Drac AC intersecting EF at G.

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in D ADC,
EG || DC (As EF || DC)
So, \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AG}}{\mathrm{GC}}\) Similarly, from D CAB,
\(\frac{\mathrm{CG}}{\mathrm{AG}}=\frac{\mathrm{CF}}{\mathrm{BF}}\)
ie., \(\frac{\mathrm{AG}}{\mathrm{GC}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)
Therefore, from (1) and (2),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)

In \(\triangle ADC\) and \(\triangle ACB\),
\(\angle ADC=\angle ACB\)  [each 90°]
\(\angle DAC=\angle CAB\) [common angle]
So, \(\triangle ADC\sim \triangle ACB\)
[by AA similarity criterion]
Then, \(\frac{AD}{AC}=\frac{AC}{BA}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \) AC² = AB x AD     ..... (i)
Similarity, \(\triangle BDC\sim \triangle BCA\)
\(\therefore \frac{BD}{BC}=\frac{BC}{AB}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow \) BC² = AB x BD     ..... (ii)
On dividing Eq (ii) by Eq (i), we get
\(\frac { { BC }^{ 2 } }{ { AC }^{ 2 } } =\frac { BD }{ AD } \)

In \(\triangle\) PQR, CA || PR
\(\therefore \quad \frac { PC }{ CQ } =\frac { RA }{ AQ } (By\quad BPT)\)
\(\Rightarrow \quad \frac { PC }{ 15 } =\frac { 20 }{ 12 } \)
\(\Rightarrow \quad PC=\frac { 20\times 15 }{ 12 } =25\)
\(\therefore \quad PC=25\quad cm\)
In \(\triangle\) PQR , CB || QR
\(\therefore \quad \frac { PC }{ CQ } =\frac { PB }{ BR } (By\quad BPT)\)
\(\Rightarrow \quad \frac { 25 }{ 15 } =\frac { 15 }{ BR } \)
\(\Rightarrow \quad BR=\frac { 15\times 15 }{ 25 } =9\quad cm\) 

Since the diagonals of a trapezium divide each other proportionally, we have
\(\therefore \frac { OA }{ OC } =\frac { BO }{ OD } \)
\(\Rightarrow \frac { x+5 }{ x+3 } =\frac { x-1 }{ x-2 } \)
\(\Rightarrow\) (x + 5)(x - 2) = (x - 1)(x + 3)
\(\Rightarrow\) x²- 2x + 5x - 10 = ~ + 3x - x - 3
\(\Rightarrow\) 3x-2x = 10-3
\(\therefore\) x = 7.

OA. OB = OC . OD    (Given)
So, \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}\)     (1)
Also, we have  \(\angle\)AOD = \(\angle\)COB   (Vertically opposite angles) (2)
Therefore, from (1) and (2),  \(\Delta\)AOD \(\sim\) \(\Delta\)COB (SAS similarity criterion)
So, \(\angle\)A = \(\angle\)C and \(\angle\)D = \(\angle\)B
(Corresponding angles of similar triangles)

Since, XY || QR
\(\therefore \frac { PX }{ XQ } =\frac { PY }{ YR } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { PY }{ PR-PY } \)
\(\Rightarrow\) PR-4=8
\(\Rightarrow\) PR=12 cm

In right \(\triangle\)PQR, QR²=PR²-PQ²
= 12²-6²
= 144-36=108
QR = 6\(\sqrt3\)cm

In \(\triangle\)AOB and \(\triangle\)COD,
AB || CD

OAB = DCO and OBA = ODC (Alternate angles)
\(\triangle\)AOB ~\(\triangle\)COD    (AA similarity)
\(\therefore \frac { AO }{ BO } =\frac { CO }{ DO } .\\ \) (Corresponding sides of similar triangles)
or \(\frac { AO }{ CO } =\frac { BO }{ DO } \)

From given figures.
\(\frac { PQ }{ ZY } =\frac { 4.2 }{ 8.4 } =\frac { 1 }{ 2 } ;\)
\(\frac { PR }{ ZX } =\frac { 3\sqrt { 3 } }{ 6\sqrt { 3 } } =\frac { 1 }{ 2 } ;\)
\(\frac { QR }{ ZY } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } ;\)
\(\Rightarrow \quad \frac { PQ }{ ZY } =\frac { PR }{ ZX } =\frac { QR }{ ZY } \)
\(\Rightarrow \triangle PQR\)~\( \triangle ZYX\)
\(\therefore \quad \angle X=\angle R\)
\(\therefore \quad \angle X=\)180⁰-(60⁰+70⁰)=50⁰

