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

Let $r_1$ and $r_2$ be the radius of the two circles. Then,
$\frac{2\pi\text{r}_1}{2\pi\text{r}_2}=\frac{3}{4}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{3}{4}$
Now
Ratio of areas $=\frac{\pi\text{r}^2_1}{\pi\text{r}^2_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{3}{4}\Big)^2=\frac{9}{16}$
Thus, the required ratio is $9 : 16.$
Hence, the correct option is $(c).$

Let $r_1$ and $r_2$ be the radius of the two circles. Then,
$\frac{\pi\text{r}^2_1}{\pi\text{r}^2_2}=\frac{49}{36}$
$\Rightarrow\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{7}{6}\Big)^2$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{7}{6}$
Now
Ratio of circumferences $=\frac{2\pi\text{r}_1}{2\pi\text{r}_2}=\frac{\text{r}_1}{\text{r}_2}=\frac{7}{6}$
Thus, the required ratio is $7 : 6.$
Hence, the correct option is $(a).$

Let $r$ be the radius of the circle and $a$ be the side of the equilateral triangle. Then,
Area of circle = $154\ cm^2$
$\Rightarrow154=\frac{22}{7}\times\text{r}^2\Rightarrow\text{r}=\sqrt{\frac{154\times7}{22}}=7\text{cm}$
Therefore
Area of equilateral triangle $=\frac{\text{a}^2\sqrt{3}}{4}=\frac{7^2\sqrt{3}}{4}=\frac{49\sqrt{3}}{4}  ( \because a = r = 7\ cm)$
Hence, the correct option is $(c).$

Let $r_1$ and $r_2$ be the radius of the two circles. So
$\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}$
Now
Ratio of areas $=\frac{\pi\text{r}^2_1}{\pi\text{r}^2_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{2}{3}\Big)^2=\frac{4}{9}$
Thus, the required ratio is $4 : 9.$
Hence, the correct option is $(b).$

 Let $r$ be the radius of the circle. Then,
Area of original circle $=\pi\text{l}^2\text{cm}^2$
Radius of circle after increment $= (r + 1)\ cm$
Thus,as per the question
$\pi(\text{r}+1)^2-\pi\text{r}^2=22$
$\Rightarrow(\text{r}+1)^2-\text{r}^2=\frac{22}{\frac{22}{7}}=7$
$\Rightarrow\text{r}^2+2\text{r}+1-\text{r}^2=7$
$\Rightarrow\text{r}=3\text{cm}$
Thus, the original radius of the circle is $3\ cm.$
Hence, the correct option is $(b).$

Let $r_1$ and $r_2$ be the radius of the two circles. Then,
$2\pi\text{r}-\text{r}=37$
$\Rightarrow2\times\frac{22}{7}\times\text{r}-\text{r}=37$
$\Rightarrow\frac{44\text{r}-7\text{r}}{7}=37$
$\Rightarrow\text{r}=\frac{37\times7}{37}=7\text{cm}$
Now
Area of circle $=\pi\text{r}^2=\frac{22}{7}\times7\times7=154\text{cm}^2$
Hence, the correct option is $(c).$

Let $r$ be the radius of the circle. Then,
$44=2\pi\text{r}\Rightarrow\text{r}=\frac{44}{2\times\frac{22}{7}}=7\text{cm}$
Therefore
Area of circle $=\pi\text{r}^2=\frac{22}{7}\times7^2=154\text{cm}^2$
Hence, the correct option is $(b).$