Triangles

TRIANGLES Triangles are closed figures containing three angles and three sides. General Properties of Triangles: 1. The sum of the two sides is greater than the third side: a + b > c, a + c > b, b + c > a The two sides of a triangle are 12 cm and 7 cm. If the third side is an integer, find the sum of all the values of the third side. Answer: Let the third side be of x cm. Then, x + 7 > 12 or x > 5. Therefore, minimum value of x is 6. Also, x < 12 + 7 or x < 19. Therefore, the highest value of x is 18. The sum of all the integer values from 6 to 18 is equal to 156. 2. The sum of the three angles of a triangle is equal to 180°: In the triangle below ∠A + ∠B + ∠C = 180° Also, the exterior angle α is equal to sum the two opposite interior angle A and B, i.e. α = ∠A + ∠B. Find the value of a + b in the figure given below: Answer: In the above figure, ∠CED = 180° − 125° = 55°. ∠ACD is the exterior angle of ∆ABC. Therefore, ∠ACD = a + 45°. In ∆CED, a + 45° + 55° + b = 180° ⇒ a + b = 80° 3. Area of a Triangle: 1 1 Area of a triangle = × base × height = × a × h 2 2 1 1 1 Area of a triangle = bc sin A = ab sin C = ac sin B 2 2 2 a+b+c Area of a triangle = s(s a)(s b)(s − c) where s = 2 abc Area of a triangle = where R = circumradius 4R a+b+c Area of a triangle = r × s where r = inradius and s = 2 4. More Rules: • • SinA SinB SinC = = a b c 2 2 b + c − a2 a2 + c2 − b2 b2 + a2 − c2 , CosB = , CosC = Cosine Rule: CosA = 2bc 2ac 2ab Sine Rule: Let D and E be on sides AB and AC of triangle ABC such that AD x AE l = and = . Then, area DB y EC m 1 1 lx sin A and area triangle ABC = (l + m)(x + y) sin A .Therefore, 2 2 Area ∆ADE lx = Area ∆ABC (x + y)(l + m) triangle ADE = Points D, E and F divide the sides of triangle ABC in the ratio 1: 3, 1: 4, and 1: 1, as shown in the figure. What fraction of the area of triangle ABC is the area of triangle DEF? Answer: Area ∆ADE 1 × 3 Area ∆BDF 3 1×1 1 Area ∆CFE 4 ×1 2 , = = = = , = = Area ∆ABC 4 × 5 20 Area ∆ABC 4 × 2 8 Area ∆ABC 5 × 2 5 Therefore, Area ∆DEF 1 2 ⎞ 13 ⎛ 3 =1−⎜ + + ⎟= Area ∆ABC 20 8 5 ⎠ 40 ⎝ 5. Medians of a triangle: The medians of a triangle are lines joining a vertex to the midpoint of the opposite side. In the figure, AF, BD and CE are medians. The point where the three medians intersect is known as the centroid. O is the centroid in the figure. • The medians divide the triangle into two equal areas. In the figure, area ∆ABF = area ∆AFC = Area ∆ABC area ∆BDC = area ∆BDA = area ∆CBE = area ∆CEA = 2 AO BO CO • The centroid divides a median internally in the ratio 2: 1. In the figure, = = =2 OF OD OE 2 2 2 2 2 2 2 2 2 2 • Apollonius Theorem: AB + AC = 2(AF + BF ) or BC + BA = 2(BD + DC ) or BC + AC = 2(EC2 + AE2) ABCD is a parallelogram with AB = 21 cm, BC = 13 cm and BD= 14 cm. Find the length of AC. Answer: The figure is shown below. Let AC and BD intersect at O. O bisects AC and BD. Therefore, OD is the median in triangle ADC. ⇒ AD2 + CD2 = 2(AO2 + DO2) ⇒ AO = 16. Therefore, AC = 32. 