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

 The length of the room $= l$
The breadth of the room $= b$
The height of the room $= h$
Therefore, the area of four walls will be
$= 2(l × h + h × b)$
$= 2h(l + b)$

 Here, radius$(r) = 7\ cm$
Let the height be $= h\ cm$
We Know that the total surface area of a cylinder is given by the formula
$=2\pi\text{r}(\text{h + r)}$
Putting the Values
$\Rightarrow 2\times\frac{22}{7}\times7\times(\text{h + 7)}=1144$
$= 44\times(\text{h + 7)} = 1144$
$= \text{h} = 26 - 7 = 19\text{cm}$

 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.
Hence, each side of the hexagon is equal to the radius of the hexagon and the perimeter of the hexagon is $6r.$

 Perimeter $= 4 × 2 = 8\ cm$

Given, a cylindrical hat with base radius $R$ and ris radius of the top surface.
Now, total surface area of hat $=$ Curved surface area $+$ Top surface area $+$ Base surface area
$=2\pi\text{rh}+\pi\text{r}^2+\pi(\text{R}^2-\text{r}^2)$
$=2\pi\text{rh}+\pi\text{r}^2+\pi\text{R}^2-\pi\text{r}^2$
$=2\pi\text{rt}+\pi\text{R}^2$

 $\pi\text{r}^2=16\pi$
$\Rightarrow \text{r}=4\text{cm}$
$\therefore$ Curved surface area,
$=2\times\pi\times4\times5 = 40\pi\text{cm}^2$

 Perimeter $= 2(4 + 3) = 14\ cm.$

 Let the new height and radius be $\frac{\text{h}}{4}$ and $2r$ respectively, where r and h are original radius and original height respectively of the cyinder.
We know that, curved surface area of cylinder $=2\pi\text{rh}$
Then, curved surface of the new cylinder
$=2\pi(2\text{r})\times\frac{1}{4}\text{h}=4\pi\text{r}\times\frac{1}{4}\text{h}=\pi\text{rh}$
$=\frac{1}{2}\times2\pi\text{rh}$ $[$multiplying and dividing by $2]$
$=\frac{1}{2}$ original curved surface area
Hence, the curved surface area of the cylinder will be halved.

 $\frac{\text{r}_1}{\text{r}_1} = \frac{\pi(1)^24}{\pi(2)^21} = 1:1$

 Perimeter $= 3 + 3 + 2 + 4 = 12\ cm$

 Here, length$(I) =13 m$
Breadth $(b)=9 m$
And, height( $h$ ) $=5 m$
We need to find the area of wall individually,
Area of the two opposite walls will be given by,
$2(b \times h)=2(9 \times 5)=90 m^2$
Area of the other two walls will be,
$2(I \times h)=2(13 \times 5)=130 m^2$
Area of the ceiling will be given by,
$2(I \times b)=2(13 \times 9)=234 m^2$
The required area will be,
Required area $=90 m^2+130 m^2+234 m^2$
$=454 m^2$
Total cost $=454 \times 7=$ $Rs. 3118$

 Given, $r = h$
Then, volume of cylinder $=\pi\text{r}^2\text{h}=\pi\text{r}^2\text{r}=\pi\text{r}^3$

Volume $= \frac{1}{2}\pi \times3\times3\times8 $
$=36\pi\text{cm}^3$

 $\frac{\pi\text{r}_1^2\text{h}}{\pi\text{r}_2^2\text{h}} = \frac{16\pi}{81\pi}$
$\Rightarrow \frac{\text{r}_1}{\text{r}_2} = \frac{4}{9}$

$ 1\ cm = 10\ mm$

 Perimeter $= 2(4 + 1) = 10\ cm.$

 Given,
Dimension of the room are $2m, 3m, 4m$
Dimension of the cubes are $\frac{1}{2}\text{m}, \frac{1}{3}\text{m}, \frac{1}{4}\text{m}$
Number of cubes placed in the room,
$=\frac{\text{Capacity of the room}}{\text{Volume of a cube} }$
$=\frac{2\times3\times4}{\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}}$
$2 × 3 × 4 × 2 × 3 × 4 = X$
$X = 576$

 Perimeter $= 4 + 3 + 5 = 12\ cm$

 Square formula $= 4\ ×$ side
$4 × 1 = 4$
Whereas perimeter of square $= 1 + 1 + 1 + 1 = 4$