Geometry concepts for MBA entrance exams

1. 08/22/11 GEOMETRY & MENSURATION Concepts To be remembered The lines are parallel and a transversal cuts these lines. If Angle 1 = Ѳ , A ngle2 = [180 – Ѳ ] 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1

2. 08/22/11 GEOMETRY & MENSURATION Concepts To be remembered Lines AB & CD are parallel. E is a point such that Ľ BAE = 45 0 and Ľ ECD = 30 0 . Find Ľ AEC Draw a line GF passing through E parallel to AB F G 45 0 30 0 Ľ AEG = 45 0 Ľ GED = 30 0 Ľ AEC = 75 0 A D B C E 45 0 30 0

3. 08/22/11 GEOMETRY & MENSURATION Concepts To be remembered Lines AB & CD are parallel. AE and CE are internal angular bisectors . Find Ľ AEC Let ĽFAB = 2x 0 Ľ ACD = 2x 0 . Ľ BAC = 2y 0 2x+2y = 180 0 Ľ CAE = y 0 Ľ ACE = x 0 X + Y = 90 From ∆ AEC, Ľ AEC = 90 0 2x A D B C E F y y x x

5. 08/22/11 Sum of all internal angles = (n-2) π Sum of all External angles = 2 π In a regular n sided figure, Each internal angle = (n-2) π /n Each external angle = 2 π /n GEOMETRY & MENSURATION N Sum of all internal Angles Average 3 180 60 4 360 90 5 540 108 6 720 120 7 900 900/7 8 1080 135 9 1260 140 10 1440 144

15. 08/22/11 GEOMETRY & MENSURATION Certain Basic Notations & Facts. BC = a , CA = b & AB = c a + b > c b + c > a c + a > b ABC is called a right angled ∆ , if either Ĺ A or Ĺ B or Ĺ C equals 90 0 . b 2 = a 2 + c 2 ABC is called an acute angled ∆ , if Ĺ A , Ĺ B & Ĺ C are less than 90 0 . a 2 < b 2 + c 2 b 2 < c 2 + a 2 c 2 < a 2 + b 2 ABC is called an obtuse angled ▲ , if either Ĺ A or Ĺ B or Ĺ C is more than 90 0 . a 2 > b 2 + c 2 or b 2 > c 2 + a 2 or c 2 > a 2 + b 2 C A B a b c a b c C A B a b c C A B C A B c b a

16. 08/22/11 GEOMETRY & MENSURATION How many obtuse angled ∆ can be drawn if two of the sides are 8 & 15 respectively. Let the third side be “x”. 8 ≤ x ≤ 22. Case 1: 15 is not the longest side x 2 > 8 2 +15 2 > 289 :: x 2 > 289 x > 17 :: x < 23 X can take values 18,19,20,21,22– 5 values. Case 2 : 15 is the longest side 225 > 8 2 + x 2 x 2 < 161 x < 13 x > 7 X can take values 8,9,10,11& 12 – 5 values Totally 10 triangles can be drawn. C A B 8 15 C A B 8 15

17. 08/22/11 GEOMETRY & MENSURATION How many obtuse angled ∆les can be drawn if two of the sides are 6 & 10 respectively. Let “X” be the third side. 5 ≤ x ≤ 15. Case 1: 10 is the longest side 100 > 6 2 + x 2 x 2 < 64 x < 8 x > 4 X can take values 5,6 & 7 – 3 values Case 2: 10 is not the longest side x 2 > 6 2 +10 2 > 136 x > 11 x < 16 X can take values 12,13,14&15 – 4 values. Totally 7 triangles can be drawn. C A B 6 10 C A B 6 10

18. 08/22/11 GEOMETRY & MENSURATION ∆ ABC & ∆ DEF are similar as Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ F Then AB / DE = BC / EF = CA / FD ∆ ABC & ∆ DEF are similar as AB / DE = BC / EF = CA / FD = ¾ Then Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ F C A B ǿ δ θ F D E ǿ δ θ C A B 6 8 12 F D E 3 4 6

