Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{1.5}{3}=\frac{1}{x}\)
\(x=\frac{3 \times 1}{1.5}\)
x =  2
∴ EC = 2 cm

Total number of cards = 52
Number of aces and kings = 4 + 4 = 8
Number of cards which are neither ace nor king = 44
Probability that the card drawn is neither an ace nor a king = \(\frac{44}{52}=\frac{11}{13}\)

Given, P(E) = 0.05
we know that P(E) + P(\(\bar{E}\)) = 1
\(\therefore\) P(\(\bar{E}\)) = 1 - P(E)  \(\Rightarrow\) P(\(\bar{E}\)) = 1 - 0.05 = 0.95

Hint Given \(\triangle N S Q \cong \triangle M T R\) and \(\angle 1=\angle 2\)
To prove \(\triangle P T S \sim \triangle P R Q\)
Proof Since, \(\triangle N S Q \cong \triangle M T R\)
\( \therefore  S Q=T R \)
\( \text { Also, } \ \triangle=\angle 2 \)
\(\Rightarrow \quad P T=P S\)
[since sides opposite to equal angles are also equal] From Eqs. (i) and (ii),
\(\frac{P S}{S Q}=\frac{P T}{T R} \)
\(\Rightarrow S T \| Q R\)
[by converse of basic proportionality theorem]
\( \therefore \quad \angle 1=\angle P Q R\)
\(\text { and } \quad \angle 2=\angle P R Q \quad \text { [corresponding angles] }\)
In ΔPTS and ΔPRQ,
\(\angle T P S=\angle Q P R \quad \text { [common angle] }\)
\(\triangle=\angle P Q R \quad \text { [proved above] }\)
\(\text { and } \quad \angle 2=\angle P R Q\)
\(\therefore \quad \triangle P T S-\triangle P R Q \text { [by AAA similarity criterion] }\)
Hence proved. 

Total cards = 5
(i) P(Queen) = \(\frac{1}{5}\)
(ii) P(Ace) = \(\frac{1}{4}\) (Since, Queen was kept aside, 5 - 1 = 4).

P(sum is 8) = \(\frac{5}{36}\)

In \(\triangle\)POQ, AB || PQ,   (Given)
\(\Rightarrow \frac { AO }{ AP } =\frac { OB }{ BQ } \quad ...(i)\quad (BPT)\)
In \(\triangle\)OPR,   AC || PR,   (Given)
\(\therefore\frac { OA }{ AP } =\frac { OC }{ CR } \quad ...(ii)\quad (BPT)\)
From equations (i) and (ii),
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \)
\(\therefore\) BC || QR, (By converse of BPT)

Given,  \(\triangle ABE\cong \triangle ACD\)
\(\Rightarrow \)   AB = AC and AE = AD    [by CPCT]
\(\Rightarrow \)   \(\frac{AB}{AC} =1\)
and \(\frac{AD}{AE}=1\)
\(\Rightarrow\frac { AB }{ AC } =\frac { AD }{ AE } \)    .... (i)
In \(\triangle ADE\) and \(\triangle ABC\) , we have
\(\frac{AD}{AE}=\frac{AB}{AC}\)  [from Eq.(i)]
\(\Rightarrow\frac{AD}{AB}=\frac{AE}{AC}\)
and \(\angle DAE=\angle BAC\)   [common angle]
\(\therefore \) \(\triangle ADE\sim \triangle ABC\)  [by SAS similarity criterion]
Hence proved.

In \(\triangle PQR, \angle 1=\angle 2\)  [given]
\(\Rightarrow \) PR = PQ  [since, sides opposite to equal angles of a triangle are not equal]
Given that, \(\frac { QR }{ QS } =\frac { QT }{ PR } \Rightarrow \quad \frac { QR }{ QS } =\frac { QT }{ PQ } \)
[\(\because \)  PQ = PR, proved above]
\(\Rightarrow \frac { QS }{ QR } =\frac { PQ }{ QT } \)  
[on taking of the terms reciprocals]  ... (i)
In \(\triangle PQS\) and \(\triangle TQR\), we have
\(\angle PQS=\angle TQR\)     [common]
and \(\Rightarrow \frac { QS }{ QR } =\frac { QP }{ QT } \)  [from Eq.(i)]
\(\therefore \triangle PQS\sim \triangle TQR\)    [by SAS similarity criterion]
Hence proved.

Possible outcomes 
(0,0),(0,1),(0,1),(0,1),(0,6),(0,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)​​​​​​​
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)​​​​​​​
Different total scores are 0, 1, or 12
Let A = getting a total of 7
No. of favourable outcomes are = 12
\(\therefore\) P(A) = \(\frac{12}{36}=\frac{1}{3}\)​​​​​​​