Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

PQ II RS


Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

Distance between the insect and the foot of the lamp post BD = 8 m
The height of the lamp post, AB = 6 m
After moving a distance of x m, let the insect be at C
Let, AC = CD = x . Then BC = BD − CD = 8 − x
In \(\triangle\)ABC, \(\angle\)B = 90^o
AC² = AB² + BC² gives x² = 62 + (8 - x)2
x² = 36 + 64 − 16x + x²
16x = 100 then x = 6.25
Then, BC = 8 − x = 8 − 6.25 = 1.75m
Therefore the insect is 1.75 m away from the foot of the lamp post.

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

\(\Delta \ CAB\) and \(\Delta CED\),\(\angle C\) is common, \(\angle A=\angle CED\)
Therefore, \(\Delta CAB\sim \Delta CED\)
Hence, \(\frac { CA }{ CE } =\frac { AB }{ DE } =\frac { CB }{ CD } \)
\(\frac { AB }{ DE } =\frac { CB }{ CD } \quad \frac { 9 }{ x } =\frac { 10+2 }{ 8 } ,x=\frac { 8\times 9 }{ 12 } =6\) cm.

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

Let the first aeroplane starts from O and goes upto A towards north, (Distance=Speed × time)
where \(OA=\left( 100\times \frac { 3 }{ 2 } \right) km=1500km\)
Let the second aeroplane starts from O at the same time and goes upto B towards west,
where \(OB=(1200\times \frac { 3 }{ 2 } )=1800km\)
The required distance to be found is BA.
In right angled triangle AOB, AOB, AB² = OA² + OB²
AB² = (1500)² + (1800)² = 100² (15² +18²)
= 100² x 549 = 100² x 9 x 61
\(AB=100\times 3\times \sqrt { 61 } =300\sqrt { 61 } kms.\)

Since, \(\triangle\)QAQC is a right triangle at C, AQ² = AC² + QC²  ...(1)
Also, \(\triangle\)BPC is a right triangle at C, BP² = BC²+ CP²        ...(2)
\(\triangle\) ABCC is a right triangle at C, AB² = AC² + BC²              ....(3)
From (1) and (2), AQ²+  BP² = AC²+ QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)²+ ABC² + (2CP)²
=  4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²) (From equation (3))
4(AQ² + BP²) = 5AB²

Given, Speed = 1.2 m/s,
time = 4 seconds
Distance = speed x time
= 1.2 x 4
= 4.8 m
Let x be the length of the shadow after 4 seconds
\(\Delta ABE\sim \Delta CDE,\frac { BE }{ DE } =\frac { AB }{ CD } \) gives \(\frac { 4.8+x }{ x } =\frac { 3.6 }{ 0.9 } =\frac { 3.6 }{ 0.9 } =4\) (since 90 cm = 0.9 m)
4.8 + x = 4x, gives 3x = 4.8 so, x = 1.6m
The length of his DE = 1.6m