Triangles — Part 1 MCQs with Explanations

Since, ΔABC and ΔPQR are similar triangles. then, ∠B = ∠Q = 83° Thus, in ΔABC, ∠C = 180° - (∠A + ∠ B) or, ∠C = 180° - (47° + 83°) ∠C = 50°

In Triangle ABD,∠BAD + ∠B + 90° = 180° or, ∠BAD + ∠B = 90° - - - - - - - (1) Now, in Triangle ABC, ∠ACB + ∠B + ∠A = 180° ∠ACB + ∠B = 180° - 90° ∠ACB = 90° - ∠B - - - - - (2) From (1) and (2), ∠BAD = ∠ACB So, AC = CD

In δ PQR, QR + 2 = 2PQ QR = 2PQ - 2 - - - - - - - (1) PR = PQ + 10 - - - - - - (2) Perimeter = 40 m PQ + QR + Rp = 40 Putting the value of PQ and QR from equation (1) and (2), PQ + 2PQ - 2 + PQ + 10 = 40 4PQ = 32 PQ = 8 cm which is the smallest side of the triangle.

Since, AB and CD bisects each other at O, Hence, BC = AD = 6 cm.

∠BAL + ∠B + 90° = 180° or, ∠BAL + ∠B = 90° or, ∠BAL = 90° - ∠B - - - - - - - - (1) Now, in ΔABC, ∠ACB + ∠B + ∠A = 180° ∠ACB = 90° -∠B - - - - - (2) From, (1) and (2), ∠BAL = ∠ACB

Here, ∠ACB = c + 180 - (2c - b) = 180 - (b + c) So, We can say that ΔBCD and ΔABC will be similar. According to property of similarity, $$\frac{{{\text{AB}}}}{{12}} = \frac{{12}}{9}$$ Hence, AB = 16 $$\frac{{{\text{AC}}}}{6} = \frac{{12}}{9}$$ AC = 8 Hence, AD = 7 and AC = 8 Now, $$\eqalign{ & \frac{{{\text{Perimeter of Delta ADC}}}}{{{\text{Perimeter of Delta BDC}}}} \cr & = \frac{{6 + 7 + 8}}{{9 + 6 + 12}} \cr & = \frac{{21}}{{27}} \cr & = \frac{7}{9} \cr} $$

Let BC = x and Ad = y, then as per bisector theorem, $$\frac{{{\text{BD}}}}{{{\text{DC}}}} = \frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{4}{3}$$ Hence, BD = $$\frac{{4{\text{x}}}}{7}$$ and DC = $$\frac{{3{\text{x}}}}{7}$$ Now, in ΔABD using cosine rule, $$\cos {30^ \circ } = \frac{{{4^2} + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$ $${\text{or,}}\,4\sqrt {3{\text{y}}} = {16 + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} } $$ - - - - - - (i) Similarly in ΔADC, $$\cos {30^ \circ } = \frac{{{3^2} + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$ $${\text{or,}}\,3\sqrt {3{\text{y}}} = {9 + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} } $$ - - - - - - - - (ii) From equation (i) and (ii), we get $${\text{y}} = \frac{{12\sqrt 3 }}{7}$$

Orthocentre.

∠1 + ∠2 = ∠3 [PS is bisector.] - - - - - - (1) ∠Q = 90° - ∠1 ∠R = 90° -∠2 - ∠3 So, ∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3) ∠Q - ∠R = ∠2 + ∠3 - ∠1 ∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1 [using equation 1] ∠Q - ∠R = 2∠2 $$\frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS

Two right angled triangle are congruent if the hypotenuse of one triangle is equal to the hypotenuse of the other and a side of one triangle is equal to the corresponding side of the other triangle.

Every equilateral triangle is necessarily an isosceles triangle. A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.

Statement II is false, as in case of equilateral triangles of different sides of the triangle are similar.

In ΔADC AB 2 = AD 2 + BD 2 - - - - - - - (1) In Right angled Δ ACD, AC 2 = AD 2 + BD 2 - - - - - - - (2) By (1) - (2) AB 2 - AC 2 = BD 2 - CD 2 AB 2 - BD 2 = AC 2 - CD 2

In ΔGEF and ΔGCD, we have ∠EFG = ∠GCD (Alternative angle) ∠EFG = ∠CGD (Vertically opposite angles) ΔGEF ~ ΔGCD Thus, $$\eqalign{ & \frac{{{\text{GE}}}}{{{\text{CG}}}} = \frac{{{\text{EF}}}}{{{\text{CD}}}} \cr & {\text{or, }}\frac{5}{{10}} = \frac{{{\text{EF}}}}{{18}} \cr & {\text{or,}}\,{\text{EF}} = 9\,{\text{cm}} \cr} $$

Addition any two side of a triangle is always greater then another side then only triangle formation is possible lets e.g a, b, c is a side of a triangle then a + b > c, b + c > a, c + a > b So option B is not satisfying the condition 3 + 4 ≯ 8

According to question, AO = I R = Circumradius DO = I r = Inradius = 3 cm Median AD = 3 units 1 unit = 3 cm 3 units = 3 × 3 = 9 cm ∴ AD = 9 cm

According to question, Let sides of the triangle be 3x, 4x, 6x Now check the square of biggest side and sum of square of two smallest side and check which is greater. ∴ (3x) 2 + (4x) 2 < (6x) 2 ⇒ 25x 2 < 36x 2 ∴ The triangle will be obtuse angled triangle.

