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

Proof:
In \(\triangle \)ABC and \(\triangle \)DAC we have,
\(\angle \)ADC =\(\angle \)BAC [ ∵ given ]
\(\angle \)L =\(\angle \)C [ ∵ common]
\(\angle \)ABC =\(\angle \)DAC [∵ Remaining angle]
\(\triangle \)ABC ~ \(\triangle \)DAC [∵  By AAA similarity]
⇒ \(\frac { AB }{ DA } =\frac { BC }{ AC } =\frac { AC }{ DC } \) [sides are proportional]
⇒ \( \frac { CB }{ CA } =\frac { CA }{ CD }\)
or CA² = CB x CD.

Given that  ΔPQR ～ ΔXYZ
∴ Their corresponding sides are proportional
⇒ \(\frac { PQ }{ XY } =\frac { QR }{ YZ } =\frac { PR }{ XZ } \)
⇒ \(\frac { 8 }{ a } =\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 8 }{ a } =\frac { 10 }{ 16 } \)
⇒ \(a=\frac { 8\times 16 }{ 10 } =\frac { 128 }{ 10 } \)
a = 12.8 cm
Also, \(\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 14\times 16 }{ 10 } =\frac { 224 }{ 10 } \)
∴ b = 22.4 cm

Extend MO and let it meet RS at Y. Extend NO and let it meet PQ at X.
As PQ || RS,
∠OYN = ∠OMX = 55^o (alternate angles are equal)
∴ ∠MON = ∠OYN +∠ONY (exterior angle of ΔOYN = sum of interior opposite angles)
= 55^o + 30^o = 85^o
⇒∠MOX = 180^o− 85^o = 95^o (∠MON, ∠MOX are linear pair)

Given:
PQ = 5 cm, QR = 5 cm, ∠QPR = 50^o, ∠PRS = 40^o and ∠RPS = 80^o


Steps:
1. Draw a line segment PQ = 5 cm.
2. At P on PQ, make ∠QPX = 50^o
3. With Q as centre, draw an arc of radius 5 cm. Let it cut PX at R.
4. At R on PR, make ∠PRS = 40^o and at P on PR, make  ∠RPS = 80^o.Let them intersect at S.
5. PQRS is the required quadrilateral
Calculation of Area:
Area of the quadrilateral PQRS = \(\frac12\) ×d×(h₁+ h₂) sq. units
= \(\frac12\) x 6.4 x (4.7 + 3.8)
= 3.2 x 8.5 = 27.2 cm²

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

Given:
MA = 4 cm, AT = 3.6 cm,
TH = 4.5 cm, MH = 5 cm and ∠A = 85°


Steps:
1. Draw a line segment MA = 4 cm.
2. Make ∠A = 85°.
3. With A as centre, draw an arc of radius 3.6 cm. Let it cut the ray AX at T.
4. With M and T as centres, draw arcs of radii 5 cm and 4.5 cm respectively and let them cut at H.
5. Join MH and TH.
6. MATH is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MATH = \(\frac12\) × d × (h₁+ h₂) sq.units
= \(\frac12\) x 5.1 x (3.9 + 2.8)
= 2.55 x 6.7 = 17.09 cm²

Given:
NI = 4.5 cm, IC = 4.3 cm,
NE = 3.5 cm and two diagonals,
NC = 5.5 cm and IE = 5 cm


Steps:
1. Draw a line segment NI = 4.5 cm.
2. With N and I as centres, draw arcs of radii 5.5 cm and 4.3 cm respectively and let them cut at C.
3. Join NC and IC.
4. With N and I as centres, draw arcs of radii 3.5 cm and 5 cm respectively and let them cut at E.
5. Join NE, IE and CE .
6. NICE is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral NICE = \(\frac12\)× d × (h₁+ h₂) sq. units
= \(\frac12\) x 5 x (2.4 + 3.1)
= 2.5 x 5.5 = 13.75 cm²

Given: DE = 6 cm, EA = 5 cm, AR = 5.5 cm,
RD = 5.2 cm and a diagonal DA = 10 cm


Steps:
1. Draw a line segment DE = 6 cm.
2. With D and E as centres, draw arcs of radii 10 cm and 5 cm respectively and let them cut at A.
3. Join DA and EA.
4. With D and A as centres, draw arcs of radii 5.2 cm and 5.5 cm respectively and let them cut at R.
5. Join DR and AR.
6. DEAR is the required quadrilateral
Calculation of Area:
Area of the quadrilateral DEAR = \(\frac12\) x d x (h₁+ h₂) sq. units
= \(\frac12\) x 10 x (1.9+ 2.3)
= 5 x 4.2 = 21 cm²