6. Altitudes of a Triangle: The altitudes are the perpendiculars dropped from a vertex to the opposite side. In the figure, AN, BF, and CE are the altitudes, and their point of intersection, H, is known as the orthocenter. Triangle ACE is a right-angled triangle. Therefore, ∠ECA = 90° − ∠A. Similarly in triangle CAN, ∠CAN = 90° − ∠C. In triangle AHC, ∠CHA = 180° − (∠HAC + ∠HCA) = 180° − (90° − ∠A + 90° − ∠C) = ∠A + ∠C = 180° − ∠B. Therefore, ∠AHC and ∠B are supplementary angles. 7. Internal Angle Bisectors of a Triangle: In the figure above, AD, BE and CF are the internal angle bisectors of triangle ABC. The point of intersection of these angle bisectors, I, is known as the incentre of the triangle ABC, i.e. centre of the circle touching all the sides of a triangle. A ⎛B C⎞ ⎛B + C⎞ ⎛ 180 − A ⎞ ∠BIC = 180° − (∠IBC + ∠ICB) = 180 − ⎜ + ⎟ = 180 − ⎜ • ⎟ = 180 − ⎜ ⎟ = 90 + 2 2 ⎝2 2⎠ ⎝ 2 ⎠ ⎝ ⎠ • AB BD (internal bisector theorem) = AC CD 8. Perpendicular Side Bisectors of a Triangle: In the figure above, the perpendicular bisectors of the sides AB, BC and CA of triangle ABC meet at O, the circumcentre (centre of the circle passing through the three vertices) of triangle ABC. In figure above, O is the centre of the circle and BC is a chord. Therefore, the angle subtended at the centre by BC will be twice the angle subtended anywhere else in the same segment. Therefore, ∠BOC = 2∠BAC. 9. Line Joining the Midpoints: In the figure above, D, E and F are midpoints of the sides of triangle ABC. It can be proved that: FE // BC, DE // AB and DF // AC. BC AB AC • FE = , DE = , FD = 2 2 2 Area ∆ABC • Area ∆DEF = Area ∆AFE = Area ∆BDF = Area ∆DEC = 4 Corollary: If a line is parallel to the base and passes through midpoint of one side, it will • pass through the midpoint of the other side also. • In the figure given below: AG = GE and GF // ED, EF //BD and ED // BC. Find the ratio of the area of triangle EFG to trapezium BCDE. Answer: We know that line parallel to the base and passing through one midpoint passes through another midpoint also. Using this principle, we can see that G, F, E and D are midpoints of AE, AD, AB, and AC respectively. Therefore, GF, EF, ED, and BD are medians in triangles AFE, AED, ADB and ABC. We know that medians divide the triangle into two equal areas. Let the area of triangle AGF = a. Therefore, the areas of the rest of the figures are as shown above. The required ratio = a/12a = 1/12. Types of triangles: Scalene Triangle No side equal. All the general properties of triangle apply Equilateral Triangle Isosceles Triangle Each side equal Each angle = 60° Length of altitude 3 a = 2 3 2 a Area = 4 a Inradius = 2 3 a Circumradius = 3 Two sides equal. The angles opposite to the opposite sides are equal. Right Triangle One of the angles is a right angle, i.e. 90°. 1 Area = AB × BC 2 AC2 = AB2 + BC2 AB × BC Altitude BD = AC The midpoint of the hypotenuse is equidistant from all the three vertices, i.e. EA = EB = EC In triangle DEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D = 40° then what is angle ACB in degrees? (CAT 2001) 1. 140 2. 70 3. 100 4. None of these Answer: Let ∠AEC = ∠EAC = α and ∠CBF = ∠CFB = β. We know that α + β = 180° − ∠D = 140°. ∠ACB = 180° − (∠ECA + ∠BCF) = 180° − (180° − 2α + 180° − 2β) = 100°. In the figure (not drawn to scale) given below, if AD = CD = BC, and ∠BCE = 96°, how much is ∠DBC? (CAT 2003) 1. 