CBSE Class 11 Mathematics Solution of Triangles Notes Set 02

Circle Connected with Triangle

(1) Circumcircle of a Triangle and its Radius

(i) Circumcircle: The circle which passes through the angular points of a triangle is called its circumcircle. The centre of this circle is the point of intersection of perpendicular bisectors of the sides and is called the circumcentre. Its radius is always denoted by R. The circumcentre may lie within, outside or upon one of the sides of the triangle.
(ii) Circum-radius: The circum-radius of a \( \triangle ABC \) is given by
(a) \( \dfrac{a}{2\sin A} = \dfrac{b}{2\sin B} = \dfrac{c}{2\sin C} = R \)    (b) \( R = \dfrac{abc}{4\Delta} \) [Δ = area of \( \triangle ABC \)]

(2) Inscribed Circle or Incircle of a Triangle and its Radius

(i) In-circle or inscribed circle: The circle which can be inscribed within a triangle so as to touch each of the sides is called its inscribed circle or in circle. The centre of this circle is the point of intersection of the bisectors of the angles of the triangle. The radius of this circle is always denoted by r and is equal to the length of the perpendicular from its centre to any one of the sides of triangle.
(ii) In-radius: The radius r of the inscribed circle of a triangle ABC is given by
(a) \( r = \dfrac{\Delta}{s} \)    (b) \( r = 4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} \)
(c) \( r = (s - a)\tan\dfrac{A}{2},\ r = (s - b)\tan\dfrac{B}{2},\ r = (s - c)\tan\dfrac{C}{2} \)
(d) \( r = \dfrac{a\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A}{2}},\ r = \dfrac{b\sin\frac{A}{2}\sin\frac{C}{2}}{\cos\frac{B}{2}},\ r = \dfrac{c\sin\frac{B}{2}\sin\frac{A}{2}}{\cos\frac{C}{2}} \)
(e) \( \cos A + \cos B + \cos C = 1 + \dfrac{r}{R} \)

(3) Escribed Circles of a Triangle and their Radii

(i) Escribed circle: The circle which touches the side BC and two sides AB and AC produced of a triangle ABC is called the escribed circle opposite to the angle A. Its radius is denoted by \( r_1 \). Similarly, \( r_2 \) and \( r_3 \) denote the radii of escribed circles opposite to the angles B and C respectively.
The centres of the escribed circles are called the ex-centres. The centre of the escribed circle opposite to the angle A is the point of intersection of the external bisectors of angles B and C. The internal bisector of angle A also passes through the same point. The centre is generally denoted by \( I_1 \).

(ii) Radii of ex-circles: In any \( \triangle ABC \), we have
(a) \( r_1 = \dfrac{\Delta}{s - a},\ r_2 = \dfrac{\Delta}{s - b},\ r_3 = \dfrac{\Delta}{s - c} \)
(b) \( r_1 = s\tan\dfrac{A}{2},\ r_2 = s\tan\dfrac{B}{2},\ r_3 = s\tan\dfrac{C}{2} \)
(c) \( r_1 = \dfrac{a\cos\frac{B}{2}\cos\frac{C}{2}}{\cos\frac{A}{2}},\ r_2 = \dfrac{b\cos\frac{C}{2}\cos\frac{A}{2}}{\cos\frac{B}{2}},\ r_3 = \dfrac{c\cos\frac{A}{2}\cos\frac{B}{2}}{\cos\frac{C}{2}} \)
(d) \( r_1 + r_2 + r_3 - r = 4R \)
(e) \( \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} = \dfrac{1}{r} \)
(f) \( \dfrac{1}{r^2} + \dfrac{1}{r_1^2} + \dfrac{1}{r_2^2} + \dfrac{1}{r_3^2} = \dfrac{a^2 + b^2 + c^2}{\Delta^2} \)
(g) \( \dfrac{1}{bc} + \dfrac{1}{ca} + \dfrac{1}{ab} = \dfrac{1}{2Rr} \)
(h) \( r_1r_2 + r_2r_3 + r_3r_1 = s^2 \)
(i) \( \Delta = 2R^2\sin A\sin B\sin C = 4Rr\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2} \)
(j) \( r_1 = 4R\sin\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2},\ r_2 = 4R\cos\dfrac{A}{2}\sin\dfrac{B}{2}\cos\dfrac{C}{2},\ r_3 = 4R\cos\dfrac{A}{2}\cos\dfrac{B}{2}\sin\dfrac{C}{2} \)

(4) Centroid (G)

Common point of intersection of medians of a triangle. Divides every median in the ratio 2 : 1. Always lies inside the triangle.

