By the exterior angle property of a cyclic quadrilateral, we get, y°=100° and
x° + 30°= 60⁰ and so x° = 30°

PQRS is a cyclic quadrilateral
Given ㄥPSR = 70°
ㄥPSR+ㄥPQR = 180° (state reason________)
70° +ㄥPQR = 180°
ㄥPQR = 180°− 70°
ㄥPQR = 110°
In ΔPQR we have, ㄥPQR+ㄥPRQ+ㄥQPR = 180° (state reason_________)
110° +ㄥPRQ + 40° = 180°
ㄥPRQ = 180°− 150°
ㄥPRQ = 30°

Draw OX perpendicular to QP
In ΔADP
MN = \(\frac { 1 }{ 2 } \)AP
In ΔBCQ
MN =\(\frac { 1 }{ 2 } \) QB
So, AP = BQ (or) AB+BP = AB+QA  ∴ PB = QA
∴ QA = AB = BP (or) QP = QA + AB + BP = 3 AB
Area of ΔOQP =\(\frac { 1 }{ 2 } \)\(\times\) QP \(\times\) OX
\(=\frac{1}{2} \times 3 \mathrm{AB} \times \mathrm{OX} \)
\( =\frac{3}{2} \times \mathrm{AB} \times \mathrm{OX} \)
\(=\frac{3}{2} \mathrm{AB}(\mathrm{OY}+\mathrm{YX}) \)
\( =\frac{3}{2} \times \mathrm{AB} \times \mathrm{OY}+\frac{3}{2} \times \mathrm{AB} \times \mathrm{YX} \quad(\mathrm{AB}=\mathrm{MN}) \)
\(=\frac{3}{2} \times \mathrm{MN} \times \mathrm{OY}+\frac{3}{2} \times \mathrm{AB} \times \mathrm{YX} \)
\(=3 \text { Area of } \triangle \mathrm{OMN}+\frac{3}{2} \text { Area of ABNM } \)
= 3\(\left[ \frac { 1 }{ 4 } Area\quad MNCD \right] +\frac { 3 }{ 2 } \left[ \frac { 1 }{ 2 } Area\quad ABCD \right] \)
= \(\frac { 3 }{ 4 } \left[ \frac { 1 }{ 2 } Area\quad ABCD \right] +\frac { 3 }{ 4 } \left[ \frac { 1 }{ 2 } Area\quad ABCD \right] \)
= \(\frac { 3 }{ 8 } \) Area ABCD + \(\frac { 3 }{ 4 } \) Area ABCD
= Area ABCD \(\left[ \frac { 3 }{ 8 } +\frac { 3 }{ 4 } \right] \) 
= Area ABCD \(\left[ \frac { 3+6 }{ 8 } \right] \)
= Hence area of \(\Delta\) QPO is \(\frac{9}{8}\) of the area of the parallelogram ABCD.

(i) Angle between band c = 30° (vertically opposite angles)
(ii) Angle between d and e = 180° - 75° = 105° (sum of the adjacent angles of a parallelogram is 180°)
(iii) Angle between d and f = 75° (opposite angles of a parallelogram)
(iv) Angle between c and f = 180° - 75° = 105° (Adjacent angles of a parallelogram)

\(\angle x=35^0\)(alternate angles are equal)
\(\angle y=180^0-x^0\)(angle of a straight line)
= 180° - 35°
= 145°​​​​​​​
The value of \(\angle x=35^0 \)and \(\angle y=145^0\).

\(\angle A=72^0\)
\(\angle C=180^0-72^0\) (\(\angle A\) and \(\angle C\) are supplementary)
= 108⁰
\(\angle A+\angle B+\angle C+\angle D=360^0\) (Total angles of a quadrilateral)
72^o+ 2x -10 +108^o+ x + 4 = 360^o
3x + 174^o = 360^o
3x = 360^o-174^o = 186⁰
\(x=\frac{186^0}{3}\)
x = 62^o
The value of x = 62°
\(\angle B=2x-10\)
= 2(62^o) -10
= 124^o-10^o = 114^o
\(\angle D=x+4\)
= 62^o+ 4 = 66^o
The other angles are 72°, 114°, 108° and 66°.

Given: AB = 2, BC = 6, AE = 6, BF = 8, CE = 7 and CF = 7
Consider ΔAEC and ΔBCF.
In ΔAEC, AE = 6, EC = 7 and AC = 8 (2 + 6 = 8)
In ΔBCF, BC = 6, CF = 1 and BF = 8
∴ ΔAEC ≅ BCF
∴ Area of ΔAEC = Area of ΔBCF (Two triangles are similar areas are equal)
Subtract area of ΔBDC on both sides we get,
Area of ΔAEC - Area of ΔBDC = Area of ΔBCF - Area of ΔBDC
Area of quadrilateral ABDE = Area of ΔCDF.
The answer is 1 : 1.

Given ∠BAD = 120° and AC bisects ∠BAD
\(\angle{BAC}=\frac{1}{2}\times 120^{0}=60^{0}\)
∠1 = ∠2 = 60°
AD || BC and AC is the traversal
∠2 = ∠4 = 60°
Δ ABC is isosceles triangle    [∴ ∠1 = ∠4 = 60 °]
⇒ AB = BC
Parallelogram ABCD is a rhombus.

Step 1 : Draw the ΔABC with AB = 6cm, ㄥB = 65° and AC = 7cm

Step 2 : Construct the angle bisectors of any two angles (A and B) and let them meet at I.
Then I is the incentre of ΔABC. Draw perpendicular from I to any one of the side (AB) to meet AB at D

Step 3: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally

Step 4: Measure inradius 
In radius = 1.9 cm

Step 1: 
Draw the ΔABC with the given measurements.

Step 2:
Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at S which is the circumcentre.
Step 3:
S as centre and SA = SB = SC as radius, draw the Circumcircle to passes through A, B and C. Circumradius = 3.9 cm.