Triangles — Part 2 MCQs with Explanations

According to question, AB + BC = 12 cm BC + CA = 14 cm CA + AB = 18 cm 2(AB + BC + CA) = 44 cm AB + BC + CA = $$\frac{{44}}{2}$$ cm AB + BC + CA = 22 cm Perimeter of triangle = 22 cm Perimeter of triangle = Perimeter of circle 22 = 2πr 2 × $$\frac{{22}}{7}$$ × r = 22 r = $$\frac{7}{2}$$ cm

According to question, AB = AC, ∠B = ∠C ∠A + ∠B + ∠C = 180° ∠A + 2∠B = 180° ∠A = 180° - 70° ∠A = 110° Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle. ∠BAD = $$\frac{{\angle A}}{2}$$ ∠BAD = $$\frac{{{{110}^ \circ }}}{2}$$ ∠BAD = 55°

According to question, Given : ∠A = 40° AB = AC ∴ ∠B = ∠C In ΔABC ∠A + ∠B + ∠C = 180° 40° + 2∠B = 180° 2∠B = 180° - 40° 2∠B = 140° ∠B = 70° ∴ External angle at B = 180° - 70° = 110°

According to question, ΔABC ∼ ΔACD $$\eqalign{ & \therefore \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{B{C^2}}}{{A{D^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{{8^2}}}{{{6^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{64}}{{36}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{16}}{9} \cr} $$ area ΔABC : area ΔACD = 16 : 9

According to question, Given : ∠BOC = 60° As we know that ∴ ∠O = 90 - $$\frac{1}{2}$$ ∠A ⇒ $$\frac{1}{2}$$ ∠A = 90° - 60° ⇒ $$\frac{1}{2}$$ ∠A = 30° ⇒ ∠A = 60°

ar ADE = ar DEBC So, ar ΔADE = 1 unit 2 and ar ABC = 2 unit 2 $$\eqalign{ & \frac{{{\text{ar}}\,\Delta ADE}}{{{\text{ar}}\,\Delta ABC}} = \frac{{A{D^2}}}{{A{B^2}}} \cr & \frac{1}{2} = {\left( {\frac{{AD}}{{AB}}} \right)^2} \cr & \frac{1}{{\sqrt 2 }} = \frac{{AD}}{{AB}} \cr & \therefore \frac{{AD}}{{DB}} = \frac{1}{{\sqrt 2 - 1}} \cr & \left( {\therefore DB = AB - AD = \sqrt 2 - 1} \right) \cr & {\text{So,}}\,AD:BD = 1:\sqrt 2 - 1 \cr} $$

According to question, Let AB = AC = 2x ∵ AQ = QC = x ∴ AB is a secant ∴ AP × AB = AQ 2 AP × 2x = x 2 $$\eqalign{ & AP = \frac{x}{2} \cr & \frac{{AP}}{{AB}} = \frac{x}{{2 \times 2x}} = \frac{1}{4} \cr & \frac{{AP}}{{AB}} = \frac{1}{4} \cr & AP:AB = 1:4 \cr} $$

The sum of two sides of a triangle should be greater than the third side. There are only two possible pairs (2, 5, 6) and (3, 5, 6)

According to question, If the median drawn on the base of a triangle is half of its base of the triangle then the triangle will be right angled triangle.

According to question, Given : ∠ABC = 60° ∠BCA = 80° ∠BIC = ? ∠BAC = 40° ∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 40° ∠BIC = 110°

If three altitudes are equal then the triangle is equilateral

According to question, Given : AB = AC y = 2(x + x) y = 4x ∴ x + x + y = 180° 2x + 4x = 180° 6x = 180° x = $$\frac{{{{180}^ \circ }}}{6}$$ x = 30°

According to question, Area of Δ(OBA + OAC + OBC) = area of ΔABC $$\frac{1}{2}$$ × 4x × 3 + $$\frac{1}{2}$$ × 5x × 3 + $$\frac{1}{2}$$ × 6x × 3 = $$\frac{1}{2}$$ × (6x × AD) $$\frac{1}{2}$$ × 3(4x + 5x + 6x) = $$\frac{1}{2}$$ × (6x × AD) 45x = 6x × AD AD = $$\frac{{15}}{2}$$ AD = 7.5 cm

According to question, ABC is a right angled triangle ∴ By using Pythagoras theorem AC 2 = AB 2 + BC 2 AC 2 = (5) 2 + (12) 2 AC 2 = 25 + 144 AC 2 = 169 AC = $$\sqrt {169} $$ AC = 13 cm BD = I R = Circumradius = $$\frac{{AC}}{2}$$ ∴ I R = $$\frac{{13}}{2}$$ I R = 6.5 cm

According to question, In ΔPQR x + y + 40° = 180° x + y = 140° . . . . . . . (i) In ΔAQC x + x + ∠C = 180° ∠C = 180° - 2x . . . . . . . . (ii) In ΔBCR y + y + ∠C = 180° ∠C = 180° - 2y . . . . . . . . (iii) But ∠ACB = 180° - 180° + 2x - 180° +2y ∠ACB = 2x + 2y - 180° ∠ACB = 2(x + y) - 180° . . . . . . . . . (iv) Put the value of equation (i) in equation (iv) ∠ACB = 2 × 140° - 180° ∠ACB = 280° - 180° ∠ACB = 100°

