The ratio of areas of two similar triangles is equal to ratio of squares of corresponding sides.
$\frac{49}{64}=\frac{(\text{Side of first 1st triangle}{)}^2}{(\text{Corresponding side of 2nd triangle}{)}^2}$
 
$\frac{7}{8}=\frac{\text{Side of first 1st triangle}}{\text{Corresponding side of 2nd triangle}}$ 
$\Rightarrow$ Ratio of corresponding sides $=7:8$

Consider $\mathrm{â–³}AEF$
AE = AF (As $\mathrm{â–³}ABC$ is equilateral and E and F are midpoints of sides AC and AB respectively)
$⇒\mathrm{âˆ}AFE=\mathrm{âˆ}AEF=\mathrm{âˆ}A={60}^o$
$⇒\mathrm{△}AEF$ is equilateral.
Similarly we can prove that $\mathrm{△}DEC$  and $\mathrm{△}BFD$ are equilateral.
Since $\mathrm{â–³}DEF$ share a common side with each of the other three triangles, $\mathrm{â–³}DEF$ is also equilateral.
$⇒\mathrm{â–³}DEFâ‰\dots \mathrm{â–³}AEF$
$⇒\mathrm{â–³}DEFâ‰\dots \mathrm{â–³}DEC$
$⇒\mathrm{â–³}DEFâ‰\dots \mathrm{â–³}BFD$ by using SSS postulate.

In $\mathrm{â–³}$ABC,

It's given LM $âˆ\yen$ AB,
So, $\frac{AL}{CA}=\frac{BM}{CB}$      [By BPT]
$\frac{(x−3)}{2x}=\frac{(x−2)}{(2x+3)}$

$(x−3)×(2x+3)$ = $(x−2)×2x$

$2x^2$ - $3x$ - 9 = $2x^2$ - $4x$

$x=9$
$So\text{Â}AC=2x=18$

By applying Pythagoras theorem:

$A{B}^2$+$B{C}^2$=$A{C}^2$

$6^2$+${2.5}^2$=$A{C}^2$

We get, AC = 6.5m

Consider  $\mathrm{△}ADB$ and  $\mathrm{△}ABC$

∠BAD = ∠BAC                [common angle]

∠BDA = ∠ABC                [ ${90}^{∘}$]

Therefore by AA similarity criterion, $\mathrm{△}ADB$ and  $\mathrm{△}ABC$ are similar.

So,  $\frac{AB}{AC}=\frac{AD}{AB}$  ⇒  ${AB}^2$ = AC.AD   ---(I)

Similarly, $\mathrm{△}BDC$ and  $\mathrm{△}ABC$ are similar.

So, $\frac{BC}{AC}$ =  $\frac{DC}{BC}$  ⇒  ${BC}^2$ = AC.DC   ---(II)

Dividing (I) and (II) and cancelling out AC, we get,   ${\left(\frac{AB}{BC}\right)}^2$ =   $\frac{AD}{DC}$  ----(III)

Also, In $\mathrm{△}ADB$, ${AB}^2$ = ${AD}^2$ + ${DB}^2$ and in  $\mathrm{△}BDC$, ${CB}^2$ = ${CD}^2$ + ${DB}^2$ [Pythagoras theorem]

Subtracting these two equations above and cancelling off ${DB}^2$ on both sides, we get 

${AB}^2$ - ${BC}^2$ = ${AD}^2$ - ${CD}^2$ ⇒    ${AB}^2$ + ${CD}^2$ = ${AD}^2$  + ${BC}^2$  --------------(IV)

Dividing this equation with ${BC}^2$ on both sides,  $\frac{{AB}^2}{{BC}^2}$ +  $\frac{{CD}^2}{{BC}^2}$ =  $\frac{{AD}^2}{{BC}^2}$ +  $\frac{{BC}^2}{{BC}^2}$   ⇒  $\frac{{AB}^2}{{BC}^2}$ + $\frac{{CD}^2}{{BC}^2}$ = $\frac{{AD}^2}{{BC}^2}$ + 1

