**Definitions**

**Regular Polygon**A polygon that is both equiangular and equilateral is called a regular polygon. Familiar examples of regular polygons are the equilateral triangle and the square.

**Inscribed Polygon**A rectilinear figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.

**Circumscribed Polygon**A rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed figure touches the circumference of the circle.

**Inscribed Circle**Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.

**Circumscribed Circle**A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.

**Radius**The radius of the circle circumscribed about a regular polygon is called the radius of the polygon.

**Apothem**The radius of the circle inscribed in a regular polygon is called the apothem of the polygon. The apothem is evidently perpendicular to the side of the regular polygon.

**Center**The common center of the circles circumscribed about and inscribed in a regular polygon is called the center of the polygon.

**Angle at the Center**The angle between the radii drawn to the extremities of any side of a regular polygon is called the angle at the center of the polygon. THEOREMS

**Theorem 1**A circle may be circumscribed about, and a circle may be inscribed in any regular polygon.

**Corollary**

- The radius drawn to any vertex of a regular polygon bisects the angle at the vertex.
- The angles at the center of any regular polygon are equal, and each is supplementary to an interior angle of the polygon.
- An equilateral polygon inscribed in a circle is a regular polygon.
- An equiangular polygon circumscribed about a circle is a regular polygon.

**Theorem 2**If a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regular circumscribed polygon.

**Corollary**

- Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides.
- Tangents to a circle at the mid-points of the arcs subtended by the sides of a regular inscribed polygon form a regular circumscribed polygon, whose sides are parallel to the sides of the inscribed polygon and whose vertices lie on the radii (produced) of the inscribed polygon.
- Lines drawn from each vertex of a regular polygon to the mid-points

- Tangents at the mid-points of the arcs between adjacent points of

- The perimeter of a regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides; and the perimeter of a regular circumscribed polygon is greater than that of a regular circumscribed polygon of double the number of sides.

**Theorem 3**Two regular polygons of the same number of sides are similar.

**Corollary**The areas of two regular polygons of the same number of sides are to each other as the squares on any two corresponding sides.

**Theorem 4**The perimeters of two regular polygons of the same number of sides are to each other as their radii, and also as their apothems.

**Corollary**The areas of two regular polygons of the same number of sides are to each other as the squares on the radii of the circumscribed circles, and also as the squares on the radii of the inscribed circles.

**Theorem 5**If the number of sides of a regular inscribed polygon is indefinitely increased, the apothem of the polygon approaches the radius of the circle as its limit.

**Corollary**If the number of sides of a regular inscribed polygon is indefinitely increased, the square on the apothem approaches the square on the radius of the circle as its limit.

**Theorem 6**An arc of a circle is less than a line of any kind that envelops it on the convex side and has the same extremities.

**Corollary**A circle is less than the perimeter of any polygon circumscribed about it.

**Theorem 7**Two circumferences have the same ratio as their Radii.

**Corollary**

- The ratio of any circle to its diameter is constant. The constant ratio of a circle to its diameter is represented by the Greek letter ???? (pi).
- In any circle circumference = 2 ???? r.

**Problem 1**

**To inscribe a triangle equiangular with a given triangle in a given circle.**Let ABC be the given circle, and DEF the given triangle. It is required to inscribe a triangle equiangular with the triangle DEF in the circle ABC. Draw tangent line GH touching the circle ABC at A. Construct the angle HAC equal to the angle DEF on the straight line AH and at the point A on it, and construct the angle GAB equal to the angle DFE on the straight line AG and at the point A on it. Join BC. Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC equals the angle ABC in the alternate segment of the circle. But the angle HAC equals the angle DEF, therefore the angle ABC also equals the angle DEF. For the same reason the angle ACB also equals the angle DFE, therefore the remaining angle BAC also equals the remaining angle EDF. Therefore a triangle equiangular with the given triangle has been inscribed in the given circle. It is worth mentioning that the above generalized construction can be used to inscribe an equilateral triangle in a circle.

**Problem 2**

**To inscribe a square in a given circle.**Let

*ABCD*be the given circle. It is required to inscribe a square in the circle

*ABCD.*Draw two diameters

*AC*and

*BD*of the circle

*ABCD*at right angles to one another, and join

*AB, BC, CD,*and

*DA.*Then, since

*BE*equals

*ED,*for

*E*is the center, and

*EA*is common and at right angles, therefore the base

*AB*equals the base

*AD.*For the same reason each of the straight lines

*BC*and

*CD*also equals each of the straight lines

*AB*and

*AD.*Therefore the quadrilateral

*ABCD*is equilateral. It is also right-angled. For, since the straight line

*BD*is a diameter of the circle

*ABCD,*therefore

*BAD*is a semicircle, therefore the angle

*BAD*is right. For the same reason each of the angles

*ABC, BCD,*and

*CDA*is also right. Therefore the quadrilateral

*ABCD*is right-angled. But it was also proved equilateral, therefore it is a square, and it has been inscribed in the circle

*ABCD.*Therefore the square

*ABCD*has been inscribed in the given circle.