According to question, By internal bisector property $$\eqalign{ & \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}} \cr & \frac{{AB}}{{AC}} = \frac{5}{{2.5}} \cr & \frac{{AB}}{{AC}} = \frac{2}{1} \cr & \therefore AB:AC = 2:1 \cr} $$

According to question, Circumradius of an equilateral triangle I R = 10 cm AO = I R = 10 cm DO = I r = ? 2 units → 10 cm 1 unit → 5 cm ∴ DO = I r = 5 cm

According to question, BI and CI are the angle bisector ∴ ∠CBI = 30° and ∠BCI = 25° In ΔBIC ∠CBI + ∠BCI + ∠BIC = 180° 30° + 25° + ∠BIC = 180° ∠BIC = 180° - 55° ∴ ∠BIC = 125°

According to question, Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle. Note: In right angle triangle median divides the hypotenuse in two equal parts ∴ BD = $$\frac{{\text{H}}}{2}$$ BD = $$\frac{{10}}{2}$$ BD = 5 cm

According to question, Given: ∠BAC = 85° ∠BCA = 75° ∠OAC = ? ∠ABC + ∠BCA + ∠CAB = 180° ∠ABC = 20° ∴ ∠COA = 2 × ∠ABC ∠COA = 2 × 20 = 40° In ΔAOC We know OC = OA ∴ ∠OAC = ∠OCA ∴ ∠OAC + ∠OCA + ∠COA = 180° 2∠OAC = 180° - 40° 2∠OAC = 140° ∠OAC = 70°

Inradius = $$\frac{a}{{2\sqrt 3 }}$$ (a = side of Δ) 3 = $$\frac{a}{{2\sqrt 3 }}$$ a = 6$$\sqrt 3 $$ cm

According to question, Given: AC = 9 cm area of ΔABC = 40 cm 2 area of ΔADC = 10 cm 2 ΔABC ∼ ΔADC $$\frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ADC}} = \frac{{A{B^2}}}{{A{D^2}}} = \frac{{B{C^2}}}{{A{C^2}}}$$ (In similar Δratio of their area is square of ratio of corresponding sides) $$\eqalign{ & \frac{{40}}{{10}} = \frac{{B{C^2}}}{{{{\left( 9 \right)}^2}}} \cr & \frac{{40}}{{10}} \times 81 = B{C^2} \cr & BC = 18\,{\text{cm}} \cr} $$

The orthocentre of a right angled triangle lies at the right angular vertex

According to question, Given: ∴ ∠BOC = 90° + $$\frac{1}{2}$$ ∠A ∠BOC = 90° + $$\frac{1}{2}$$ × 30° ∠BOC = 90° + 15° ∠BOC = 105°

According to question, Given: ∠C = 90° BC = AC = 5 cm (Isosceles triangle) By Pythagoras theorem AB 2 = AC 2 + BC 2 AB 2 = 5 2 + 5 2 AB 2 = 25 + 25 AB 2 = 50 AB = 5$$\sqrt 2 $$ cm

According to question, Given: BAC is a right angle triangle AD ⊥ BC AD = 6 cm BD = 4 cm BC = ? In ΔBAD $$\eqalign{ & AB = \sqrt {B{D^2} + A{D^2}} \cr & AB = \sqrt {{4^2} + {6^2}} \cr & AB = \sqrt {52} \,cm \cr} $$ ΔBAC ∼ ΔBDA $$\eqalign{ & \therefore \frac{{BC}}{{AB}} = \frac{{AB}}{{BD}} \cr & \therefore \frac{{BC}}{{\sqrt {52} }} = \frac{{\sqrt {52} }}{4} \cr & BC = \frac{{52}}{4} \cr & BC = 13\,cm \cr} $$ Alternate : $$\eqalign{ & A{B^2} = BD.BC \cr & {\left( {\sqrt {B{D^2} + A{D^2}} } \right)^2} = BD.BC \cr & {\left( {\sqrt {{4^2} + {6^2}} } \right)^2} = 4.BC \cr & \frac{{52}}{4} = BC, \cr & \therefore BC = 13\,cm \cr} $$

According to question, Given : ∠BOD = 15° ∴ ∠BDO + ∠DOB + ∠DBO = 180° ∠DBO = 75° ∠ABC = 2 × ∠DBO ∠ABC = 2 × 75° ∠ABC = 150°

Circumcentre of a triangle lies outside then triangle is obtuse angled triangle.