$⟹$  $\frac{{AB}^2}{{BC}^2}$ - 1 = $\frac{{AD}^2}{{BC}^2}$ - $\frac{{CD}^2}{{BC}^2}$

$⟹$ $\frac{AD}{DC}$ - 1 =  $\frac{{AD}^2−{CD}^2}{{BC}^2}$    [Using (III)]

$⟹$$\frac{AD−DC}{DC}$ = $\frac{(AD−CD)(AD+CD)}{{BC}^2}$

$⟹$$\frac{1}{DC}$ = $\frac{AC}{{BC}^2}$ 

$⟹$ ${BC}^2$ = AC.DC

Alternatively,

Consider $\mathrm{△}ABC$ and $\mathrm{△}BDC$,

$\mathrm{âˆ}ABC=\mathrm{âˆ}BDC={90}^{∘}$

$\mathrm{âˆ}C=\mathrm{âˆ}C$    [Common angle]

Therefore by AA similarity criterion, $\mathrm{△}ABC$ and  $\mathrm{△}BDC$ are similar.

$\frac{AC}{BC}$ = $\frac{BC}{DC}$

$B{C}^2$ = $AC×DC$

Since ΔADB and ΔDBC are right triangles, using Pythagoras theorem, we can write:

${AB}^2={AD}^2+{BD}^2\text{Â}\text{Â}...\left(1\right)$

${BC}^2={CD}^2+{BD}^2\text{Â}\text{Â}...\left(2\right)$

Subracting (2) from (1), we get,
${AB}^2={AD}^2+{BD}^2\underset{â}{{BC}^2={CD}^2+{BD}^2(−)\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}(−)\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}(−)}{AB}^2−{BC}^2={AD}^2−{CD}^2$
Rearranging, we get,
$⇒{AB}^2+{CD}^2={AD}^2+{BC}^2$

Let the other two sides of the triangle be a and b. Since the largest side is the hypotenuse, we have, by Pythagoras theorem,

${Hypotenuse}^2$ = $a^2$ +  $b^2$

${13}^2$ = $a^2$ + $b^2$^ = 169           ------------------- (I)

Now,

Sum of the areas of the three squares = ${\left(side\text{Â}of\text{Â}yellow\text{Â}square\right)}^2$ + ${\left(side\text{Â}of\text{Â}brownish\text{Â}square\right)}^2$  +

${\left(side\text{Â}of\text{Â}blue\text{Â}square\right)}^2$

Sum of the areas of the three squares =$a^2$  + $b^2$^ + ${13}^2$ = ${13}^2$ + ${13}^2$ (from (I))

Sum of the areas of the three squares = 169 + 169 = 338.

 

Let O be the centre of the circle. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. (here that point is P). 
$\mathrm{âˆ}AOB={180}^{∘}$
$⇒\mathrm{âˆ}APB=\frac{\mathrm{âˆ}AOB}{2}={90}^{∘}$

$∴\text{Â}\mathrm{â–³}APB$ is a triangle, right angled at $P$.
$⇒$ By pythagoras theorem,
${AB}^2={AP}^2+{PB}^2={\left(2PB\right)}^2+{PB}^2\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}\text{Â}=5\text{Â}{PB}^2$
$⇒AB=\sqrt{5}\text{Â}PB$

$⇒PB=\frac{AB}{\sqrt{5}}=\frac{d}{\sqrt{5}}$

 

The ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding heights
So,
$\frac{(Heightofsmaller△{)}^2}{(Heightofbigger△{)}^2}=\frac{12}{48}=\frac{1}{4}$

Let the height of bigger triangle be  x
$\therefore$$\frac{(2.1{)}^2}{x^2}=\frac{1}{4}\iff x=\sqrt{4\times (2.1{)}^2}$ 
                           $=(2\times 2.1)=4.2cm$

All congruent figures are similar but the similar figures need not be congruent as in case of similar figures only shape is considered whereas, in the case of congruent figures, both shape and sizes are considered. Hence, the statement is correct.