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 $= 5x$
$= 5 \times 4$
$= 20\ cm$