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In rhombus ABCD,
AO = OP + PA = 2 + 2 ⇒ AO = 4
and OB = OQ + QB = 2 + 1 ⇒ OB = 3
In ΔOAB,
(AB)² =(OA)² + (OB)²
[by Pythagoras theorem]
⇒ (AB)² =(4)² + (3)² = 25
⇒  AB=5
Since, AB is diameter of semi-circle.
∴ Radius = \(\frac { 5 }{ 2 } \)=2.5
Hence, radius of the semi-circle is 2.5. The value depict here is participation unity and creativity.

Let ABGD is a rhombus

Then, BD = 8 cm, AC =15 cm
and AB = BG = CD = DA
We know that, in a rhombus, diagonals bisect each other at right angles.
So, DO = OB and AO = OC
∴ DO =4 cm and OC = 7.5 cm
Now, we see that a /lDOC is formed such that
ㄥDOC = 90°
∴ (DC)² = (DO)² + (OC)²
[∵ in a right angled triangle, the square of the side opposite to the right angle is equal to the sum of the squares of other two sides]
(DC)² =(4)² + (7.5)² = 16+ 56.25
= 72.25 cm²
or DC = \(\sqrt { 72.25 } \) = 8.5 cm
Thus, side of the rhombus is 8.5 cm.

Let the adjacent angles are 3x and 7x respectively. We know that, in a parallelogram sum of two adjacent angles is 180°.

∴ 3x+7x=180⁰   ⇒ 10x=180⁰
⇒ x=180⁰ x \(\frac { 1 }{ 10 } \) ⇒ x=18⁰
So, the adjacent angles are
3 x 180⁰ =54⁰  and 7 x 18⁰ =126⁰
Also, we know that, in a parallelogram opposite angles are equal.
Thus, all the angles of the parallelogram are 54°, 126°,54°, 126°.

Let perpendicular bisector of PQ is TS.

Such that  ㄥPTS = ㄥSTQ = 90°
Also, PT = TQ and TS = ST
So, we see that
ΔPTS  ≅ ΔQTS       [by SAS]
So, we can say that,
P S= SQ
But PQRS is a rhombus.
So, PQ = PS = SQ
So, ΔPQS is an equilateral triangle.
ㄥSPQ = 60° or ㄥR = 60°
[opposite angles of rhombus are equal]
Also, ㄥQ = 180° - 60° = 120°
[∵ adjacent angles of a rhombus are supplementary]
and ㄥQ = ㄥS = 120°
Thus, ㄥP = 60° , ㄥR = 60°
ㄥQ = 120°, ㄥS = 120°

In a regular pentagon,
Exterior angle = \(\frac { 360^{ 0 } }{ 5 } \)=72⁰

and interior angle =180°- 72° =108°
ㄥEAB = 2 ㄥEAM
[∵  AM is the bisector of ㄥA]
⇒ 108° = 2 ㄥEAM
⇒ ㄥEAM = \(\frac { 108^{ 0 } }{ 2 } \)=54⁰
Now, in quadrilateral EAMD,
ㄥMAE + ㄥAED + ㄥEDM + ㄥDMA = 360°
⇒ 54⁰ + 108⁰ + 108⁰ + ㄥDMA=360⁰
⇒ 270° + ㄥDMA = 360°
⇒ ㄥDMA = 360° - 270° = 90°
⇒ ㄥAMC = 180° ㄥDMA
= 180° - 90° = 90°.

Let the regular polygon have n sides.
∴ Exterior angle =\(\frac { { 360 }^{ 0 } }{ n } \)
and interior angle = \({ 180 }^{ 0 }-\frac { { 360 }^{ 0 } }{ n } =\frac { 180^{ 0 }n-{ 360 }^{ 0 } }{ n } \)
According to the question,
\(\frac { { 360 }^{ 0 } }{ n } ;\frac { { 180 }^{ 0 }n-{ 360 }^{ 0 } }{ n } \)=1:3
⇒ \(\frac { { 360 }^{ 0 } }{ n } \times \frac { n }{ { 180 }^{ 0 }n-{ 360 }^{ 0 } } =\frac { 1 }{ 3 } \)
⇒ 360⁰ x 3 =180⁰n-360⁰
⇒ 1080⁰+360⁰=180⁰n
⇒ \(\frac { { 1440 }^{ 0 } }{ 180^{ 0 } } \)=n  ⇒ n=8
∴ Exterior angle = \(\frac { { 360 }^{ 0 } }{ n } =\frac { 360^{ 0 } }{ 8 } \)=45⁰
and interior angle = 180⁰-45⁰=135⁰

(i) Given, ABCD is a parallelogram in which ㄥB = 100°.
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
ㄥA + ㄥB = 180°
[∴ ㄥA and ㄥB are adjacent angles]
⇒ z+100⁰=180° ⇒ z=180⁰-100° =80°
Also, opposite angles of a parallelogram are of equal measure.
∴  ㄥD=ㄥB ⇒ y=100⁰
and ㄥC=ㄥA ⇒ x=z=80°
Hence, the measure of x, y and z are 80°,100° and 80°, respectively.