**Problem 3**

**To inscribe an equilateral and equiangular pentagon in a given circle.**Let ABCDE be the given circle. It is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Set out the isosceles triangle FGH having each of the angles at G and H double the angle at F. Inscribe in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angles CAD, ACD, and CDA equal the angles at F, G, and H respectively. Therefore each of the angles ACD and CDA is also double the angle CAD. Now bisect the angles ACD and CDA respectively by the straight lines CE and DB, and join AB, BC, DE, and EA. Then, since each of the angles ACD and CDA is double the angle CAD, and they are bisected by the straight lines CE and DB, therefore the five angles DAC, ACE, ECD, CDB, and BDA equal one another. But equal angles stand on equal circumferences, therefore the five circumferences AB, BC, CD, DE, and EA equal one another. But straight lines that cut off equal circumferences are equal, therefore the five straight lines AB, BC, CD, DE, and EA equal one another. Therefore the pentagon ABCDE is equilateral. It is also equiangular. For, since the circumference AB equals the circumference DE, add BCD to each, therefore the whole circumference ABCD equals the whole circumference EDCB. And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB, therefore the angle BAE also equals the angle AED. For the same reason each of the angles ABC, BCD, and CDE also equals each of the angles BAE and AED, therefore the pentagon ABCDE is equiangular. But it was also proved equilateral, therefore an equilateral and equiangular pentagon has been inscribed in the given circle.

**Problem 4**

**To inscribe a regular hexagon in a given circle.**Let

*ABCDEF*be the given circle. It is required to inscribe an equilateral and equiangular hexagon in the circle

*ABCDEF.*Draw the diameter

*AD*of the circle

*ABCDEF.*Take the center

*G*of the circle. Describe the circle

*EGCH*with center

*D*and radius

*DG.*Join

*EG*and

*CG*and carry them through to the points

*B*and

*F.*Join

*AB,*

*BC, CD, DE, EF,*and

*FA.*The hexagon

*ABCDEF*is equilateral and equiangular. For, since the point

*G*is the center of the circle

*ABCDEF, GE*equals

*GD.*Again, since the point

*D*is the center of the circle

*GCH, DE*equals

*DG.*But

*GE*was proved equal to

*GD,*therefore

*GE*also equals

*ED.*Therefore the triangle

*EGD*is equilateral, and therefore its three angles

*EGD,*

*GDE,*and

*DEG*equal one another, inasmuch as, in isosceles triangles, the angles at the base equal one another. And the sum of the three angles of the triangle equals two right angles, therefore the angle

*EGD*is one-third of two right angles. Similarly, the angle

*DGC*can also be proved to be one third of two right angles. And, since the straight line

*CG*standing on

*EB*makes the sum of the adjacent angles

*EGC*and

*CGB*equal to two right angles, therefore the remaining angle

*CGB*is also one-third of two right angles. Therefore the angles

*EGD, DGC,*and

*CGB*equal one another, so that the angles vertical to them, the angles

*BGA, AGF,*and

*FGE,*are equal. Therefore the six angles

*EGD, DGC, CGB, BGA, AGF,*and

*FGE*equal one another. But equal angles stand on equal circumferences, therefore the six circumferences

*AB, BC, CD, DE, EF,*and

*FA*equal one another And straight lines that cut off equal circumferences are equal, therefore the six straight lines equal one another. Therefore the hexagon

*ABCDEF*is equilateral. It is also equiangular. For, since the circumference

*FA*equals the circumference

*ED,*add the circumference

*ABCD*to each, therefore the whole

*FABCD*equals the whole

*EDCBA.*And the angle

*FED*stands on the circumference

*FABCD,*and the angle

*AFE*on the circumference

*EDCBA,*therefore the angle

*AFE*equals the angle

*DEF.*Similarly it can be proved that the remaining angles of the hexagon

*ABCDEF*are also severally equal to each of the angles

*AFE*and

*FED,*therefore the hexagon

*ABCDEF*is equiangular. But it was also proved equilateral, and it has been inscribed in the circle

*ABCDEF.*Therefore an equilateral and equiangular hexagon has been inscribed in the given circle. SECTION – C (AREAS OF POLYGONS)

**DEFINITIONS**

**Unit of Surface**A square the side of which is a unit of length is called a unit of surface.

**Area of a Surface**The measure of a surface, expressed in units of surface, is called its area. THEOREMS

**TRIANGLE**

**Theorem 1**The area of a triangle is equal to half the product of its base by its altitude.

**Corollary**

- Triangles having equal bases and equal altitudes are equivalent.
- Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the products of their bases by their altitudes.
- The product of the sides of a right triangle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle.

**RECTANGLES**

**Theorem 2**Two rectangles having equal altitudes are to each other as their bases.

**Corollary**Two rectangles having equal bases are to each other as their altitudes.

**Theorem 3**Two rectangles are to each other as the products of their bases by their altitudes.

**Theorem 4**The area of a rectangle is equal to the product of its base by its altitude.

**PARALLELOGRAM**

**Theorem 5**The area of a parallelogram is equal to the product of its base by its altitude.

**Corollary**

- Parallelograms having equal bases and equal altitudes are equivalent.
- Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases; any two parallelograms are to each other as the products of their bases by their altitudes.

**RHOMBUS**

**Theorem 6**The area of a rhombus is equal to half the product of its diagonals.

**TRAPEZOID**

**Theorem 7**The area of a trapezoid is equal to half the product of the sum of its bases by its altitude.

**Corollary**The area of a trapezoid is equal to the product of the line joining the mid-points of its nonparallel sides by its altitude.

**Area of an Irregular Polygon**The area of an irregular polygon may be found by dividing the polygon into triangles, and then finding the area of each of these triangles separately.

**CIRCLE**

**Theorem 8**The area of a circle is equal to half the product of its radius by its circumference.

**Corollary**

- The area of a circle is equal to ???? times the square on its radius.
- The areas of two circles are to each other as the squares on their radii.
- The area of a sector is equal to half the product of its radius by its arc.

**Miscellaneous Theorems**

**Theorem 9**The areas of two similar triangles are to each other as the squares on any two corresponding sides.

**Theorem 10**The areas of two similar polygons are to each other as the squares on any two corresponding sides.

**Theorem 11**The area of a regular polygon is equal to half the product of its apothem by its perimeter.

**Theorem 12**The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle.