Tag: mapping space around us

$\triangle LMN$ is reduced to $\triangle L'M'N'$,
Thus, $\dfrac{L'M'}{LM} = 0.8$
$\Rightarrow \dfrac{5.4}{LM} = 0.8$
$\Rightarrow LM = \dfrac{5.4}{0.8}$
$\Rightarrow LM = 6.75$ cm 

$\triangle LMN$ is reduced to $\triangle L'M'N'$,
Thus, $\dfrac{M'N'}{MN} = 0.8$
$\Rightarrow \dfrac{M'N'}{8} = 0.8$
$\Rightarrow M'N' = 0.8 \times 8$
$\Rightarrow M'N' = 6.4 $ cm 

$\triangle ABC$ is enlarged to $\triangle A'B'C'$,
Thus, $\dfrac{B'C'}{BC} = 3$
$\Rightarrow \dfrac{B'C'}{BC} = 3$
$\Rightarrow BC = \dfrac{15}{3}$
$\Rightarrow BC = 5$ cm 

For similarity of triangles we have SSS criteria. So $S _1$ is true. 
But for polygons to be similar, the corresponding sides must be in equal ratio as well as the corresponding angles must be congruent.

Since, there is nothing mentioned about the angles of the pentagons, so $S _2$ is false.

$\triangle ABC$ is enlarged to $\triangle A'B'C'$,
Thus, $\dfrac{A'B'}{AB} = 3$
$\Rightarrow \dfrac{A'B'}{4} = 3$
$\rightarrow A'B' = 4 \times 3$
$\Rightarrow A'B' = 12 $ cm 

$\triangle ABC$ is enlarged to $\triangle A'B'C'$,
Thus, $\dfrac{OC'}{OC} = 3$
$\Rightarrow \dfrac{OC'}{21} = 3$
$\Rightarrow OC' = 21 \times 3$
$\Rightarrow OC' = 63 $ cm

Flagstaff and shade forms right triangle with height $17.5$ m and base $40.25$ m

Ratio of the length of the sides of the two triangle $=2:5$

If image is enlarged, $k>1.$
If image size does not change, then $k=1.$
If image size is reduced, $k<1.$

Maps must have a scale and a legend