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),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

Given OP = 5 cm, radius r = 3 cm
To find the length of tangent PT.
In right angled \(\triangle\)OTP
OP² = OT² + PT² gives PT² = 25 - 9 = 16
Length of the tangent PT = 4cm

Let x be the length of the ladder. BC = 4 ft, AC = 7 fit.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65, Hence \(x=\sqrt { 65 } \)
The number \(\sqrt { 65 } \) is between 8 and 8.1.
82 = 64 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1ft

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.

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

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 \(\triangle\)PST and \(\triangle\)PQR,
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } ,\frac { PT }{ PR } =\frac { 4 }{ 4+2 } =\frac { 2 }{ 3 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is common
Therefore, by SAS similarity,
\(\triangle\) PST~\(\triangle\)PQR

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.

Since OP is the radius and PT tangent
\(\angle O P T=90^{\circ}\)
Applying Pythagoras theorem in \(\triangle\)OPT, we have
OT² = OP² + PT²
13² = 5² + PT²
PT² = 169 - 25
PT² = 144
PT = 12cm
Since the lengths of tangents drawn from an exterior point to a circle are equal.
AP = AE = x(say)
AT = PT - AP = (12-x)cm
Since AB is the tangent to the circle at E
\( \therefore \mathrm{OE} \perp \mathrm{AB} \)
\(\ \Rightarrow \angle O E A =90^{\circ} \)
\(\\ \Rightarrow \angle A E T =90^{\circ}\)
\( \\ \mathrm{AT}^{2} =\mathrm{AE}^{2}+\mathrm{ET}^{2}\)
[Applying Pythagoras theorem in \(\triangle\)AET ]
(12 - x)² = x² +(13 -5)²
144 - 24 + x² = x² +64
24x = 144 - 64
24x = 80
3x = 10
\( x=\frac{10}{3} \mathrm{~cm} \)
\(Similarly\ B E=\frac{10}{3} \mathrm{~cm}\)
AB = AE + BE
\( =\left(\frac{10}{3}+\frac{10}{3}\right) \mathrm{cm} \)
\(A B =\frac{20}{3} \mathrm{~cm}\)

Given: ABCD is a trapezium in which DC || AB and EF || AB
To prove that \(\frac { AE }{ ED } =\frac { BF }{ FC } \)
Construction : join AC meeting EF at G
Proof:
In ADC, we have
EG || DC
\(\Rightarrow \frac{A E}{E D}=\frac{A G}{G C}\) [By Thales theorem]               ...(1)
In ABC , we have
\(\frac{A G}{G C}=\frac{B F}{F C}\) [By Thales theorem]        ....(2)
From (1) and (2), we get
\(\frac{A E}{E D}=\frac{B F}{F C}\)

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9