Answer: Option B. -> 9: 16
:
B
Let the triangles be $△ABC$ and $△DEF$
Then $AB=3$ units and $DE=4$ units.
Each angle of both $△ABC$ and $△DEF$ is ${60}^{\circ }$.
$△ABE\sim △$ACF
(AAA similarity)
So,
$\frac{ar\left(ABC\right)}{ar\left(DEF\right)}=\frac{B{C}^2}{E{F}^2}=\frac{3^2}{4^2}=\frac{9}{16}$

Answer: Option B. -> 2:1
:
B

Let AB be a units long.
We know that diagonal of a square = $\sqrt{2}×$ side length
$⇒AC=\sqrt{2}×AB$
Now, $\mathrm{â–³}$ABEand $\mathrm{â–³}$ ACF are equilateral triangles.
Each angle of both $\mathrm{△}$ABEand $\mathrm{△}$ ACF is ${60}^{∘}$.
$∴\mathrm{△}ABE∼\mathrm{△}$ACF (by AAA similarity criterion)
So, $\frac{ar\left(ACF\right)}{ar\left(ABE\right)}=\frac{A{C}^2}{A{B}^2}=\frac{2a^2}{a^2}=\frac{2}{1}$

Answer: Option C. ->   a2 = b2 +  c2 + bc
:
C
In $\mathrm{â–³}CDB$,
${BC}^2={CD}^2+{BD}^2\text{[Pythagoras theorem]}$
${BC}^2={CD}^2+{(DA+AB)}^2$
${BC}^2={CD}^2+{DA}^2+{AB}^2+(2×DA×AB)...\left(i\right)$
In $\mathrm{â–³}ADC$,
${CD}^2+{DA}^2={AC}^2...\left(ii\right)\text{[Pythagoras Theorem]}$
Here, $\mathrm{âˆ}CAB={120}^{∘}$ (given)
$⇒\mathrm{âˆ}CAD={60}^{∘}$ (since$\mathrm{âˆ}CAD$ and$\mathrm{âˆ}CAB$ form a linear pair of angles)
Also, $\cos {60}^{∘}=\frac{AD}{AC}$
$AC=2AD...\left(iii\right)$
Substituting the values from $\left(ii\right)\&\left(iii\right)\text{in}\left(i\right)$ we get,
${BC}^2={AC}^2+{AB}^2+(AC×AB)$
$a^2=b^2+c^2+bc$
Alternatively,
Since $\mathrm{âˆ}A$ is an obtuse angle in $\mathrm{â–³}ABC$so,
${BC}^2={AB}^2+{AC}^2+2AB.AD={AB}^2+{AC}^2+2×AB×\frac{1}{2}×AC[âˆ\mu AD=AC\cos {60}^{∘}=\frac{1}{2}AC]={AB}^2+{AC}^2+AB×AC$
$⇒a^2=b^2+c^2+bc$

Answer: Option A. -> 1.4 cm
:
A
By Basic Proportionality Theorem,
$\frac{BD}{AD}=\frac{CE}{AE}$
$\frac{BD}{AD}+1=\frac{CE}{AE}+1$
$\frac{AD+BD}{AD}=\frac{AE+CE}{AE}$
$\frac{AB}{AD}=\frac{AC}{AE}$
Subtituting values:
$\frac{3.6}{2.1}=\frac{2.4}{AE}$
$AE=\frac{2.1×2.4}{3.6}=1.4cm$

:

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 =$2x^2$ -$4x$
$x=9$
$SoAC=2x=18$

:
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

Answer: Option B. -> BC2
:
B
Consider $\mathrm{â–³}ADB$ and$\mathrm{â–³}ABC$
∠BAD =∠BAC [common angle]
∠BDA=∠ABC [ ${90}^{∘}$]
Therefore by AA similarity criterion,$\mathrm{â–³}ADB$ and$\mathrm{â–³}ABC$ are similar.
So,$\frac{AB}{AC}=\frac{AD}{AB}$⇒${AB}^2$= AC.AD ---(I)
Similarly,$\mathrm{â–³}BDC$ and$\mathrm{â–³}ABC$ 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$\mathrm{â–³}ADB$,${AB}^2$ =${AD}^2$ + ${DB}^2$ and in$\mathrm{â–³}BDC$,${CB}^2$ = ${CD}^2$ +${DB}^2$[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 $\mathrm{â–³}ABC$ and$\mathrm{â–³}BDC$,
$\mathrm{âˆ}ABC=\mathrm{âˆ}BDC={90}^{∘}$
$\mathrm{âˆ}C=\mathrm{âˆ}C$ [Common angle]
Therefore by AA similarity criterion,$\mathrm{â–³}ABC$ and$\mathrm{â–³}BDC$ are similar.
$\frac{AC}{BC}$ =$\frac{BC}{DC}$
$B{C}^2$ = $AC×DC$

Answer: Option C. -> AB2 + CD2 = BC2 +  AD2
:
C

Since ΔADB and ΔDBC are right triangles, using Pythagoras theorem, we can write:
${AB}^2={AD}^2+{BD}^2...\left(1\right)$
${BC}^2={CD}^2+{BD}^2...\left(2\right)$
Subracting(2)from (1), we get,
${AB}^2={AD}^2+{BD}^2\underset{â}{{BC}^2={CD}^2+{BD}^2(−)(−)(−)}{AB}^2−{BC}^2={AD}^2−{CD}^2$
Rearranging, we get,
$⇒{AB}^2+{CD}^2={AD}^2+{BC}^2$

Answer: Option B. -> d√5
:
B

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).
$\mathrm{âˆ}AOB={180}^{∘}$
$⇒\mathrm{âˆ}APB=\frac{\mathrm{âˆ}AOB}{2}={90}^{∘}$
$∴\mathrm{△}APB$ is a triangle, right angled at $P$.
$⇒$ By pythagoras theorem,
${AB}^2={AP}^2+{PB}^2={\left(2PB\right)}^2+{PB}^2=5{PB}^2$
$⇒AB=\sqrt{5}PB$
$⇒PB=\frac{AB}{\sqrt{5}}=\frac{d}{\sqrt{5}}$

Answer: Option C. -> 15 m 
:
C
Let $AB$ and $CD$ be two trees such that $AB=20m,CD=28m\&BD=17m$
Draw $BE$ parallel to $AC$. Then,
$ED=8m$.
By applying Pythagoras' theorem,
$B{E}^2+D{E}^2=B{D}^2$.
$∴BE=\sqrt{{BD}^2−{DE}^2}=\sqrt{{\left(17\right)}^2−8^2}$
$=\sqrt{289−64}=\sqrt{225}=15m$
$∴AC=BE=15m$