(5) Orthocentre of a Triangle

The point of intersection of perpendicular drawn from the vertices on the opposite sides of a triangle is called its orthocentre.

Pedal Triangle

Let the perpendiculars AD, BE and CF from the vertices A, B and C on the opposite sides BC, CA and AB of \( \triangle ABC \) respectively, meet at O. Then O is the orthocentre of the \( \triangle ABC \). The triangle DEF is called the pedal triangle of the \( \triangle ABC \).
Orthocentre of the triangle is the incentre of the pedal triangle.
If O is the orthocentre and DEF the pedal triangle of the \( \triangle ABC \), where AD, BE, CF are the perpendiculars drawn from A, B, C on the opposite sides BC, CA, AB respectively, then
(i) \( OA = 2R\cos A,\ OB = 2R\cos B \) and \( OC = 2R\cos C \)
(ii) \( OD = 2R\cos B\cos C,\ OE = 2R\cos C\cos A \) and \( OF = 2R\cos A\cos B \)

(1) Sides and angles of a pedal triangle: The angles of pedal triangle DEF are: 180° − 2A, 180° − 2B, 180° − 2C and sides of pedal triangle are:
\( EF = a\cos A \) or \( R\sin 2A \) ;  \( FD = b\cos B \) or \( R\sin 2B \) ;  \( DE = c\cos C \) or \( R\sin 2C \).
If given \( \triangle ABC \) is obtuse, then angles are been represented by 2A, 2B, 2C − 180° and the sides are \( a\cos A,\ b\cos B,\ -c\cos C \).

(2) Area and circum-radius and in-radius of pedal triangle
Area of pedal triangle = \( \dfrac{1}{2} \) (Product of the sides) × (sine of included angle)
\( \Delta = \dfrac{1}{2}R^2\sin 2A\sin 2B\sin 2C \)
Circum-radius of pedal triangle = \( \dfrac{EF}{2\sin FDE} = \dfrac{R\sin 2A}{2\sin(180° - 2A)} = \dfrac{R}{2} \).
In-radius of pedal triangle = \( \dfrac{\text{area of }\Delta DEF}{\text{semi-perimeter of }\Delta DEF} = \dfrac{\frac{1}{2}R^2\sin 2A\sin 2B\sin 2C}{2R\sin A\sin B\sin C} = 2R\cos A\cos B\cos C \).

Important Tips

1. Circum-centre, Centroid and Orthocentre are collinear.
2. In any right angled triangle, the orthocentre coincides with the vertex containing the right angled.
3. The mid-point of the hypotenuse of a right angled triangle is equidistant from the three vertices of the triangle.
4. The mid-point of the hypotenuse of a right angled triangle is the circumcentre of the triangle.
5. The length of the medians AD, BE, CF of \( \triangle ABC \) are given by

\( AD = \dfrac{1}{2}\sqrt{2b^2 + 2c^2 - a^2} = \dfrac{1}{2}\sqrt{b^2 + c^2 + 2bc\cos A},\ BE = \dfrac{1}{2}\sqrt{2c^2 + 2a^2 - b^2} = \dfrac{1}{2}\sqrt{c^2 + a^2 + 2ca\cos B} \)
\( CF = \dfrac{1}{2}\sqrt{2a^2 + 2b^2 - c^2} = \dfrac{1}{2}\sqrt{a^2 + b^2 + 2ab\cos C} \)