19. 08/22/11 ∆ ABC & ∆ DEF are right angled triangles and are similar. AB / DE = BC / EF = CA / FD = 2 / 1 Perimeter of ∆ ABC / Perimeter of ∆ DEF = 2 / 1 In radius of ∆ ABC / In radius of ∆ DEF = 2 / 1 Circum radius of ∆ ABC / Circum radius of ∆ DEF = 2 / 1 Area of ∆ ABC / Area of ∆ DEF = 4 / 1 Area = 24 :: Perimeter =24 In radius = 2 Circum radius = 5 Area = 6 :: Perimeter =12 In radius = 1 Circum radius =2. 5 F D E C A B 3 4 5 6 8 10

20. 08/22/11 GEOMETRY & MENSURATION ∆ ABC & ∆ DEF are similar as Ĺ A = Ĺ D :: Ĺ B = Ĺ E :: ĹC = Ĺ F AB / DE = BC / EF = CA / FD = m / n Perimeter of ∆ ABC / Perimeter of ∆ DEF = m / n In radius of ∆ ABC / In radius of ∆ DEF = m / n Circum radius of ∆ ABC / Circum radius of ∆ DEF = m / n Area of ∆ ABC / Area of ∆ DEF = m 2 / n2 C A B ǿ δ θ F D E ǿ δ θ

22. 08/22/11 GEOMETRY & MENSURATION ▲ ABD = ⅟ 2 (BD) (Alt. from A to BC) ▲ ADC =⅟ 2 (DC) (Alt. from A to BC) ▲ ABD / ▲ADC = BD / DC = m / n If the base is divided in the ratio m : n, area of the 2▲ les so formed are in the ratio m : n. 1600 If area of triangle ABC is 100 sq units, what is the area of triangle ADE given that BD = 3 AB & CE =4 AC? Let us join DC. Consider ▲ ACD. AB:BD:: 1 : 3 ▲ BCD = 300 sq units ▲ ACD = 400 sq units Consider ▲ CDE. AC : CE :: 1 : 4 ▲ CDE= 1600 Sq.units ▲ ADE=2000 Sq.units A B C D m n A B C D E 3 100 1 300 4 1

23. 08/22/11 GEOMETRY & MENSURATION If area of triangle ABC is 20 sq units, what is the area of triangle ADE given that BD = 9 AB & CE =19AC? Let us join DC. Consider ∆ACD. AB:BD::1:9 ∆ BCD= 180 sq units. ∆ACD =200 sq units AC: CE:: 1:19 ∆ CDE= 3800 Sq.units ∆ ADE=4000 Sq.units If area of triangle ABC is “x” sq units, what is the area of triangle ADE given that BD = k AB & CE =m AC? Let us join DC. Consider ∆ACD. AB:BD::1:k ∆ BCD = k x sq units ∆ ACD = x + kx sq units AC: CE:: 1:m ∆ CDE= m(x+kx )Sq.units ∆ ADE=(k+1)(m+1)x 180 9 20 1 A B C D E 19 1 3800 A B C D E 1 1 K m x kx m(x+kx)

25. GEOMETRY & MENSURATION 08/22/11 AB 2 + AC 2 = 2(AD 2 +BD 2 ) 16 2 + 8 2 = 2(AD 2 +6 2 ) AD 2 = 124 In ∆ ABC, AB = 16,BC= 12, AC = 8. Find AD, BE & CF (Medians) BC 2 + BA 2 = 2(BE 2 +CE 2 ) 16 2 + 12 2 = 2(BE 2 +4 2 ) BE 2 = 184 CB 2 + CA 2 = 2(CF 2 +BF 2 ) 12 2 + 8 2 = 2(BE 2 +8 2 ) CF 2 = 40 BE > AD > CF BE 2 + AD 2 + CF 2 = 348 AB 2 + BC 2 + CA 2 = 464 [ BE 2 + AD 2 + CF 2 ] /[ AB 2 + BC 2 + CA 2 ] = ¾ D G A B E F C

26. GEOMETRY & MENSURATION 08/22/11 AD 2 = 6 2 + 4 2 = 52 CF 2 = 8 2 + 3 2 = 73 BE 2 = 25 In a right angled  ABC, AB = 6 BC = 8 & AC = 10. Let AD , BE & CF be medians. CF > AD > BE BE 2 + AD 2 + CF 2 = 150 AB 2 + BC 2 + CA 2 =200 [ BE 2 + AD 2 + CF 2 ] / [ AB 2 + BC 2 + CA 2 ] = ¾ D G A B E F C

