Let given polygon be
ABCDE It is clear that, AB is a straight line.
90° + ㄥ1 = 180°
[by linear pair angle]
⇒ ㄥ1 = 180⁰ - 90⁰ = 90⁰
Now, x +90° +60° +90° + 70° = 360°
The sum of the exterior angles of any polygon is 360°.
⇒ x + 310⁰ = 360°
⇒ x = 360⁰-310⁰=  50°
Hence, the measure of angle x is 50°.

If a quadrilateral is not convex, then it will be concave. In the quadrilateral ABCD, join BD. Then, quadrilateral split into two triangles.

In ΔADB, we have
mㄥ1 + mㄥ2 + mㄥ3 = 180°
[using angle sum property of a traingle]    ... (i)
In ΔBDC, we have
mㄥ4 + mㄥ5 + mㄥ6 = 180°
[using angle sum property of a triangle]     ... (ii)
On adding Eqs. (i) and (ii), we get
mㄥ1 + mㄥ2 + mㄥ3 + mㄥ4 + mㄥ5 +mㄥ6 = 180°+180°
⇒ mㄥ1 + mㄥ2 + mㄥ6+ mㄥ5 + (mㄥ3 + mㄥ4) = 360°
∴ mㄥA + mㄥB + mㄥC + mㄥD = 360°
Hence, the sum of the interior angles of a quadrilateral is 360°.

Given,ABCDis a quadrilateral and AC is one of its diagonal.
Clearly, ㄥ1 + ㄥ4 = ㄥA             ......(i)
and ㄥ2 + ㄥ5 = ㄥC                  ......(ii)
Also, diagonal AC divides quadrilateral ABCD into two triangles i.e. ΔABC and ΔADC.
We know that, the sum of all angles of a triangle is 180°.
Therefore,
In ΔABC, we have ㄥ4+ㄥ5+ㄥ6=180⁰    .....(iii)
In ΔACD, we have ㄥ1+ㄥ2+ㄥ3=180⁰    .....(iv)
On adding Eqs. (iii) and (iv), we get
ㄥ1 + ㄥ2 + ㄥ3 + ㄥ4 + ㄥ5 + ㄥ6 = 180° +180°
⇒ (ㄥ1 + ㄥ4) + (ㄥ2 + ㄥ5) + ㄥ3 + ㄥ6 = 360°
∴ ㄥA + ㄥC + ㄥD + ㄥB = 360° [from Eqs. (i) and (ii)]
Hence, the sum of the measures of ㄥA, ㄥB, ㄥC and ㄥD is 360°.

(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).

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°.

80^o, 100^o, 80^o, 100^o

Sides are 12 cm, 5 cm and diagonal is 13 cm.

2: 3

Given, ABC is a right angled triangle and O is the mid-point of the side AC. Now, produce BO to D such that OB = OD and then join AD and CD. Thus, ABCD is a
rectangle, BD and AC are its diagonals, which intersect each other at O. We know that in a rectangle, the diagonals are equal in length and bisect each other.
∴ OA = OC and OB = OD
Also, AC = BD
Therefore, OA = OB = OC = OD
Thus, O is equidistant om A, B and C.

To find the mid-point of \(\bar { AC } \) we fold it by placing C on A. Then, we get the point O, which is the intersecting point of diagonals \(\bar { AC } \) and .\(\bar { DB } \)
Thus, it shows that diagonal \(\bar { DB } \) bisects the diagonal \(\bar { AC } \) at the point O.

No, it cannot be an interior angle of a regular polygon, because if interior angle is 22°, then each exterior angle is (180° - 22°) = 158°, which is not a divisor of 360°.

(a) We know that, the sum of the exterior angles of any polygon is 360°.
∴ x+125°+125°=360°
⇒ x + 250° = 360°
⇒ x = 360° - 250° = 110°
Hence, the measure of angle x is 110°.

1620°