6. The distance between the circumcentre O and centroid G of \( \triangle ABC \) is given by
\( OG = \dfrac{1}{3}OH = \dfrac{1}{3}R\sqrt{1 - 8\cos A\cos B\cos C} \), where H is the orthocentre of \( \triangle ABC \).
7. The distance between the orthocentre H and centroid G of \( \triangle ABC \) is given by \( HG = \dfrac{2}{3}R\sqrt{1 - 8\cos A\cos B\cos C} \).
8. The distance between the circumcentre O and the incentre I of \( \triangle ABC \) is given by \( OI = R\sqrt{1 - 8\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}} \).
9. If \( I_1 \) is the centre of the escribed circle opposite to the angle A, then \( OI_1 = R\sqrt{1 + 8\sin\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}} \).
Similarly, \( OI_2 = R\sqrt{1 + 8\cos\dfrac{A}{2}\sin\dfrac{B}{2}\cos\dfrac{C}{2}},\ OI_3 = R\sqrt{1 + 8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\sin\dfrac{C}{2}} \).
10. Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as "Nine point circle".
11. Circumcentre of the pedal triangle of a given triangle bisects the line joining the circum-centre of the triangle to the orthocentre.

Ex-central Triangle

Let ABC be a triangle and I be the centre of incircle. Let \( I_1,\ I_2 \) and \( I_3 \) be the centres of the escribed circles which are opposite to A, B, C respectively then \( I_1I_2I_3 \) is called the Ex-central triangle of \( \triangle ABC \).
\( I_1I_2I_3 \) is a triangle, thus the triangle ABC is the pedal triangle of its ex-central triangle \( I_1I_2I_3 \). The angles of ex-central triangle \( I_1I_2I_3 \) are \( 90° - \dfrac{A}{2},\ 90° - \dfrac{B}{2},\ 90° - \dfrac{C}{2} \) and sides are \( I_1I_3 = 4R\cos\dfrac{B}{2} \) ; \( I_1I_2 = 4R\cos\dfrac{C}{2} \) ; \( I_2I_3 = 4R\cos\dfrac{A}{2} \).

Area and circum-radius of the ex-central triangle
Area of triangle = \( \dfrac{1}{2} \) (Product of two sides) × (sine of included angles)
\( \Delta = \dfrac{1}{2}\left(4R\cos\dfrac{B}{2}\right)\cdot\left(4R\cos\dfrac{C}{2}\right) \times \sin\left(90° - \dfrac{A}{2}\right) \)
\( \Delta = 8R^2\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2} \)
Circum-radius = \( \dfrac{I_2I_3}{2\sin I_2I_1I_3} = \dfrac{4R\cos\frac{A}{2}}{2\sin\left(90° - \frac{A}{2}\right)} = 2R \).

Example 31. In a \( \triangle ABC \), \( r_1 < r_2 < r_3 \), then [EAMCET 2003]
(A) a < b < c   (B) a > b > c   (C) b < a < c   (D) a < c < b
Answer:
Ans. (A)
In a \( \triangle ABC \), \( r_1 < r_2 < r_3 \)
\( \Rightarrow \dfrac{1}{r_1} > \dfrac{1}{r_2} > \dfrac{1}{r_3} \Rightarrow \dfrac{s - a}{\Delta} > \dfrac{s - b}{\Delta} > \dfrac{s - c}{\Delta} \Rightarrow (s - a) > (s - b) > (s - c) \Rightarrow -a > -b > -c \Rightarrow a < b < c \).

Example 32. Which of the following pieces of data does NOT uniquely determine an acute-angled triangle ABC (R being the radius of the circum-circle) [IIT Screening 2002]
(A) a, sin A, sin B   (B) a, b, c   (C) a, sin B, R   (D) a, sin A, R
Answer:
Ans. (D)
\( \dfrac{a}{\sin A} = 2R \) and \( b = 2R\sin B \). So two sides and two angles are known. So \( \angle C \) is known. Therefore, two sides and included angle is known. So, Δ is uniquely known in case (A).
If a, b, c are known the Δ is uniquely known in case (B). \( b = 2R\sin B \), \( \sin A = \dfrac{a}{2R} \). So, sides a, b and angle A, B are known. So \( \angle C \) is known. Therefore two sides and included angle is known. So, Δ is uniquely known in case (C).
\( \dfrac{a}{\sin A} = 2R \). So, only a side and an angle is known. So, Δ is not uniquely known in case (D).

Example 33. In a triangle ABC, let \( \angle C = \dfrac{\pi}{2} \). If r is the in radius and R is the circum-radius of the triangle, then 2(r + R) is equal to [IIT Screening 2000]
(A) a + b   (B) b + c   (C) c + a   (D) a + b + c
Answer:
Ans. (A)
\( \dfrac{c}{\sin C} = 2R \), ∴ \( c = 2R\sin 90° = 2R \)
Also \( r = (s - c)\tan\dfrac{C}{2} = (s - c) \) [∵ tan 45° = 1]
\( 2r = 2s - 2c = a + b - c = a + b - 2R \), 2(r + R) = a + b.

