To prove: Ratio of the areas of two triangles of the same bases is equal to the ratio of their heights.
Proof:
Let us take triangle ${ T } _{ 1 }$ with height  ${ h } _{ 1 }$ and base ${ b } _{ 1 }$ and Triangle ${ T } _{ 2 }$ with height  ${ h } _{ 2 }$ and base ${ b } _{ 2 }$.

Area of Triangle ${ T } _{ 1 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 }$

Area of Triangle ${ T } _{ 2 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 }$

Ratio of area of two triangles,
$\dfrac { \text{Area of triangle} \ { T } _{ 1 } }{\text{ Area of triangle} \ { T } _{ 2 } } =\frac { \frac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 } }{ \frac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 } }$      
$ =\dfrac { h _{ 1 } }{ { h } _{ 2 } } $                                      (Base of two triangles are equal. So, $ { b } _{ 1 } = { b } _{ 2 }$) 
Proved.              

Areas of two similar triangles are $12 cm^2$ and $48 cm^2$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
Hence, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h _2)^2}$

$(h _2)^2= 4 \times (2.1)^2$

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$

The length of stick corresponds to the length of tower and shadow of stick corresponds to the shadow of tower.
Thus, $\dfrac{Length\ Stick}{Shadow\ Stick} = \dfrac{Length\ Tower}{Shadow\ Tower}$

Area of similar triangles are in the ratio of square of the corresponding sides.
Hence, $\dfrac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \frac{2^2}{3^2}$
$\Rightarrow \dfrac{12}{\text{Area of larger triangle}} = \dfrac{4}{9}$
$\text{Area of larger triangle}  = 27$

In $\displaystyle \Delta ABC$ and $\displaystyle \Delta EDA$,
We have
$\displaystyle \angle ABC=\angle ADE$ [Each equal to $\displaystyle { 90 }^{ o }$]
$\displaystyle \angle ACB=\angle EAD$ [Alternate angles]
$\displaystyle \therefore $ By AA Similarity
$\displaystyle \Delta ABC\sim \Delta EDA$
$\displaystyle \Rightarrow \frac { BC }{ AB } =\frac { AD }{ DE } $
$\displaystyle \Rightarrow DE.BC=AD.AB$.
Hence proved.

In two similar triangles, if the corresponding sides are in a particular ratio, then altitudes will also be in the same ratio.
Hence the ratio of the altitudes will be 2 : 3.

If two sides of a triangle are proportional to the corresponding two sides in another triangle, and their included angles are equal, then the two triangles are similar by SAS rule.

If $\quad \dfrac { AB }{ DE } = \dfrac { BC }{ FD } $, then for two triangles ABC and DEF to be similar, the included angle must be equal. In this case, the included angles are $\quad \angle B\quad and\quad \angle D$.

Ratio of areas of two similar triangles is equal to ratio of squares of the corresponding altitudes and ratio of squares of corresponding medians. This means that if the ratio of either altitude or median is given and asked to find ratio of areas , then it will be the ratio of squares of the corresponding altitudes or ratio of squares of corresponding medians.

Two similar triangles have equal angles.

If two triangles are similar to each other, then the ratio of the area of this triangle will be equal to the square of the ratio of the corresponding sides of this triangle.

In $\displaystyle \Delta ABC\sim \Delta DEF$
$\displaystyle \frac { ar.\left( \Delta ABC \right)  }{ ar.\left( \Delta DEF \right)  } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } $
$\displaystyle \frac { 36 }{ 64 } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } $
$\displaystyle \frac { 6 }{ 8 } =\frac { 3 }{ DE } $
$\displaystyle DE=\frac { 8\times 3 }{ 6 } =\frac { 24 }{ 6 } =4cm$
Therefore, D is the correct answer.

Given the areas of two similar triangles are $121$ sq cm and $81$ sq cm

Altitudes of similar triangles are in ratio $4:9$
Hence, area of these triangles
$=$ square of the ratio of their heights or altitudes
$=(4:9)^2=16:81$
Option $(4)$.

