In \(\triangle \)APB and \(\triangle \)DPC
\(\angle \)A =\(\angle \)D = 90° [given]
\(\angle \)APB =\(\angle \)DPC [Vertically opposite angles]
\(\angle \)ABP =\(\angle \)DCP [Remaining angle]
 ∴ \(\triangle \)APB ~ \(\triangle \)DPC [AAA criteria]
 \( \frac {AP }{ DP } =\frac { BP }{ PC }\) [Corresponding sides are proportional]
AP x PC = BP x DP

In  \(\triangle \)EPB and \(\triangle \)DPC,
\(\angle \)PEB =\(\angle \)PDC = 90° [given]
\(\angle \)EPB =\(\angle \)DPC [Vertically opposite angles]
\(\angle \)EPB =\(\angle \)PCD [∵ Remaining angles]
Thus, \(\triangle \)EPB ~ \(\triangle \)DPC [∵ By AAA criteria]
\(\frac { EP }{ DP } =\frac { PB }{ PC } \)
BP x PD = EP x PC

Given that ΔPEN ～ ΔPAD,
∴ \(\frac { PE }{ PA } =\frac { EN }{ AD } \Rightarrow \frac { 4 }{ 7 } =\frac { 6 }{ x } \Rightarrow x=\frac { 42 }{ 4 } \)
Also, \(\frac { PE }{ PA } =\frac { PN }{ PD } =\frac { y }{ y+5 } \)
i.e. \(\frac { 4 }{ 7 } =\frac { y }{ y+5 } \) ⇒ 4y + 20 =7y ⇒ 7y - 4y = 20
3y = 20 ⇒ \(y=\frac { 20 }{ 3 } \)cm

Now, from Fig in ΔABC \(\angle\)A = x (vertically opposite angles)
Similarly \(\angle\)B = \(\angle\)C = \(\angle\)x (Why?)
⇒ \(\angle\)A + \(\angle\)B + \(\angle\)C = 180^o (angle sum property in ΔABC)
⇒ 3x = 180^o
⇒ x = 60^o
⇒ y = 180^o − 60^o = 120^o

Now, from Fig PQ = PR
⇒\(\angle\)Q = \(\angle\)R (angles opposite to equal sides are equal)
⇒ \(\angle\)x = \(\angle\)y
⇒\(\angle\)x+ \(\angle\)y + 50^o = 180^o (angle sum property in ΔPQR)
⇒ 2\(\angle\)x = 130^o
⇒\(\angle\)x = 65^o
⇒ \(\angle\)y = 65^o

In triangles \(\triangle \)AOB and \(\triangle \)COD
\(\angle \)A =\(\angle \)C ( given) 
\(\angle \)AOB =\(\angle \)COD [∵ Vertically opposite angles] 
\(\angle \)ABO =\(\angle \)CDO l∵ Remaining angles of MOB and ~COD]
∴\(\triangle \)AOB ~\(\triangle \)COD [∵ AAA similarity]
∴ \(\triangle \)AOB ~\(\triangle \)COD [∵ AAA similarity]

In \(\triangle \)AOB and \(\triangle \)COD, we have 
\(\angle \)AOB =\(\angle \)COD [∵ Vertically opposite angles]
\(\frac { AO }{ OC } =\frac { BO }{ OD } \) [given]
So by SAS criteria of similarity we have \(\triangle \)AOB ~\(\triangle \)COD
\(\frac { AO }{ OC } =\frac { BO }{ OD } =\frac { AB }{ DC } \)
\(\frac { 1 }{ 2 } =\frac { 5 }{ DC } \)  [∵ AB = 5cm]
DC = 2 x 5
DC = 10 cm.

