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

Perimeter of rectangle $=480\text{m}$
$\therefore\text{l + b}=\frac{480}{2}=240\text{m}$
$\text{l : b}=5:3$
$\therefore\text{length}=\frac{240\times5}{5+3}=\frac{240\times5}{8}=150$
and breadth $=\frac{240\times3}{5+3}=\frac{240\times3}{8}=90\text{m}$
and area $=\text{l}\times\text{b}=150\times90=13500\text{m}^2$

Let the radius of the circle be $r\ cm$.
Circumference $=(2\pi\text{r})\text{cm}$
$(2\pi\text{r})=44$
$\Rightarrow\Big(2\times\frac{22}{7}\times\text{r}\Big)=44$
$\Rightarrow\text{r}=\Big(\frac{44\times7}{2\times22}\Big)=7\text{cm}$
$\therefore$ Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times7\times7\Big)\text{cm}^2=154\text{cm}^2$

Perimeter of room $= 18\ m$
and height $= 3\ m$
Area of $4$ walls = Perimeter $\times $ height
$=18 \times 3=54 \mathrm{~m}^2$

Given that the area of the square is $50\ cm^2$
We know:
Area of a square
$=\Big\{\frac{1}{2}\times(\text{Diagonal})^2\Big\}\text{ sq. units}$
$\therefore$ Diagonal of the square $=\sqrt{2\times\text{Area of the square}}$
$=(\sqrt{2\times50})\text{cm}=(\sqrt{100})\text{cm}=10\text{cm}$
Hence, the diagonal of the square is $10\ cm$.

Length of rectangle $AB = 16\ cm$
and diagonal $BD = 20\ cm$

But, in right $\triangle\text{ABD}$
$B D^2=A B^2+A D^2$
$\Rightarrow(20)^2=(16)^2+A D^2$
$\Rightarrow 400=256+A D^2$
$\Rightarrow A D^2=400-256=144=(12)^2$
$\Rightarrow A D=12 \mathrm{~cm}$
$\text { Area }=l \times b=16 \times 12=192 \mathrm{~cm}^2$

Let $a = 13\ cm, b = 14\ cm$ and $c = 15\ cm$
$\text{s}=\frac{\text{a + b + c}}{2}$
$=\Big(\frac{13+14+15}{2}\Big)\text{cm}$
$=21\text{cm}$
$\therefore$ Area of the triangle
$=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}\text{ sq. units}$
$=\sqrt{21(21-13)(21-14)(21-15)}\text{cm}^2$
$=\sqrt{21\times8\times7\times6}\text{cm}^2$
$=\sqrt{3\times7\times2\times2\times2\times7\times2\times3}\text{cm}^2$
$=(2\times2\times3\times7)\text{cm}^2=84\text{cm}^2$

Let $r$ be the radius of the circle Then
$\text{c}=2\pi\text{r}$
$2\pi\text{r}-\text{r}=37$
$\Rightarrow\text{r}\Big(2\times\frac{22}{7}-1\Big)=37$
$\Rightarrow\text{r}\Big(\frac{44}{7}-1\Big)=37$
$\Rightarrow\text{r}\Big(\frac{37}{7}\Big)=37$
$\therefore\text{r}=\frac{37\times7}{37}=7\text{cm}$
and area $=\pi\text{r}^2=\frac{22}{7}\times7\times7=154\text{cm}^2$

Side of a square $= a$
Area = $a^2$
Side of equilateral triangle $= a$
$\therefore\text{Area}=\frac{\sqrt{3}}{4}\text{a}^2$
$\therefore$ Ratio in their areas $=\frac{\text{a}^2}{\frac{\sqrt{3}}{4}\text{a}^2}$
$=\frac{4\text{a}^2}{\sqrt{3}\text{a}^2}=\frac{4}{\sqrt{3}}$
$=4:\sqrt{3}$

Length of diagonals of a rhombus are $24\ cm$ and $18\ cm$
$\therefore\text{Area}=\frac{\text{Product of diagonals}}{2}$
$=\frac{24\times18}{2}=216\text{cm}^2$

Sides are $13\ cm, 14\ cm, 15\ cm$
$\therefore\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{13+14+15}{2}=\frac{42}{2}=21$
and area $=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21\times8\times7\times6}$
$=\sqrt{3\times7\times2\times2\times2\times7\times2\times3}$
$=2\times2\times3\times7=84\text{cm}^2$

Let original radius $=\text{r}$
Then its area $=\pi\text{r}^2$
Radius of increased circle $=\text{r}+1$
$\therefore\text{Area}=\pi(\text{r + 1})^2$
Now $\pi(\text{r}+1)^2-\pi\text{r}^2=22$
$\Rightarrow\pi(\text{r}^2+2\text{r}+1)-\pi\text{r}^2=22$
$\Rightarrow\pi\text{r}^2+2\pi\text{r}+\pi-\pi\text{r}^2=22$
$\Rightarrow\pi(2\text{r}+1)=22$
$\Rightarrow\frac{22}{7}(2\text{r}+1)=22$
$\Rightarrow2\text{r + 1}=\frac{22\times7}{22}$
$\Rightarrow2\text{r}=7-1=6$
$\Rightarrow\text{r}=\frac{6}{2}=3$
Radius of original circle $=3\text{cm}$

Let the length of the rectangular park be $4x.$
Breadth $= 3x$
Perimeter of the park $= 2 (l + b) = 56m$ (given)
$\Rightarrow 56 = 2 (4x + 3x)$
$\Rightarrow 56 = 14x$
$\Rightarrow x = 4$
Length $= 4x = (4 × 4) = 16m$
Breadth $= 3x = (3 × 4) = 12m$
Area of the rectangular park $= 16m \times 12m=$ $192 m^2$