According to question, In equilateral triangle 'AD' bisects the BC in two equal parts. Let side of equilateral triangle is 2 cm $$\eqalign{ & \therefore \frac{{AB}}{{BD}} = \frac{2}{1} \cr & AB:BD = 2:1 \cr} $$

According to question, Given : AB = AC = 2a BC = a AD ⊥ BC In isosceles triangle perpendicular sides bisects the opposite side of the length ∴ BD = $$\frac{{BC}}{2}$$ BD = $$\frac{a}{2}$$ In ΔADB using Pythagoras theorem AB 2 = BD 2 + AD 2 (2a) 2 = $${\left( {\frac{a}{2}} \right)^2}$$ + AD 2 4a 2 = $$\frac{{{a^2}}}{4}$$ + AD 2 AD 2 = 4a 2 - $$\frac{{{a^2}}}{4}$$ AD = $$\frac{{\sqrt {15} }}{2}$$ a units

According to question, Given : ∠D = 50° ∠BAC = 2∠BDC (property) ∴ ∠BAC = 2 × 50° ∠BAC = 100°

According to question, AD is the median and 'C' is the centroid ∴ AO = 10 cm 2 units = 10 1 units = 5 ∴ OD = 5 cm

According to question, Given : PQR is an equilateral triangle QR = RS PR = RS ∠SRP = 180° - 60° (Exterior ∠) ∠SRP = 120° ∴ ∠RPS = ∠RSP ∴ ∠RPS + ∠PRS + ∠RSP = 180° 2∠PSR = 180° - 120° ∠PSR = $$\frac{{{{60}^ \circ }}}{2}$$ ∠PSR = 30°

According to question from ΔAQD, ∠A = $$\frac{{180 - 90}}{2}$$ ∠A = 45° ∠DAQ = 30° sin 60° = $$\frac{{AQ}}{{AD}}$$ $$\frac{{\sqrt 3 }}{2}$$ = $$\frac{b}{{AD}}$$ AD = $$\frac{{2b}}{{\sqrt 3 }}$$ From ΔAPD $$\eqalign{ & \sin {75^ \circ } = \frac{{AP}}{{AD}} \cr & \sin {75^ \circ } = \frac{a}{{2b}} \times \sqrt 3 \cr & \sin {75^ \circ } = \frac{{\sqrt 3 a}}{{2b}} \cr} $$

According to question, Given : ∠BAD = 109° ∠ACB = 72° ∴ ∠ACD = 180° - 72° ∠ACD = 108° ∴ AC = CD ∠CAD = ∠CDA In ΔCDA ∠CAD + ∠CDA + ∠DCA = 180° 2∠CAD + 108° = 180° 2∠CAD = 180° -108° 2∠CAD = 72° ∠CAD = $$\frac{{{{72}^ \circ }}}{2}$$ ∠CAD = 36° ∴ ∠CAB = 109° - 36° ∠CAB = 73° In ΔABC ∠ABC + ∠ACB + ∠CAB = 180° ∠ABC + 72° + 73° = 180° ∠ABC + 145° = 180° ∠ABC = 180° - 145° ∠ABC = 35°

The equidistant point from the vertices of a triangle is called circumcenter

ABC is an equilateral triangle and AX, BY and CZ be the altitude so AX = BY = CZ

According to question, In ΔABC Exterior angle CAD = ∠ABC + ∠ACB CAD = 2∠ABC (∵ ∠ABC = ∠ACB) CAD = 2 × 30° CAD = 60° In ΔCAD, ∠ACD = ∠ADC = $$\frac{{180 - \angle {\text{CAD}}}}{2}$$ = 60° ⇒ ∠BCD = ∠ACD + ∠BCA ⇒ ∠BCD = 60° + 30° ⇒ ∠BCD = 90°

According to question, To see in the figure AB > AP BC > BQ AC > CR ∴ AP + BQ + CR < AB + BC + AC

According to question, Given : ∠B = 60° ∠C = 40° As we know that ∠A + ∠B + ∠C = 180° ∠A = 180° - 60° - 40° ∠A = 80° ∴ ∠BAD = $$\frac{{{{80}^ \circ }}}{2}$$ = 40° In ΔAEB ∠A + ∠B + ∠E = 180° ∠A = 180° - 60° - 90° ∠A = 30° Then, ∠DAE = ∠DAB - ∠EAB ∠DAE = 40° - 30° ∠DAE = 10° By Trick $$\eqalign{ & \angle DAE = \frac{{\angle B - \angle C}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{60}^ \circ } - {{40}^ \circ }}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^ \circ } \cr} $$

