeter is $4y$ hence,

Perimeter of a square $= 4 \times $ side
$= 4 \times 4 = 16m$

Area of rectangle is $36$ units we have,
$\Rightarrow 36 = 6 \times 6$
$= 2 \times 3 \times 3 \times 2$
$= 4 \times 9$
the sides of a rectangle are $4\ cm$ and $9\ cm$
Perimeter $= 2(l + b)$
$= 2(4 + 9)$
$= 13 \times 2$
$= 26$ units

Let the perimeter of circle and square is $1$Then perimeter of circle $={2}\pi\text{r}$
$= 1$ (wherer is redius of circle) $\Rightarrow\text{r}=\frac{1}{2\pi}$
Then area of circle $ = \pi\text{r}^{2} = \pi(\frac{1}{2\pi})^{2} = \frac{1}{4\pi} = 0.0789$
perimeter of square $= 4l = 1$ then l $= \frac{1}{4}$ (where l id the side of square)
Then area of square $=\frac{1}{4}\times\frac{1}{4} = \frac{1}{16} = 0.0625$
Then area of circle is greater then that of squre

Let the length and breadth of rectangle be $l$ and $b .$
Given that length is 6m less than three times its breadth $\Rightarrow l = 3b − 6 ............................ (i)$
Given its perimeter is 148m.we k.n.t perimeter of a rectangle is $2(l + b) \Rightarrow 2(l + b)$
$=148\Rightarrow(1+\text{b})=\frac{148}{2}={74}............\text{(ii)}$
Substitute $(i)$ in $(ii) l + b = (3b - 6) + b = 744b$
$= 74 + 6$
$={80}\text{b}=\frac{80}{4}=20$
substituting value of $b$ in $(i) l = 3b - 6 = 3(20) - 6 = 60 - 6 = 54$
thus, length and breadth of given rectangle are $54m, 20m$

Side of the square $= 16\ cm$
Perimeter of the square$ = (4 \times $ side)
$= (4 \times 16)cm$
$= 64\ cm$

Breadth $= x$
Length $= 3x$
Perimeter$ = 2$(length + breadth)
Perimeter $\Rightarrow 2(x + 3x) = 400$
$\Rightarrow 2(4x) = 400$
$\Rightarrow x = 50$
Length $= 3x = 150m$
Breadth $= x = 50m$

Length of photograph $= 39.5\ cm$
Breadth of photograph $= 31\ cm$
$\therefore$ Required length of the wooden strip
= Perimeter of photograph
$= 2(39.5 + 31) = 2(70.5) = 141\ cm$

Perimeter of a rectangle $= 2 \times $ (length $+$ breadth) Perimeter $= 2 \times (25 + 15) = 80\ cm$

Area of pond $= 3m \times 2m = 6sq m$
area of park $= 28 \times 28$
$= 784sq m$
area of the park excluding the pond$= 784 - 6$
$= 778sq m$

Wire needed would be perimeter of the playground : we know, Perimeter of rectangle :$ 2(l + b) 250 + 20 + 250 + 20 = 540m$

Perimeter of square $= 4 × s = 144$
Hence s$= 144 ÷ 4 = 36m$

Length of a rectangular room $(l) = 5m 40\ cm = 5.4m$
and breadth$ (b) = 4m 50\ cm$
$= 4.5m$
Area$ = l \times b$
$= 5.4 \times 4.5m^2$
$= 24.3m^2$

Length of a sheet $(l) = 72 \ cm$
and breadth $(b) = 48 \ cm$
Area $= l x b = 72 \times 48 \ cm^2$
Area of paper for one envelope $= 18 \times 12\ cm^2$
No. of envelopes $=\frac{72\times 48}{18\times 12}=16$

The maximum length of the side of a square sheet that can be cut off from a rectangular sheet of size $8m \times 3m$ is $3m.$

$(b) 6r$, 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 $= (22 + 2)m = 24m$

Total cost of fencing around a square field $= Rs. 2000$
and rate $= Rs. 25$ per metre
$\therefore$ Circumference $=\frac{2000}{25}=80\text{m}$
$\therefore$ Length of each side $=\frac{80}{4}=20\text{m}$

$6$ inches : The perimeter of a pentagon is the sum of its five sides :$ x + x + x + 3x + 3x = 9x$ If $x$ is $\frac{2}{3}$ of an inch, the perimeter is $9\big(\frac{2}{3}\big)$

The perimeter of the rectangle is the sum of all sides that is $2 \times $ length $+ 2 \times $ breadth So, we can say that the perimeter of a rectangle is twice
the sum of length and breadth.

We know, Perimter of a square
$= 4 \times $ Side $= 4 \times 4 = 16m$

Perimeter of the square $(P) = 48$ units
Step - by - step explanation:
Let side ofasquare $= a$ units
Diagonal$(d) = 12$ and $8730 ; 2$ units $\times $ given
Now,
Area of square $\text{(A)} = \text{a}^{2} = \frac{\text{d}^{2}}{2}$
$\text{a}^2 = \frac{(12\sqrt{2})^2}{2}$
$\Rightarrow\text{a}^2 = \frac{12^2\times2}{2}$
$\Rightarrow\text{a}^2 {(12\text{ unit})}^2$
$\Rightarrow\text{a} = \sqrt{12}^2$
$\text{a} = {12}\text{ unit}$
$\text{perimeter of the square }(\text{p}) = {4}\text{r} $
$= {4}\times{12}\text{ unit}$
$\therefore\text{p} = {48} \text{ unit}$

Perimeter of a rectangle is the sum of all its four sides.
Since, two sides measure l and the other two sides measure b, Perimeter of a rectangle whose length $(l)$ and breadth $(b)$ are given, is $l + b + l + b = 2(l + b)$

From $(I) :$ We can find the side, area and perimeter ofsquare.From
$(II) :$ Since Perimeter $\times $ rate of fencing per metre = Total cost (in rupees)
each statement alone is sufficient.

Area of figure $= 25 \times 1 = 25sq$. unit