In triangles \(\triangle \)RPQ and \(\triangle \) RTS, we have
\(\angle \)RPQ=\(\angle \)RTS [∵ given]
\(\angle \)PRQ =\(\angle \)TRS [∵ common]
\(\angle \)PQR =\(\angle \)RST l∵ Remaining angle]
\(\triangle \)RPQ ~ \(\triangle \)RTS [∵ By AAA similarity]

In triangles \(\triangle \)ABC and \(\triangle \)AED.
\(\angle \)ABC=\(\angle \)AED = 90°
\(\angle \)BAC=\(\angle \)EAD [Each equal to A]
∴ \(\angle \)ADE=\(\angle \)ACB [∵ Remaining angles]
∴ By AAA criteria of similarity  \(\triangle \)ABC ~\(\triangle \)AED.

The lower triangle cannot be constructed as the sum of two sides 5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed.

(i) in proportion
(ii) shape
(iii) equal
(iv) congruent
(v) similar

YO = 6 cm, OG = 6 cm, ∠O = 55°, ∠G = 35° and ∠A = 100°.

Steps:
1. Drawn a line segment OG = 6 cm
2. At G on DG made an angle ∠OGY = 55°
3. At G on GO made ∠GOX = 55°.
4. GY and OX meet cut A.
5. At A on OA made∠OAZ = 55°
6. Drawn an arc of radius 6 cm with center O. It cut AZ at Y.
7. Joined OY.
8. YOGA is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral YOGA = \(\frac { 1 }{ 2 } \times\ d\times({ h }_{ 1 }+{ h }_{ 2 })sq.units\) = \(\frac { 1 }{ 2 } \times5.2\times({ 5.9}+4.9)\) cm² 
=  \(\frac { 1 }{ 2 } \times5.2\times10.8\) cm² = 5.2 x 5.4 cm² = 28.08 cm²
Area of the quadrilateral = 28.08 cm²

AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.

steps:
1. Draw a line segment AG = 4.5 cm
2. At G on AG made ∠AGX = 110°
3. With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
4. At R on GR made ∠GRZ = 90°
5. At A on AG made ∠GAY = 90°
6. AY and RZ meet at I.
7. AGRI is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral AGRI
\( =1 / 2 \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right) \text { sq. units } \)
\(=1 / 2 \times 7 \times(2.4+2.7) \)
\(=1 / 2 \times 7 \times 5.1=17.85 \text { sq. cm. }\)

  Given, PL = 7 cm, LA = 6 cm, AY = 6 cm, PA = 8 cm and LY = 7 cm.

Steps:
1. Drawn a line segment PL = 7 cm
2. With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
3. Joined PA and LA.
4. With L and A as centers, drawn arcs of radii 7 crn and 6 cm respectively and let them cut at Y. 
5. Joined LY, PY and AY.
6. PLAY is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PLA Y = \(\frac { 1 }{ 2 } \times\ d \times({ h }_{ 1 }+{ h }_{ 2 })sq.units\)
\( =1 / 2 \times d \times\left(h_{1}+h_{2}\right) \text { sq. units } \)
\(=1 / 2 \times 8 \times(5.2+1) \)
\(=1 / 2 \times 8 \times 6.2 \)
= 24.8 sq. cm.

Given, KI = 5.4 cm, IT = 4.6 cm, TE = 4.5 cm, KE = 4.8 cm and IE = 6 cm.

Steps:
1. Drawn a line segment KI = 5.4 cm
2. With K and I as centers drawn arcs of radii 4.8 cm and 6 cm respectively and let them cut at E.
3. Joined KE and IE.
4. With E and I as centers, drawn arcs of radius 4.5 cm and 4.6 cm respectively and let them cut at T.
5. Joined ET and IT.
6. KITE is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral = \(\frac { 1 }{ 2 } \times\ d\times({ h }_{ 1 }+{ h }_{ 2 })sq.units\)
= \(\frac { 1 }{ 2 } \times6\times({ 3.4 }+3.9)\)cm²
= 3 x 7.3 cm² = 21.9 cm²
Area of the quadrilateral = 21.9 cm²

\( \text { In } \Delta \mathrm{GUB}, \) \(\angle \mathrm{G} =\angle \mathrm{B} \)
\(\therefore \mathrm{GU} =\mathrm{BU} \)
\(\text { In } \Delta \mathrm{MUT}, \ \therefore \mathrm{M} =\angle \mathrm{T} \)
\(\therefore \mathrm{MU} =\mathrm{TU} \)
\( In \triangle \mathrm{MUG} \ and \ \triangle TUB \)
\(G U=B U\)
\(\angle \mathrm{GUM}=\angle \mathrm{BUT}\)
\(\Delta \mathrm{MUG} \equiv \Delta \mathrm{TUB}\)
Hence proved

Given that l₁ || l₂
∴ In \(\triangle \)PQS and \(\triangle \)PRT
\(\angle \)P is common
\(\angle \)Q =\(\angle \)R  [ ∵ PR is the transversal for l₁ and l₂ corresponding angles]
\(\angle \)S =\(\angle \)T  [∵ corresponding angles]
∴ \(\triangle \)PQS ~ \(\triangle \)PRT  [ ∵ By AAA congruency]
In similar triangles, corresponding sides are proportional.
\(\frac { PQ }{ PR } =\frac { PS }{ PT } \\ \frac { 24 }{ 24+40 } =\frac { 30 }{ PT } \\ \frac { 24 }{ 64 } =\frac { 30 }{ PT } \\ PT=\frac { 30\times64 }{ 24 } \\ PT=80\)

\(\text { In } \triangle \mathrm{ODC} \text { and } \triangle \mathrm{EDC} \text {, }\)
CD = CD
\(\angle \mathrm{CDO}=\angle \mathrm{CDE}=90^{\circ}\)
OD = ED (given)
\(\therefore \triangle \mathrm{ODC} \equiv \Delta \mathrm{EDC}(\mathrm{SAS})\)
Hence proved

\(In \ \triangle\ \mathrm{HIO}, \)
\(\angle \mathrm{HIP}=\angle \mathrm{HOP}(\text { given })\)
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 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
\(\frac { WH }{ WE } =\frac { HY }{ ET } =\frac { WY }{ WT } \)
\(\\ \frac { 6 }{ 6+HE } =\frac { 4 }{ ET } =\frac { 4 }{ 16 }\)
\( \\ \frac { 6 }{ 6+HE } =\frac { 4 }{ 16 }\)
\( \\ 6+HE=\frac { 6 }{ 4 } \times16\)
\(\\ 6+HE=24\)
\(\\ \therefore HE=24-6\\ HE=18\)
\(\\ Again\quad \frac { 4 }{ ET } =\frac { 4 }{ 16 } \)
\(\\ ET=\frac { 4 }{ 4 } \times 16\)
\(\\ ET=16\)