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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 ABCD is a parallelogram and the obtuse angles of the parallelogram are ㄥA and ㄥC.

Let AM is the altitude drawn from the vertex A on DG and CN is the altitude drawn from the vertex C on AB.
∴ ㄥDAM = 45° and ㄥNCB = 45°    [given]
In ΔAMD, we know that, sum of the interior angles of a triangle is 180°.
ㄥD + 45° + 90° = 180°
 ㄥD =180° - 45° - 90°
=180° -135° = 45°
Similarly, in ΔCNB
ㄥB = 45°
Now, ㄥA + ㄥD =180°
[sum of adjacent angles In a paralleloprarn.is 180°]
∴ ㄥA + 45° = 180°          [∵ㄥD = 45°]
⇒ ㄥA =180° - 45° =135°
Also, ㄥA = ㄥC
 [∵ opposite angles of a parallelogram are equal]
ㄥC =135°

Since,in a parallelogram diagonals bisect each other and we know that, rectangle is a parallelogram.
So, RO= OS and IO= OE

or RS = 2RO and IE = 2OI
∴ RS=2 x (3x + 15) and IE=2 x (5x + 7)
or RS = 6x + 30 and IE = 10x + 14
Now.we again know that, in a rectangle diagonals are equal.
So, RS = IE ⇒ 6x + 30 = 10x + 14
⇒ 30-14=10x - 6x ⇒ 16 = 4x ⇒ 16 x \(\frac { 1 }{ 4 } \)=x
⇒  4=x  ⇒ x=4

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°

We have, two parallelograms ABDH and EFGC.
ㄥABD = ㄥAHD = 130°
[opposite angles of a parallelogram]
ㄥGHD = 180° - ㄥAHD = 180° - 130°
⇒ 50⁰= ㄥGHO
Also, ㄥEFG + ㄥFGC = 180°
[adjacent angles of a parallelogram]
⇒ 30° + ㄥFGC = 180°
⇒ ㄥFGC =180° - 30° =150°
∴ ㄥHGC =180° - ㄥFGC =180° -150°
   =30⁰ = ㄥHGO
Now, in ΔHGO, by using angle sum property,
ㄥOHG + ㄥHGO + ㄥHOG =180°
⇒ 50°+ 30°+ x⁰ =180°
⇒ x⁰ =180° - 80° = 100°

(i) Yes, RE = OM
Given, the above rectangle, opposite sides are equal. .
(ii) Yes, ㄥMYO = ㄥRXE
Here, MY and RX are perpendicular to OE.
Since, ㄥRXO = 90° ⇒  ㄥRXE = 90⁰
and ㄥMYE = 90°  ⇒ ㄥMYO = 90°
(iii) Yes,  ㄥMOY = ㄥREX
Since, these are alternate interior angles.
∵ RE || OM and EO is a transversal.
∴ ㄥMOE = ㄥOER
⇒ ㄥMOY = ㄥREX
(iv) Yes, ΔMYO ≅ RXE
In ΔMYO and  ΔRXE, we see that
MO=RE                   [proved]
ㄥMOY = ㄥREX     [proved]
ㄥMYO = ㄥRXE     [proved]
∴ ΔMYO ≅ ΔRXE    [by AAS]
(v) Yes, MY = RX
Since, these are corresponding part of congruent triangles.

(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.

(a) Simple curve A plane figure formed by joining a number of points without lifting a pencil from the paper and without returing any portion of the drawing other than single points is called a simple curve or curve. In the given figures, simple curves are figures (i), (ii), (v), (vi) and (vii).
(b) Simple closed curve A closed curve, which does not intersect itself, is called a simple closed curve. In the given figures, simple closed curves are figures (i), (ii), (v), (vi) and (vii).
(c) Polygon A polygon is a closed curve formed by the line segments such that
(i) no two line segments intersect except at their end points.
(ii) no two line segments with a common end points are coincide. In other words, a simple dosed curve made upto only line segments is called a polygon. In the given figures, polygons are figures (i) and (ii).
(d) Convex polygon A convex polygon is a polygon in which each interior angle has a measure less than 180°. In other words, a polygon is convex, if noportion of their diagonals in their exterior. In the given figures, convex polygon is figure (ii).
(e) Concave polygon A concave polygon is a polygon, which atleast one interior angle has measure more than 180°, i.e. atleast one segment connecting two vertices is outside the polygon. In the given figure, concave polygons are figures (i) and (iv).