$\triangle ABC\sim \triangle DEF$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$\Rightarrow \dfrac { ar(ABC) }{ ar(PQR) } ={ \left( \dfrac { AB }{ PQ }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 2.25 }{ 6.25 } ={ \left( \dfrac { AB }{ .5 }  \right)  }^{ 2 }\ \Rightarrow \dfrac { AB }{ .5 } =\dfrac { 15 }{ 25 } \ \Rightarrow AB=.3m\ \Rightarrow AB=.3\times 100=30cm$

Given $\triangle ABC\sim \triangle DEF$

In $\triangle ABC$

The areas and sides of similar triangles are related as 

Areas of two similar triangles are $49 $ cm $^2$ and $64$ cm $^2.$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding sides.
Hence, $\dfrac{A _1}{A _2} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow \dfrac{49}{64} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow\dfrac{s _1}{s _2} = \dfrac{7}{8}$

$\displaystyle\frac{16}{9}=\left[\frac{AB}{PQ}\right]^2=\left[\frac{BC}{QR}\right]^2$

$\displaystyle\Rightarrow \frac{16}{9}=\left[\frac{AB}{18}\right]^2$ and $\displaystyle\frac{16}{9}=\left[\frac{12}{QR}\right]^2$

$\displaystyle \Rightarrow \frac{4}{3}=\frac{AB}{18}$ and $\displaystyle \frac{4}{3}=\frac{12}{QR}$

$\Rightarrow AB=24$ cm, $QR=9$ cm.

$\triangle ABC$ and $\triangle DEF$ be the given triangles in which $AB=AC, DE=DF$, $\angle A=\angle D$
and $\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac{16}{25}$
Draw $AL\bot  BC$ and $DM\bot  EF$
Now, $\cfrac{AB}{BC}=1$ and $\cfrac{DE}{DF}=1$  ($\because \quad AB=AC;\quad DE=DF$)
$\Rightarrow \cfrac{AB}{AC}=\cfrac{DE}{DF}$,
$\therefore$ $\ln \triangle ABC$ and $\triangle DEF$, we have
$\cfrac{AB}{DE}=\cfrac{AC}{DF}$ and $\angle A=\angle D$
$\Rightarrow$ $\triangle ABC\sim \triangle DEF$ [By SAS similarity axiom)
But, the ratio of the areas of two similar $\triangle s$ is the same as the ratio of the squares of their corresponding heights.
$\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac {{AL}^{2}}{{DM}^{2}}$
$\Rightarrow$ $\cfrac{16}{25}={ \left( \cfrac {AL}{DM}  \right)  }^{ 2 }$
$\Rightarrow$ $\cfrac{4}{5}$
$\therefore$ $AL:DM=4:5$, i.e., the ratio of their corresponding heights$=4:5$

Given, $\triangle ABC \sim \triangle PQR$,

$\triangle ABC\sim \triangle DEF\quad $ (Given)
$\Rightarrow \cfrac { ar(ABC) }{ ar(DEF) } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $ (ratio of Areas of Similar triangles are equal to ratio of squares of corresponding sides)
$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left{ \cfrac { BC }{ EF }  \right}  }^{ 2 }={ \left{ \cfrac { 8 }{ 11 }  \right}  }^{ 2 }$
$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$
$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,
Therefore, $\displaystyle \frac{ar\left ( \triangle ABC \right )}{ar\left ( \triangle DEF \right )}=\frac{BC^{2}}{EF^{2}}$ 

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding medians. Therefore,
$\displaystyle \frac{121}{64}=\frac{\left ( 12.1 \right )^{2}}{x^{2}},$ where $x$ is the median of the other $\triangle .$
$\Rightarrow $ $\displaystyle x^{2}=\frac{\left ( 12.1 \right )^{2}\times 64}{121}\Rightarrow x=\sqrt{\frac{121}{100}\times 64}$
   $\displaystyle =\frac{11}{10}\times 8=8.8$ cm.