31. GEOMETRY & MENSURATION 08/22/11 A B C I D E F AI, BI & CI are internal angular bisectors of Ĺ A Ĺ B & Ĺ C. The point of concurrence is called the in-centre. ID,IE & IF are in radii Therefore in-centre is equidistant from the sides. AID need not be a Straight line. BIE need not be a Straight line. CIF need not be a Straight line. AB, BC & CA are tangents to in-circle. Length of tangents from a point outside the circle are equal. BD = BF :: AF = AE :: CE = CD ▲ BIC = ⅟ 2 (a)(r) ▲ CIA = ⅟ 2 (b)(r) ▲ AIB = ⅟ 2 (c)(r) ▲ ABC = ⅟ 2 (a + b + c)(r) ▲ ABC = (s) (r)

32. GEOMETRY & MENSURATION 08/22/11 A B C I D BI & CI are internal angular bisectors of Ĺ B & Ĺ C respectively. I is the in-centre & L BAC = 70 0 , find the angle BIC. Consider ∆ ABC, 2X+2Y = 110 X+Y = 55 Consider ∆ BIC, Ø+X+Y = 180 Ø = 125 0 :: Ø = 90+ ½ Ĺ A 70 ø AI, BI & CI are internal angular bisectors of Ĺ A Ĺ B & Ĺ C. ĹBIC = 90+ ⅟ 2 A ĹCIA = 90+ ⅟ 2 B ĹAIB = 90+ ⅟ 2 C X Y X Y

33. GEOMETRY & MENSURATION 08/22/11 A B C I D E F AI, BI & CI are internal angular bisectors of Ĺ A Ĺ B & Ĺ C. ID,IE & IF are in radii. If AB=20, AC=22 &BC= 24. Find BD. AB=20 AC=22 BC= 24. Let BD = X :: BF = X AF =AB – BF = 20 – X AE = 20 – X CE = AC – CE = X + 2 CD= X+2 BD+CD = 2X+2 = 24 2X = 22 X =11 BD = BF = [BC + BA – AC] / 2 CD = CE = [CB + CA – AB] / 2 AF = AE = [AB + AC – BC] / 2 X X 20 - x 20 - x 2+ x 2+ x

37. 08/22/11 E B C D A Given ABC is any triangle.AD is external angular bisector of L EAC. BD is internal angular bisector of L ABC. What is the value of L ADB - ½L ACB? Let L ABC = 2X L ACB = 2Y L BAC = 2Z. 2x+2y+2z = 180 L EAC = 2X+2Y L DAC = X+Y. From ∆ ADB, ǿ+x+2z+x+y = 180 ǿ - y = 0 L ADB - ½L ACB = 0 x x 2z 2y x + y x + y ǿ GEOMETRY & MENSURATION

40. GEOMETRY & MENSURATION 08/22/11 y xy/(z-x) z -x x x A B c D E ABC is a right angled triangle L B = 90 0. AB = x. BC = y. AD is external angular bisector, D is a point on BC Extended. Find CD. Let AD be the external angular bisector. Draw BE parallel to AD LAEB = LABE = x AE = AB = x EC = z -x ∆ CEB ||| r ∆ CAD CE /EA = CB/BD BD = x y / (z-x) ø ø ø ø

44. GEOMETRY & MENSURATION 08/22/11 A B C I D E F AI, BI & CI are internal angular bisectors of Ĺ A Ĺ B & Ĺ C. ID,IE & IF are in radii. If BF=6, CE=8 and ID = 4. Find AB+AC. ∆ ABC = √s(s-a)(s-b)(s-c) ∆ ABC = √(14+x)(x)(8)(6) ∆ ABC = “s” “r” s = 14+x ∆ ABC = (14+x) (4) (14+x)(x)(8)(6) = (14+x) 2 (16) 48x = (14+x) (16) 3x = (14+x) 2x = 14 :: x = 7 AB+AC = 28 6 8 6 8 x x