Example 34. If R is the radius of the circumcircle of the \( \triangle ABC \) and Δ is its area, then [Karnataka CET 2000]
(A) \( R = \dfrac{a + b + c}{\Delta} \)   (B) \( R = \dfrac{a + b + c}{4\Delta} \)   (C) \( R = \dfrac{abc}{4\Delta} \)   (D) \( R = \dfrac{abc}{\Delta} \)
Answer:
Ans. (C)
Area of the triangle ABC \( (\Delta) = \dfrac{bc}{2}\sin A \). From the sine formula, \( a = 2R\sin A \) or \( \sin A = \dfrac{a}{2R} \) \( \Rightarrow \Delta = \dfrac{1}{2}bc \cdot \dfrac{a}{2R} = \dfrac{abc}{4R} \) or \( R = \dfrac{abc}{4\Delta} \).

Example 35. If the length of the sides of a triangle are 3, 4 and 5 units, then R (the circum-radius) is [UPSEAT 2000]
(A) 2.0 unit   (B) 2.5 unit   (C) 3.0 unit   (D) 3.5 unit
Answer:
Ans. (B)
Given, Sides are 3, 4, 5.
Since \( 3^2 + 4^2 = 5^2 \). So, triangle is a right angle triangle. \( R = \dfrac{5}{2} = 2.5 \) unit.

Example 36. If x, y, z are perpendicular drawn from the vertices of triangle having sides a, b, and c, then the value of \( \dfrac{bx}{c} + \dfrac{cy}{a} + \dfrac{az}{b} \) will be
(A) \( \dfrac{a^2 + b^2 + c^2}{2R} \)   (B) \( \dfrac{a^2 + b^2 + c^2}{R^2} \)   (C) \( \dfrac{a^2 + b^2 + c^2}{4R} \)   (D) \( \dfrac{2(a^2 + b^2 + c^2)}{R} \)
Answer:
Ans. (A)
Let area of triangle be Δ, then according to question, \( \Delta = \dfrac{1}{2}ax = \dfrac{1}{2}by = \dfrac{1}{2}cz \)
∴ \( \dfrac{bx}{c} + \dfrac{cy}{a} + \dfrac{az}{b} = \dfrac{b}{c}\left(\dfrac{2\Delta}{a}\right) + \dfrac{c}{a}\left(\dfrac{2\Delta}{b}\right) + \dfrac{a}{b}\left(\dfrac{2\Delta}{c}\right) = \dfrac{2\Delta(b^2 + c^2 + a^2)}{abc} = \dfrac{2(a^2 + b^2 + c^2)}{abc} \cdot \dfrac{abc}{4R} = \dfrac{a^2 + b^2 + c^2}{2R} \).

Example 37. If \( r_1, r_2, r_3 \) in a triangle be in H.P., then the sides are in [EAMCET 1993]
(A) A.P.   (B) G.P.   (C) H.P.   (D) None of these
Answer:
Ans. (A)
\( \dfrac{1}{r_2} - \dfrac{1}{r_1} = \dfrac{1}{r_3} - \dfrac{1}{r_2} \Rightarrow \dfrac{s - b}{\Delta} - \dfrac{s - a}{\Delta} = \dfrac{s - c}{\Delta} - \dfrac{s - b}{\Delta} \Rightarrow \dfrac{a - b}{\Delta} = \dfrac{b - c}{\Delta} \)
\( \Rightarrow \dfrac{b - a}{\Delta} = \dfrac{c - b}{\Delta} \Rightarrow 2b = a + c \).

Example 38. If the sides of a triangle are 13, 14, 15, then the radius of its incircle is [EAMCET 1987]
(A) \( \dfrac{67}{8} \)   (B) \( \dfrac{65}{4} \)   (C) 4   (D) 24
Answer:
Ans. (C)
\( r = \dfrac{\Delta}{s} \) and \( s = \dfrac{a + b + c}{2} = \dfrac{13 + 14 + 15}{2} = 21 \)
\( \Rightarrow \Delta = \sqrt{21 \times 8 \times 7 \times 6} = 84 \Rightarrow r = \dfrac{84}{21} = 4 \).