32° 2. 84° 3. 64° 4. Cannot be determined. Answer: Let ∠DAC = ∠ACD = α and ∠CDB = ∠CBD = β. As ∠CDB is the exterior angle of triangle ACD, β = 2α. Now ∠ACD + ∠DCB + 96° = 180° ⇒ α + 180° − 2β + 96° = 180° ⇒ 3α = 96° ⇒ α = 32° ⇒ β = 64° Similarity of triangles: Two triangles are similar if their corresponding angles are equal or corresponding sides are in proportion. In the figure given above, triangle ABC is similar to triangle PQR. Then ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R and AB BC CA AI AJ = = = (altitudes) = (medians) PQ QR RP PK PL Therefore, if you need to prove two triangles similar, prove their corresponding angles to be equal or their corresponding sides to be in proportion. Ratio of Areas: If two triangles are similar, the ratio of their areas is the ratio of the squares of the length of their corresponding sides. Therefore, Area of triangle ABC AB2 BC 2 CA2 = = = Area of triangle PQR PQ2 QR2 RP2 DE 1 = . If area of triangle ADE is 10, find the area BC 4 of the trapezium BCED and the area of the triangle CED. Answer: ∆ADE and ∆ABC are similar. Therefore, Area of triangle ABC BC2 = = 16 ⇒ Area of triangle ABC = 160 ⇒ Area of trapezium BCDE = Area Area of triangle ADE DE2 ∆ABC − Area ∆ADE = 160 − 10 = 150. In triangle AC, shown above, DE // BC and To find the area of triangle CDE, we draw altitudes of triangle BDC and CDE, as shown above. Let the length of the altitudes be h. 1 1 Area of triangle BCD = × BC × h and area of triangle CDE = × DE × h 2 2 . Area of triangle BCD BC ⇒ = =4 Area of triangle CDE DE Therefore, we divide the area of the trapezium BCED in the ratio 1: 4 to find the area of triangle CDE. 1 The required area = = × 150 = 30 . 5 In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1, the ratio of CD: PQ is (CAT 2003- Leaked) 1. 1: 0.69 None of these 2. 1: 0.75 3. 1: 0.72 4. Answer: Let BQ = a and DQ = b, as shown in the figure below. Triangle ABD and triangle PQD are similar. Therefore, are similar. Therefore, PQ b . Also triangle CBD and triangle PBQ = AB a + b PQ a = CD a+b Dividing the second equality by the first, we get. AB a CD a+b 4 = = 3 . Therefore, = = = 1 : 0.75 CD b PQ a 3 In triangle ABC, lines DG, EH, and FI are parallel to the base BC. Then it can be proved that AD AG DE GH EF HI = , = , = DE GH EF HI FB IC In the figure (not drawn to scale) given below, P is a point on AB such that AP: PB = 4: 3. PQ is parallel to AC and QD is parallel to CP. In ∆ARC, ∠ARC = 90°, and in ∆PQS, D PSQ = 90°. The length of QS is 6 cm. What is ratio AP: PD? (CAT 2003) 1. 10: 3 2. 2: 1 3. 7: 3 4. 8: 3 Answer: PQ is parallel to AC ⇒ AP CQ 4 = = . PB QB 3 Let AP = 4x and PB = 3x. PD CQ 4 4PB 12x = = ⇒ PD = = QD is parallel CP ⇒ DB QB 3 7 7 12x ⇒ AP: PD = 4x: = 7: 3 7 In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points Rand S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4: 3. It is also known that the length of PO is 28 cm. (CAT 2004) What is the ratio of the length of PQ to that of QO? 1. 1 : 4 2. 1 : 3 3. 3 : 8 4. 3 : 4 What is the radius of the circle II? 1. 2 cm 2. 3 cm 3. 4 cm 4. 5cm 3. 12 3 cm 4. 14 3 cm The length of SO is 1. 8 3 cm Answer: 2. 10 3 cm