By theorem on ratio of areas of similar triangles, we get

$\dfrac {A(\triangle ADE)}{A(\triangle ABC)} = \left(\dfrac {AD}{DB}\right)^2$

$\therefore \dfrac 19 = \dfrac {AD^2}{DB^2}$

$\therefore \dfrac {AD}{DB}= \dfrac 13$.

We know that the relation between area of two similar triangle:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. 

Given two triangles are similar, then the ratio of the areas $=a^2:b^2$

Area of $\triangle ABC= \cfrac 12 \times base \times height$

Similar triangles are triangle with similar shape but can have different sizes. 

  $DE=4\,cm,\,BC=8\,cm$ and $ar(\triangle FDE)=25\,cm^2$                  [ Given ]

Given: Area of $two$ similar triangle $81{cm}^{2}$ and $49{cm}^{2}$

$\triangle ABC$ & $\triangle PQR$ are similar.

We know ratio  of areas of two similar triangles is the ratio of the square of their corresponding sides.

Given, sides of two similar triangles are in the ratio $2:3$

Given $\triangle ABC$ with $D$ a point on $BC$ such that $3BD=BC$

$\dfrac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\dfrac { 100 }{ 144 } $

Given, $DE||BC$

Let $x$ be the distance from $P$ to $AB$. By similar triangles.
$\dfrac {1}{2} = \dfrac {(1 - x)^{2}}{1^{2}}; \therefore 1 - x = \pm \dfrac {1}{\sqrt {2}}; \therefore x = \dfrac {2 - \sqrt {2}}{2}$
(negative sq. root rejected).

$\Delta ABC \Delta DEF$
$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.
We know that,
$\dfrac{Area of \Delta  ABC}{ Area of \Delta  DEF} = $ $\Rightarrow \dfrac{64}{49} = \left ( \dfrac{7}{DE} \right )^2$
$\Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \frac{7}{DE} \right )^2 \Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \dfrac{7}{DE} \right )^2 = \dfrac{49}{8}$cm 

Given $ \triangle  ABC \sim  \triangle  QRP $
Therefore, $ \frac { Area\triangle ABC\quad  }{ Area\triangle QRP\quad }=\frac { { BC }^{ 2 } }{ { PR }^{ 2 } }  $
or $ \frac { 9 }{ 4 }  = \frac { { 15 }^{ 2 } }{ { PR }^{ 2 } }  $
or $ { PR }^{ 2 }\quad =\quad \frac { { 15 }^{ 2 }\quad \times \quad 4 }{ 9 }  $ cm.
Therefore, $ { PR }=\frac { { 15 }\times \quad 2 }{ 3 }  = $ 10 cm.

Option A: This statement is correct. The ratio of the altitudes of the similar triangles  is  $2:1$

Ratio of the areas of the similar triangles $=$ Square of the ratio of  the  altitudes.

$\therefore$ Ratio  of  the  areas $ =  { \left( \dfrac { 2 }{ 1 }  \right)  }^{ 2 }=  4:1$

Option B: If  $PQ\parallel BC$,  then $\triangle APQ \sim \triangle ABC$ by AA test of similarity.

Hence, $\dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac { AP^2 }{ AB^2 }$

If $AP = x$ and $BP = 2x$, then $AB = 3x$.

$\therefore \dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac 19$

So, the given statement is false

Two triangle ABC and PQR are similar then,
$AB: BC:CA :: PQ: QR: PR$
Given $BC: CA = 1:2$ 
Hence, $QR : PR = BC : CA = 1:2$

In $\triangle XYZ$, using Pythagoras theorem,
$XY^2 = XZ^2 + YZ^2$
$\pi XY^2 = \pi XZ^2 + \pi YZ^2$ (Multiply by $\pi$)
$A _z = A _x + A _y$