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

  WO = 9 cm, OR = 6 cm, RK = 5 cm, ∠O = 100° and ∠R = 60°

Steps:
1. Drawn a line segment WO = 9 cm
2. At O on WO made an angle ∠WOR = 100° and drawn the ray OX.
3. Drawn an arc of radius 6 cm with center O. Let it intersect OX at R.
4. At R on OR, made ∠ORY = 60°, and drawn the ray RY.
5. With center R drawn an arc of radius 5 cm, let it intersect RY at K.
6. Joined WK.
7. WORK is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral WORK =  \(\frac { 1 }{ 2 } \times \ d\times({ h }_{ 1 }+{ h }_{ 2 })sq.units\) = \(\frac { 1 }{ 2 } \times11.7\times({ 0.9}+4.5)\) cm² 
= \(\frac { 1 }{ 2 } \times11.7\times5.4\) cm² = 31.59 cm²
Area of the quadrilateral = 31.59 cm²

MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°.

Steps:
1. Drawn a line segment MI = 3.6 cm
2. At M on MI made an angle ∠IMX = 50°
3. Drawn an arc with center M and radius 4 cm let it cut MX it D
4. At D on DM made an angle LMDY = 100°
5. With I as center drawn an arc of radius 4 em, let it cut DY at N.
6. Joined DN and IN.
7. MIND is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MIND
\( =1 / 2 \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right) \text { sq. units } \)
\(=1 / 2 \times 2.4 \times(3.3+2.9) \)
\(=1 / 2 \times 2.4 \times 6.2 \)
= 7.44 sq. cm.

Given in \(\triangle\) AIM
P is the mid point of MI and
N is the mid point of AI
\(\therefore \mathrm{PN} \| \mathrm{AM} \)
\(Now \)
\(\angle \mathrm{TMA}=\angle \mathrm{IAM}=\angle \mathrm{PNI} \)
\(\angle \mathrm{TAM}=\angle \mathrm{IMA}=\angle \mathrm{IPN} \)
\(\angle \mathrm{ATM}=\angle \mathrm{MIA}=\angle \mathrm{PIN} \)
\( \therefore \triangle \mathrm{PIN} \sim \triangle \mathrm{ATM} \)
Hence proved

\(\text { In } \triangle \mathrm{ODC} \text { and } \triangle \mathrm{EDC} \text {, }\)
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