According to the question,
ST || QR
\(\therefore \frac { PS }{ SQ } =\frac { PT }{ PR } \)
Given, \(\frac { PS }{ SQ } =\frac { 3 }{ 5 } and PR=28cm\)
\(\Rightarrow\frac { 3 }{ 5 } =\frac { PT }{ 28 } \)
\(\therefore PT=\frac { 3\times 28 }{ 5 } =16.8cm\)

According to the question,  (Given)
\(\angle\)OPR = 90\(\unicode{xb0} \)
\(\therefore\) QR²=QP²+PR²
\(\therefore\) PR=\(\sqrt { { 26 }^{ 2 }-{ 24 }^{ 2 } } \)
\(=\sqrt { 100 } =10\)
\(\angle\)PKR = 90\(\unicode{xb0} \)
\(PK​​=\sqrt { { 10 }^{ 2 }-8^{ 2 } } \\ =\sqrt { 100-64 } \\ =\sqrt { 36 } =6 cm.\)

Given: AD=1.5 cm, BD=3 cm and AB=AD+BD=1.5+3.0=4.5 cm.
Given, In triangle ADE and ABC,
\(\angle\)A is common and DE || BC
\(\angle\)ADE=\(\angle\)ABC  (Corresponding angles)
\(\triangle\)ADE~\(\triangle\)ABC,  (AA similarity)
\(\frac { ar(\triangle ADE) }{ ar(\triangle ABC) } =\frac { { AD }^{ 2 } }{ { AB }^{ 2 } } =\frac { { (1.5) }^{ 2 } }{ { (4.5) }^{ 2 } } =\frac { 1 }{ 9 } \)
\(\Rightarrow \quad \frac { ar(\triangle ADE) }{ ar(\triangle ABC)-ar(\triangle ADE) } =\frac { 1 }{ 9-1 } \)
\(\Rightarrow \quad \frac { ar(\triangle ADE) }{ ar(trapezium\quad BCED) } =\frac { 1 }{ 8 } \)

Let \(\triangle\)ABC be an equilateral triangle of side 24 cm and Ad is altitude which is also a perpendicular bisector of side BC.
Hence BD=\(\frac { BC }{ 2 } =\frac { 24 }{ 2 } =12\quad cm\)
AB=24 cm
\(\therefore \quad AD=\sqrt { { AB }^{ 2 }-{ BD }^{ 2 } } \)
\(=\sqrt { { (24) }^{ 2 }-{ (12) }^{ 2 } } \)
\(=\sqrt { 576-144 } \)
\(=\sqrt { 432 } \)
\(AD=12\sqrt { 3 } cm\)
The length of the altitude is \(12\sqrt { 3 } cm.\)

AB² = OA² + OB² = (3)² + (4)²  (Pythagoras the)
AB = 5 cm
AB² + AC² = (5)² + (12)² = (13)² = BC²
\(\angle\)CAB = 90\(\unicode{xb0} \) (Converse of Pythagoras the)

\({ (3\sqrt { 3 } ) }^{ 2 }={ h }^{ 2 }+{ \left( \frac { 3\sqrt { 3 } }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow 27={ h }^{ 2 }+\frac { 27 }{ 4 } \)
\(\Rightarrow { h }^{ 2 }=27-\frac { 27 }{ 4 }\)
\( \Rightarrow { h }^{ 2 }=\frac { 81 }{ 4 }\)
\(\therefore h=\frac { 9 }{ 2 } =4.5\ cm\)

Let AB be the lamp-post and ED be the position of girl after 4s.
Given, height of the girl, ED = 100 cm
and height of the lamp-post, AB = 5 m = 500 cm
Distance of the girl from lamp-post after 4 s
= 1.9 x 4 = 7.6 m = 760 cm
[\(\because\)  distance = speed x time]
i.e. BD = 760 cm
Let DC = x cm
In \(\triangle CDE\) and \(\triangle CBA\),
\(\angle DCE=\angle BCA\) [common angle]
\(\angle CDE=\angle CBA\)  [each 90°]
\(\therefore \triangle CDE\sim \triangle CBA\) [by AA similarity criterion]
So, \(\frac { CD }{ CB } =\frac { DE }{ BA } \Rightarrow \frac { x }{ x+760 } =\frac { 100 }{ 500 } \)
\(\Rightarrow\)  5x = x + 760
\(\Rightarrow\)  4x = 760
\(\Rightarrow\)  x = 190 cm
Hence, the length of her shadow 4s is 190 cm.