Example 39. In an equilateral triangle the inradius and the circumradius are connected by [EAMCET 1983]
(A) r = 4R   (B) \( r = \dfrac{R}{2} \)   (C) \( r = \dfrac{R}{3} \)   (D) None of these
Answer:
Ans. (B)
\( r = 4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} \)
For an equilateral triangle, A = B = C = 60°
∴ \( r = 4R\sin 30° \sin 30° \sin 30° = 4R \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{R}{2} \).

Cyclic Quadrilateral

A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S.
Let a cyclic quadrilateral be such that PQ = a, QR = b, RS = c and SP = d. Then \( \angle Q + \angle S = 180° \) and \( \angle P + \angle R = 180° \). Let 2s = a + b + c + d and Δ = Area of cyclic quadrilateral PQRS

Δ = Area of \( \triangle PQR \) + Area of \( \triangle PRS = \dfrac{1}{2}ab\sin Q + \dfrac{1}{2}cd\sin S \)
\( \Delta = \dfrac{1}{2}ab\sin Q + \dfrac{1}{2}cd\sin(\pi - Q) = \dfrac{1}{2}(ab + cd)\sin Q \)
\( \Delta = \dfrac{1}{2}(ab + cd)\sin Q \) ......(i)
In \( \triangle PQR \) and \( \triangle PRS \),
From cosine rule, \( PR^2 = PQ^2 + QR^2 - 2PQ \cdot QR\cos Q = a^2 + b^2 - 2ab\cos Q \) ......(ii)
and \( PR^2 = PS^2 + RS^2 - 2PS \cdot RS\cos S = d^2 + c^2 - 2cd\cos(\pi - Q) = d^2 + c^2 + 2cd\cos Q \) ........(iii)
From (ii) and (iii) we have, \( \Delta = \sqrt{(s - a)(s - b)(s - c)(s - d)} \) ........(iv)
Therefore,
(1) Area of cyclic quadrilateral = \( \dfrac{1}{2}(ab + cd)\sin Q \)
(2) Area of cyclic quadrilateral = \( \sqrt{(s - a)(s - b)(s - c)(s - d)} \), where 2s = a + b + c + d
(3) \( \cos Q = \dfrac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)} \)
(4) Circumradius of cyclic quadrilateral: Circum circle of quadrilateral PQRS is also the circumcircle of \( \triangle PQR \). Hence circumradius of cyclic quadrilateral PQRS = R = circumradius of \( \triangle PQR = \dfrac{PR}{2\sin Q} = \dfrac{PR(ab + cd)}{4\Delta} \)
But \( PR = \sqrt{\dfrac{(ac + bd)(ad + bc)}{(ab + cd)}} \)
Hence \( R = \dfrac{1}{4\Delta}\sqrt{(ac + bd)(ad + bc)(ab + cd)} = \dfrac{1}{4}\sqrt{\dfrac{(ac + bd)(ad + bc)(ab + cd)}{(s - a)(s - b)(s - c)(s - d)}} \)
(5) Ptolemy's theorem: In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS
\( PR \cdot QS = PQ \cdot RS + RQ \cdot PS \).

