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

Proof:

Proof

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

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)

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

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.

  PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120^o.

Steps:
1. Draw a line segment PQ = 3.5 cm
2. Made \(\angle \)Q = 120°. Drawn the ray QX.
3. With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
4. With Rand P as centres drawn arcs of radii 5.2 cm and 5.5 cm respectively and let them cut at S.
5. Joined PS and RS.
6. PQRS is the required quadrilateral.
Calculation of Area:
\( =1 / 2 \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right) \text { sq. units } \)
\(=1 / 2 \times 6.3 \times(1.8+4.2) \)
\(=1 / 2 \times 6.3 \times 6\)
= 18.9 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

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

\(\text { In } \triangle \mathrm{CAE} \angle \mathrm{A}=\angle \mathrm{E} \text { (given) }\)
\(\triangle\)CAE is isosceles
CA = CE     .........(1)
\(\triangle\)BCD is isosceles with base BD (given)
CB = CD         .........(2)
\((1)-(2) \Rightarrow \mathrm{CA}-\mathrm{CB} \equiv \mathrm{CE}-\mathrm{CD}\)
AB = ED
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
\( x =\frac{140^{\circ}}{4}=35^{\circ} \)
\(x\therefore \angle \mathrm{E} =\angle \mathrm{B}=x=35^{\circ} \)
\(\angle \mathrm{A} =\angle \mathrm{U}=2 x \)
\(=2 \times 35^{\circ}=70^{\circ} \)
\(\angle \mathrm{T} =\angle \mathrm{N}=x+40^{\circ} \)
\(=35^{\circ}+40^{\circ}=75^{\circ} \)

  \(\frac { GU }{ BO} \)= \(\frac { 32 }{ 8 } \) = \(\frac { 4 }{ 1 } \)
\(\frac { UM }{ OX} \)= \(\frac { 48 }{ 12 } \)= \(\frac { 4 }{ 1 } \)
\(\frac { GM }{ BX} \)= \(\frac {52 }{ 13 } \)=\(\frac { 4 }{ 1 } \)
We find that \(\frac { GU }{ BO} \)= \(\frac { UM }{ OX} \)= \(\frac { GM }{ BX} \)= \(\frac { 4 }{ 1 } \)
That is their corresponding sides are proportional.
:. By SSS similarity \(\triangle \)GUM ~ \(\triangle \)BOX.

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: 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