According to question, Given : ab = $$\frac{{{C^2}}}{2}$$ . . . . . . . . . . (i) ∴ In ΔABC Using Pythagoras theorem AC 2 = AB 2 + BC 2 c 2 = a 2 + b 2 . . . . . . . . . . . (ii) Put the value of C 2 in equation (i) 2ab = a 2 + b 2 a 2 + b 2 - 2ab = 0 (a - b) 2 = 0 ∴ a - b = 0 a = b If a = b means ABC is isosceles right angle triangle it means ∠A = 45° ∠B = 45°

According to question, Given : ∠A + ∠B + ∠C = 180° 2 + 3 + 4 = 9 units ∴ 9 units = 180° 1 unit = 20° ∴ ∠A = 2 × 20° = 40° ∠B = 3 × 20° =60° ∠C = 4 × 20° = 80° and AB || CD ∠B = ∠C ∴ ∠ACD = 180° - 60° - 80° ∠ACD = 40°

According to question, According to figure, when two medians intersect each other in a right angled triangle then we use this equation. ⇒ 4 (BL 2 + CM 2 ) = 5BC 2 ⇒ 4 × $${\left( {\frac{{3\sqrt 5 }}{2}} \right)^2}$$ + 4CM 2 = 5BC 2 ⇒ 45 + 4CM 2 = 125 ⇒ CM 2 = $$\frac{{125 - 45}}{4}$$ ⇒ CM 2 = 20 ⇒ CM = $$2\sqrt 5 $$ cm

According to question, In right angle ΔBAC ∠A + ∠B + ∠C = 180° ∠B = 180° - 55° - 90° ∠B = 35° In right angle ΔADB ∠ADB + ∠ABD + ∠BAD = 180° ∠BAD = 180° - 35° - 90° ∠BAD = 55° Alternate ΔBAC ∼ ΔBDA ∴ ∠BCA = ∠BAD = 55°

According to question, Given : ∠BIC = 120° ∠BIC = 90° + $$\frac{1}{2}$$ ∠A $$\frac{{\angle A}}{2}$$ = (120° - 90°) $$\frac{{\angle A}}{2}$$ = 30° ∠A = 60°

According to question, Given : AB = BC = CA = 10 cm G = Centroid AG = 2 units GD = 1 unit AD = 3 units = Height As we know that the height of the equilateral triangle is $$\eqalign{ & = \frac{{\sqrt 3 }}{2} \times 10 = 5\sqrt 3 \cr & \therefore 3\,{\text{units}} = 5\sqrt 3 \cr & \,\,\,\,\,\,1\,{\text{unit}} = \frac{{5\sqrt 3 }}{3} \cr & \,\,\,\,\,\,2\,{\text{units}} = \frac{{5\sqrt 3 }}{3} \times 2 \cr & \,\,\,\,\,\,2\,{\text{units}} = \frac{{10\sqrt 3 }}{3} \cr & \therefore {\text{AG}} = \frac{{10\sqrt 3 }}{3}\,cm \cr} $$

According to question, Given : AB = 6 cm, BC = 8 cm In right angle ΔABC By using Pythagoras theorem AC 2 = AB 2 + BC 2 AC 2 = 6 2 + 8 2 AC 2 = 36 + 64 AC 2 = 100 AC = 10 cm In radius $$\eqalign{ & = \frac{{a + b - c}}{2} \cr & = \frac{{8 + 6 - 10}}{2} \cr & = \frac{4}{2} \cr & = 2\,cm \cr} $$

According to question, In ΔABC AB 2 = AC 2 + BC 2 . . . . . . . (i) ΔACD AD 2 = AC 2 + CD 2 AC 2 = AD 2 - CD 2 . . . . . . . (ii) Put the value of AC 2 in equation (i) AB 2 = AD 2 - CD 2 + BC 2 AB 2 + CD 2 = AD 2 + BC 2

According to question, Given : ABC is an equilateral triangle CD is the angle bisector of ∠C AC = CE ∴ ∠CAE = ∠CEA ∠ACD = 30° ∴ ∠ECA = 180° - 30° ∠ECA = 150° In ΔCAE ∠CAE + ∠CEA + ∠ECA = 180° ∴ 2∠CAE = 180° - 150° 2∠CAE = 30° ∠CAE = 15°

According to question, In equilateral triangle ∠A + ∠B > ∠C 60° + 60° > 60° 120° > 60° In acute angle triangle ∠P + ∠Q > ∠R 60° + 40° > 80° 100° > 80°

According to question, ∠ABC = ∠DEF Note : Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angles of the other triangle (SAS criterion).

According to question, Given : ∠A = 75°, ∠B = 45° ∴ ∠ACD = ∠A + ∠B x° = ∠ACD = 120° Now, $$\frac{x}{3}$$% of 60° is = $$\frac{{120}}{3}$$ % of 60° = 40% of 60° = $$\frac{{40}}{{100}}$$ × 60° = 24°

According to question, Let angles are 2x, 3x and 7x ∠A + ∠B + ∠C = 180° 2x + 3x + 7x = 180° 12x = 180° x = 15° ∴ Smallest angle is = 2 × 15° = 30°