(ii) Let a parallelogram be ABCD in which ㄥD = 50° .
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴ ㄥA + ㄥD=50° ⇒ x + 50⁰= 180°
⇒ x = 180°-50° = 130°
Also, opposite angles of a parallelogram are of equal measure.
∴  ㄥC=ㄥA ⇒ y=x=130⁰ and ㄥB=ㄥD=50°
Now, ㄥB + exterior ㄥB = 180°
⇒ 50° + z = 180° [by linear pair angle]
⇒ z = 180⁰- 50° = 130°
Hence, the measure of angles x, y and z are 130°, 130° and 130° , respectively.

(iii) Let a parallelogram be ABCD, in which ㄥCBO = 30°. Here, AC and BD intersect each other at O and ㄥAOD=90°.
∴ ㄥCOB = ㄥAOD=90°
[vertically opposite angles]
We know that, the sum of three angles of a triangle is 180°.
In ΔOBC,
ㄥCOB + ㄥOCB + ㄥCBO = 180°
⇒ 90° +y +30° = 180° ⇒ Y + 120° = 180⁰
 y=180⁰ - 120⁰= 60⁰
As AD II BC and AC is a transversal.
 y = z = 60° [alternate interior angles]
Hence, the measure of angles x, y and z are 90°, 60 and 60° , respectively.

(iv) Let parallelogram be ABCD, in which ㄥB = 80°
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴  ㄥB + ㄥC = 180° ⇒ 80⁰+ ㄥC= 180°
⇒ ㄥC = 180° - 80°= 100°
Also, ㄥC + exterior ㄥC = 180° [by linear pair
:. exterior LC or z = 180° - 80°= 100°
Also, opposite angles of a parallelogram are of equal measure.
So, ㄥA = ㄥC ⇒ x = 100°
and ㄥD = ㄥB ⇒  y = 80°
Hence, the measure of angles x, y and z are 100°,80° and 100°, respectively.

(v) Let parallelogram be ABCD, in which
ㄥB = 112° and ㄥDAC = 40°.
We know that, the sum of any two adjacent angles of a parallelogram is 180°.
∴ ㄥA + ㄥB = 180°
⇒ (40° + z) + 112° = 180°
⇒ 40⁰ + z + 112° = 180° ⇒ z+152°=180⁰
⇒ z = 180⁰-152°=28⁰
Also, opposite angles of a parallelogram are of equal measure.
So, ㄥD = ㄥB ⇒ y=112°
Now, in ㄥACD, using angle sum property of a triangle,
ㄥCAD + ㄥD + ㄥDCA = 180⁰
⇒ 40⁰+y+x=180⁰
⇒ 40° +112° + x = 180°
⇒ 152° + x =180⁰ - x=180⁰-152°=28⁰
Hence, the values of x, y and z are 28⁰, 112° and 28°, respectively.

We know ,that, the sum of all angles of a triangle is 180°. Therefore
In ΔAPB, we have
ㄥx + ㄥ2 + ㄥ3 =180° ⇒ ㄥx=1800 - ㄥ2- ㄥ3 ... (i)
In ΔBPC, we have
ㄥy + ㄥ4 + ㄥ5 = 180° ⇒ ㄥy = 180°- ㄥ4 - ㄥ5 ... (ii)
In ΔCPD, we have
ㄥz + ㄥ6 + ㄥ7 = 180° ⇒ ㄥz = 180°- ㄥ6 - ㄥ7 ... (iii)
In ΔDPA, we have
ㄥw + ㄥ8 + ㄥ1=180° ⇒ w=1800 - ㄥ8 - ㄥ1 ... (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
ㄥx+ ㄥy+ ㄥz + ㄥw
=720° - (ㄥ1 + ㄥ2 + ㄥ3 + ㄥ4+ ㄥ5 + ㄥ6 + ㄥ7 + ㄥ8)
But, at point P, ㄥx + ㄥy + ㄥz + ㄥw = 360°
∴ 360°= 720° - [(ㄥ1 + ㄥ2) + (ㄥ3 + ㄥ4)+(ㄥ5 + ㄥ6) + (ㄥ7 + ㄥ8)]
⇒ 360°-720° = -(ㄥA + ㄥB + ㄥC + ㄥD)
⇒ -360° = -(ㄥA + ㄥB + ㄥC + ㄥD)
or ㄥA + ㄥB + ㄥC + ㄥD = 360°
Thus, this helps us to arrive at the result that the sum of all angles of a quadrilateral is 360°.

(i) A regular polygon having 7 sides are called regular heptagon.
(ii) According to the Diagonals The line segments obtained by joining vertices, which are not adjacent, are called the diagonals of the polygon.

In figure (i), PR, SO are diagonals and in figure (ii), KN, LM are diagonals.

Number of diagonals in a polygon of n sides= \(\frac { n(n-3) }{ 2 } \)
e.g. In a quadrilateral (a polygon of 4 sides), number of diagonals=\(\frac { 4(4-3) }{ 2 } =\frac { 4\times 1 }{ 2 } \)=2

So, AC and BD are diagonals in quadrilateral ABCD.