\(In \ \triangle\ \mathrm{HIO}, \)
\(\angle \mathrm{HIP}=\angle \mathrm{HOP}(\text { given })\)
Then \(\triangle\) HIO is isosceles
HI = HO
In \(\triangle\) HIP and \(\triangle\) HOP,
HI = HO
\(\angle \mathrm{HIP}=\angle \mathrm{HOP}\)
Side HP is common
\(\triangle\) HIP = \(\triangle\) HOP
Then IP = OP
Hence proved

\(Given \triangle \mathrm{EAT} \sim \triangle \mathrm{BUN} \)
\(\therefore \angle \mathrm{E}=\angle \mathrm{B}, \angle \mathrm{A}=\angle \mathrm{U}, \angle \mathrm{T}=\angle \mathrm{N}\)
Sum of the angles of a triangle is 180^o
\(\angle \mathrm{E}+\angle \mathrm{A}+\angle \mathrm{T}=180^{\circ}\)
x + 2x + x + 40 = 180^o
4x = 180^o - 40^o = 140^o
\( x =\frac{140^{\circ}}{4}=35^{\circ} \)
\(x\therefore \angle \mathrm{E} =\angle \mathrm{B}=x=35^{\circ} \)
\(\angle \mathrm{A} =\angle \mathrm{U}=2 x \)
\(=2 \times 35^{\circ}=70^{\circ} \)
\(\angle \mathrm{T} =\angle \mathrm{N}=x+40^{\circ} \)
\(=35^{\circ}+40^{\circ}=75^{\circ} \)

Given in\( \triangle \mathrm{WHY} and \ \triangle \mathrm{WET}, \)
\(\angle \mathrm{W} =\angle \mathrm{W} \)
\(\angle \mathrm{WYH} =\angle \mathrm{WTE} \)
\(\angle \mathrm{WHY} =\angle \mathrm{WET} \\ \therefore \Delta \mathrm{WHY} \sim \Delta \mathrm{WET} \)
Also \(\triangle \)WHY ~ \(\triangle \)WET
∴ Corresponding sides are proportionated
\(\frac { WH }{ WE } =\frac { HY }{ ET } =\frac { WY }{ WT } \)
\(\\ \frac { 6 }{ 6+HE } =\frac { 4 }{ ET } =\frac { 4 }{ 16 }\)
\( \\ \frac { 6 }{ 6+HE } =\frac { 4 }{ 16 }\)
\( \\ 6+HE=\frac { 6 }{ 4 } \times16\)
\(\\ 6+HE=24\)
\(\\ \therefore HE=24-6\\ HE=18\)
\(\\ Again\quad \frac { 4 }{ ET } =\frac { 4 }{ 16 } \)
\(\\ ET=\frac { 4 }{ 4 } \times 16\)
\(\\ ET=16\)

AB = AP + PB
= 3 + 6 cm = 9 cm.
AC = AQ + QC = 510 cm  =15
\(\frac { AP }{AB } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
\(\frac { AQ }{AC } =\frac { 5 }{15 } =\frac { 1 }{ 3 } \)
⇒\(\frac { AP }{AB } =\frac { AQ }{AC } \)
Thus in triangles APQ and ADC we have \(\frac { AP }{AB } =\frac { AQ }{AC } \) and \(\angle \)A = \(\angle \)A.
∴ By SAS criteria of similarity \(\triangle \)APQ ~\(\triangle \)ABC
⇒ \(\frac { AP }{AB } =\frac { PQ }{ BC } =\frac { AQ }{ AC } \)
⇒ \(\frac { PQ }{BC } =\frac { AQ }{ AC} \)
\(\frac { PQ }{BC } =\frac { 5 }{15 } =\frac { 1 }{ 3 } \)
\(\frac { PQ }{BC } =\frac { 1 }{ 3 } \)
⇒ BC = 3PQ

Given:
AB = 7 cm, AD = 5 cm, CD = 5 cm and two angles ∠BAC = 50° and ∠ABC = 60°


Steps:
1. Draw a line segment AB = 7 cm.
2. At A on AB, make ∠BAY = 50° and at B on AB, make ∠ABX = 60°. Let them intersect at C.
3. With A and C as centres, draw arcs of radius 5 cm. each. Let them intersect at D.
4. Join AD and CD.
5. ABCD is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral ABCD = \(\frac12\) ×d×(h₁+ h₂) sq.units
= \(\frac12\) x 6.4 x (3.8+5.3)
= 3.2 x 9.1 = 29.12 cm²