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

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 $3m.$