=

= 2

sin B. sin C

91. The area of the triangle also asin (B+ C)

since Sin A = sin {180° – (B + C)} = sin (B + C).

92. To find the radii of the circles described within and about a regular polygon of any number of sides.

Let AB be a side of a regular polygon of n sides.

Since the polygon is regular, it may

have a circle inscribed in and about it, and each of the sides will subtend the same angle at the common centre C of these circles.

Draw CD perpendicular to AB.

D

B

Then AD=DB, and CD is the radius of the inscribed circle.

Let r = CD, and R = AC.

Now the sum of all the angles which the sides subtend at C

93. To find the area of a regular polygon of any number of sides which is inscribed within or about a circle of given radius.

Let AB be an arc of the circle whose centre is C; AB a side of the inscribed regular polygon of n sides; CE at right angles to AB, and therefore bisecting it; FG a tangent through E; then FG is a side of the circumscribed polygon of n sides. Let CA = r.

ADB = 2 AD = 2. r. sin ACD = 2r. sin

F

360°

E

2n

n

B

94. To find the area of a regular polygon of n sides in

terms of a side of the polygon. (Fig. Art. 92.)

AB being a side of the polygon, we have,

95. To find the radii of the circles described in and about a triangle whose sides are given.

Let the lines bisecting the angles A and B meet in O, and from O draw OD, OE, OF perpendiculars to the sides.

Then, Euclid Iv. 4, O is the centre of the inscribed circle, and r its radius = OD OE = OF

B

E

Now, Area of ▲ ABC = ▲ AOB + ▲ BOC + ▲ COA;

·· √/{S. (S − a) . (S − b) . ( S − c)} = r.

r =

Again; bisect the sides of the triangle in D, E, F, and draw perpendiculars which will meet in a point O which is the centre of the circumscribed circle; and R, its radius, = OA = OB = OC. Euclid IV. 5.

Area of ABC AC. AB. sin A, Art. 90.

=

BE a
во

/ {S. (S − a). (S − b). (S′ – c)}

abc

=

R

=

R

4 √ {S. (S − a). (S − b). (S − c)}

96. To find the area of a quadrilateral figure whose opposite angles are supplements to each other.

Now from AABC,

a2 + b2 - AC2 = 2 ab. cos B.

And from AABC,

c2 + ď2 — AC2 = 2cd. cos D = 2cd.cos B;

=

=

1

16

1

16

1

16

1

=—

16

2 (ab + cd)

}

And (area ABCD)2 = 1 (ab + cd)2. (sin B)2;

̧· { 4 (ab + cd)2 − (a2 + b2 − c2 − d2)2}

· {2(ab+cd)+(a2+b2—c2-d2)}. {2(ab+cd) − (a2+b2—c2 —d2) }

; · {(a + b)2 − (c − d)2} . {c + d)2 − (a − b)2}

·(a+b+c−d).(a+b+d−c).(c+d+a−b). (c+d+b−a).

If S = (a+b+c+d), this equation becomes

Area ABCD = {(S − a) . (S − b) . (S − c) . (S − d)}.