Maps and Scale - Class X

The size of the rectangle become double when it is scaled up $2$ times.

Radius of circle $=7 \ \ cm$

Resizing leads to change in scale factor and if the scale factor remains equal, then the figures will be similar to each other.

If the size of the figure does not get affected by rotation or reflection, then the figure will remain same and it will be congruent.

Area of square whose side is $a =a^2$ 

Let orignal length $=l$ and scales length $=sl$

When the triangle is scaled $1.5$ times then length of each side become $1.5$ times but the angle remains the same.

$\triangle ABC$ is enlarged to $\triangle A'B'C'$,
Thus, $\dfrac{C'A'}{CA} = 2.5$
Therefore, $\dfrac{C'A'}{4} = 2.5$
$\Rightarrow C'A' = 4 \times 2.5$
$\Rightarrow C'A' = 10$ cm 

$\triangle ABC$ is enlarged to $\triangle A'B'C'$,
Thus, $\dfrac{OA'}{OA} = 3$
$\Rightarrow \dfrac{6}{OA} = 3$
$\Rightarrow OA = \dfrac{6}{3}$
$\Rightarrow OA = 2 $ cm

$\triangle LMN$ 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

Let us take the point of start A and point at a distant $2$km be B.

Let the first Tree be at point A, second at point B and  third one a C.

Area occupies by the map=$(2/5)2530cm^{2}=300cm^{2}$
Area of $300cm^{2}$ on the map represent $10800 m^{2}$ real area of the map.
So,Area of $1cm^{2}$ on the map represent =$10800m^{2}/300cm^{2} 
                                                                       =  36m^{2}:1 cm^{2}$
Ratio of length=
$R=\sqrt { \frac { Real\quad area\quad  }{ Map\quad area } \quad  } =\sqrt { \frac { 36m^{2} }{ 1cm^{2} }  } =\frac { 6m }{ 1cm } $
Scale =1cm=6m
Answer (D) 1 cm = 6 m

In right triangles ALC and BMC, we have ∠ALC = ∠BMC = 90° and ∠ACM = ∠BCL (same angle at C). By AA similarity, ΔALC ~ ΔBMC. This gives the proportion: CM/CA = CL/CB, which can be rewritten as CM·CA = CL·CB. This is a standard power of a point result.

Scale $= k =$ $\dfrac{1}{400}$
Now, $\dfrac{\text{Length of model}}{\text{Length of aeroplane}} = k$
$\text{Length of model }= \dfrac{40 \times 100}{400}$
$\text{Length of model} =10$ cm

Given: $\triangle ABC$, D, E and F are mid points of AB, BC, CA respectively.

In $\triangle ABC$
F is mid point of AC and D is mid point of AB. 
Thus, By Mid point theorem
$FD = \dfrac{1}{2} CB$ and $FD = CB$ or $FD = CE$ and $FD \parallel CE$ (1)

Given: $\triangle ABC$, $D, E \ and \ F$ are mid points of $AB, BC, CA$ respectively.

In $\triangle ABC$

In $\triangle ABC$,
AD bisects $\angle A$
By angle bisector theorem,
$\dfrac{AB}{AC} = \dfrac{BD}{DC}$
$\dfrac{BD}{DC} = \dfrac{3}{4}$

Scale $= k =$ $\dfrac{1}{400}$
Now, $\dfrac{\text{Length of model}}{\text{Length of aeroplane}} = k$
$\text{Length of aeroplace} = 400 \times 16$
$\text{Length of model }= 6400$ cm
$\text{Length of model }= 64$ m

Given: $AB = 3 cm$, $BC = 2 cm$ and $CA = 2.5 cm$ and $EF = 4 cm$ 

Ar. ($\triangle ABC$) = $32 cm^2$
$BC = 4 cm$
$EF = 5 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \frac{BC^2}{EF^2}$
$\frac{32}{A(\triangle DEF)} = \frac{4^2}{5^2}$
$A(\triangle DEF) = \frac{32 \times 25}{16}$
$A(\triangle DEF) = 50 cm^2$

Ar. ($\triangle _{1}$) = $48 cm^2$
Ar. ($\triangle _{2}$) = $75 cm^2$
$a _{1} = 3.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of their altitudes.
Thus, $\frac{A(\triangle _{1})}{A(\triangle _{2})} = \frac{(a _1)^2}{(a _2)^2}$
$\frac{48}{75} = \frac{(3.6)^2}{(a _2)^2}$
$(a _2)^2 = \frac{12.96 \times 75}{48}$
$(a _2)^2 = 20.25$
$a _2 = 4.5$cm

Ar. ($\triangle ABC$) = $49 cm^2$
Ar. ($\triangle PQR$) = $25 cm^2$
$AB = 5.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\dfrac{A(\triangle ABC)}{A(\triangle PQR)} = \dfrac{AB^2}{PQ^2}$
$\dfrac{49}{25} = \dfrac{(5.6)^2}{PQ^2}$
$PQ^2 = \dfrac{31.36 \times 25}{49}$
$PQ^2 = 16$
$PQ = 4$cm

Scaling factor = $\frac{Length \quad of \quad model}{Length \quad of \quad building}$
$\frac{1}{30} = \frac{1.2}{Length \quad of \quad building}$
Length of building = $1.2 \times 30$
Length of building = $36$ m
Similarly, Breadth of building = $75 \times 30$ = 22.5 m
Height of building = $30 \times 2$ = 60 m
hence, dimensions of building are : $36 \times 22.5 \times 60$

Given $0.7$cm represents $8.4$ km 

$\triangle ABC = \triangle PQR$, then
$\dfrac {\text {Perimeter of}\triangle ABC}{\text {Perimeter of}\triangle PQR} = \dfrac {AB}{PQ} = \dfrac {BC}{QR} = \dfrac {AC}{PR}$
$\dfrac {36}{24} = \dfrac {AB}{10}; AB = \dfrac {36}{24}\times 10 = 15\ cm$.