Given TA = TP
∴ ∠P = ∠A (if angles then sides)

Now, \(\frac { PQ }{ PR } =\frac { 20 }{ 15 } =\frac { 4 }{ 3 } \)
\(\frac { PR }{ PS } =\frac { 15 }{ 11.25 } =\frac { 4 }{ 3 } \)
Also, \(\frac { QR }{ RS } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
We find \(\frac { PQ }{ PR } =\frac { PR }{ PS } =\frac { QR }{ RS } \)
That is, their corresponding sides are proportional.
∴ By SSS Similarity, ΔPQR ～ ΔPRS

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

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

\(In \ \triangle\ \mathrm{HIO}, \)
\(\angle \mathrm{HIP}=\angle \mathrm{HOP}(\text { given })\)
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:
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