Example 40. A cyclic quadrilateral ABCD of area \( \dfrac{3\sqrt{3}}{4} \) is inscribed in a unit circle. If one of its sides AB = 1 and the diagonal \( BD = \sqrt{3} \) then the lengths of the other sides are [Roorkee 1995]
(A) 2, 1, 1   (B) 2, 1, 2   (C) 3, 1, 2   (D) None of these
Answer:
Ans. (A)
By sine formula in \( \triangle ABD \), \( \dfrac{\sqrt{3}}{\sin A} = 2R \Rightarrow \dfrac{\sqrt{3}}{\sin A} = 2 \Rightarrow \sin A = \dfrac{\sqrt{3}}{2} \Rightarrow A = \dfrac{\pi}{3} \)
Now, AB = x = 1
By cosine formula in \( \triangle ABD \) \( \cos\dfrac{\pi}{3} = \dfrac{x^2 + y^2 - 3}{2xy} \Rightarrow \dfrac{1}{2} = \dfrac{1 + y^2 - 3}{2y} \Rightarrow y = y^2 - 2 \)
\( \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y - 2)(y + 1) = 0 \Rightarrow y = 2 \) [∵ \( y \neq -1 \)] ∴ AD = 2
Since \( \angle A = 60° \) ∴ \( \angle C = 120° \)
In \( \triangle BDC \), \( 3 = p^2 + q^2 - 2pq\cos 120° \Rightarrow 3 = p^2 + q^2 + pq \) .........(i)
Also area of quadrilateral ABCD = \( \dfrac{3\sqrt{3}}{4} \)
∴ \( \dfrac{3\sqrt{3}}{4} = \Delta ABD + \Delta BCD = \dfrac{1}{2} \cdot 1 \cdot 2\sin 60° + \dfrac{1}{2}pq\sin 120° = \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{4}pq \)
\( \Rightarrow \dfrac{\sqrt{3}}{4}pq = \dfrac{3\sqrt{3}}{4} - \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3} - 2\sqrt{3}}{4} = \dfrac{\sqrt{3}}{4} \Rightarrow pq = 1 \)
∴ (i) gives, \( 3 = p^2 + q^2 + 1 \Rightarrow p^2 + q^2 = 2 \), [p, q > 0]
∴ \( p^2 + \dfrac{1}{p^2} = 2 \Rightarrow p^4 - 2p^2 + 1 = 0 \Rightarrow (p^2 - 1)^2 = 0 \Rightarrow p^2 = 1,\ \therefore p = 1, q = 1 \)
∴ AB = 1, AD = 2, BC = CD = 1.

Example 41. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the third side is 3, the remaining fourth side is [UPSEAT 1994]
(A) 2   (B) 3   (C) 4   (D) 5
Answer:
Ans. (A)
Since ABCD is cyclic quadrilateral and \( \angle ABC = 60° \) ∴ \( \angle ADC = 180° - 60° = 120° \)
Let AB = 2, BC = 5 and CD = 3
In \( \triangle ABC \), \( AB^2 + BC^2 - 2AB \cdot BC\cos 60° = AC^2 \) or \( 4 + 25 - 2 \times 2 \times 5 \times \dfrac{1}{2} = AC^2 \)
∴ \( AC^2 = 19 \) ; In \( \triangle ACD \), \( AD^2 + CD^2 - 2AD \cdot CD\cos 120° = AC^2 \)
or \( AD^2 + 9 + 2AD \cdot 3 \cdot \dfrac{1}{2} = 19 \) or \( AD^2 + 3AD - 10 = 0 \) or \( AD^2 + 5AD - 2AD - 10 = 0 \)
or AD(AD + 5) − 2(AD + 5) = 0 or (AD − 2)(AD + 5) = 0. Therefore, fourth side is AD = 2.

Example 42. Two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the area of the quadrilateral is \( 4\sqrt{3} \), then the remaining two sides are [Roorkee 1991]
(A) 2, 3   (B) 3, 4   (C) 4, 5   (D) 5, 6
Answer:
Ans. (A)
Let a = PQ = 2, b = QR = 5, RS = c, SP = d
Given, Area of quadrilateral PQRS = \( 4\sqrt{3} \)
\( \Rightarrow \) Area of (\( \triangle PQR + \triangle PRS \)) = \( \dfrac{1}{2}[2 \times 5\sin 60° + c \times d\sin 120°] \)
\( \Rightarrow 4\sqrt{3} = \dfrac{5\sqrt{3}}{2} + \dfrac{\sqrt{3}}{4}cd \Rightarrow cd = 6 \) ........(i)
Now by cosine formula \( RS^2 + SP^2 - 2RS \cdot SP\cos 120° = PR^2 = PQ^2 + QR^2 - 2PQ \cdot QR\cos 60° \)
\( \Rightarrow c^2 + d^2 - 2cd\left(-\dfrac{1}{2}\right) = 4 + 25 - 2(2)(5)\left(\dfrac{1}{2}\right) \Rightarrow c^2 + d^2 + cd = 19 \Rightarrow c^2 + d^2 = 19 - 6 = 13 \) ........(ii)
Solving (i) and (ii) we get c = 2 and d = 3.