According to question, Given : ∠ABC = 90° $$\frac{{AN}}{{NC}} = ?$$ ΔABC ∼ ΔBNC ΔABC ∼ ΔANB ∴ ΔABC ∼ ΔBNC ∼ ΔANB AB = 3, BC = 4, AC = 5 $$\eqalign{ & \frac{{AB}}{{BN}} = \frac{{AC}}{{BC}} \cr & BN = \frac{{AB \times BC}}{{AC}} = \frac{{3 \times 4}}{5} = 2.4 \cr & \frac{{BC}}{{NC}} = \frac{{AB}}{{NB}} \cr & \frac{4}{{NC}} = \frac{3}{{2.4}} \cr & NC = 3.2 \cr & \frac{{AB}}{{AN}} = \frac{{BC}}{{NB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{{AN}} = \frac{4}{{2.4}} \cr & AN = 1.8 \cr & \frac{{AN}}{{NC}} = \frac{{1.8}}{{3.2}} \cr & \frac{{AN}}{{NC}} = \frac{9}{{16}} \cr & \therefore AN:NC = 9:16 \cr} $$

According to question, Given : ∠BOC = 110° ∠BOC = 90° + $$\frac{1}{2}$$ ∠A 110° = 90° + $$\frac{{\angle A}}{2}$$ $$\frac{{\angle A}}{2}$$ = 20 ∠A = 40°

According to question, Given : ∠ACB = 55° ∠ABC = 65° ∠BIC = ? ∴ ∠ACB + ∠ABC + ∠BAC = 180° ∠BAC = 180° - 55° - 65° ∠BAC = 60° We know that ∠BIC = 90 + $$\frac{1}{2}$$ ∠A ∠BIC = 90 + $$\frac{1}{2}$$ × 60 ∠BIC = 90 + 30 ∠BIC = 120° Alternate: In ΔBIC, $$\frac{1}{2}$$ ∠B + $$\frac{1}{2}$$ ∠C + ∠BIC = 180° $$\frac{1}{2}$$ (65° + 55°) + ∠BIC = 180° ∠BIC = 180° - 60° ∠BIC = 120°

According to question, Given : ABC is a right angle triangle BC = 6$$\sqrt 3 $$ ∴ sin 30° = $$\frac{P}{H}$$ = $$\frac{{AB}}{{6\sqrt 3 }}$$ AB = 3$${\sqrt 3 }$$ sin 60° = $$\frac{P}{H}$$ = $$\frac{{AN}}{{AB}}$$ $$\frac{{\sqrt 3 }}{2}$$ = $$\frac{{AN}}{{3\sqrt 3 }}$$ AN = $$\frac{9}{2}$$ AN = 4.5 cm

According to question, PX || BD [mid point theorem] ∴ PX = $$\frac{1}{2}$$BD Similarly, QY || BD ∴ QY = $$\frac{1}{2}$$BD ∴ PX : QY, $$\frac{1}{2}$$BD : $$\frac{1}{2}$$BD PX : QY = 1 : 1

According to question, Given : BAC is right angle triangle AB = $$\frac{1}{2}$$BC $$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$

According to question, ACB is a right angle triangle ∴ area of ΔACB $$\frac{1}{2}$$ × AC × BC = $$\frac{1}{2}$$ × AB × PC $$\frac{1}{2}$$ × b × a = $$\frac{1}{2}$$ × c × p c = $$\frac{{ab}}{p}$$ . . . . . . . (i) By using pythagoras theorem AB 2 = AC 2 + BC 2 c 2 = b 2 + a 2 . . . . . . . . . . (ii) Put the value of C in equation (ii) $$\eqalign{ & {\left( {\frac{{ab}}{p}} \right)^2} = {a^2} + {b^2} \cr & \Rightarrow \frac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{{{a^2}}}{{{a^2}{b^2}}} + \frac{{{b^2}}}{{{a^2}{b^2}}} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$ Alternate : From figure, $$\eqalign{ & P = \frac{{ab}}{c} \cr & P = \frac{{ab}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\because {a^2} + {b^2} = {c^2}} \right) \cr & {P^2} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}} \cr & \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$

In a right angled triangle orthocentre lies vertex

According to question, ABC is a right angle triangle ∴ Apply Pythagoras theorem AC 2 = AB 2 +BC 2 (x + 2) 2 = (x - 2) 2 + x 2 x 2 + 4 + 4x = x 2 + 4 - 4x + x 2 x 2 = 8x x = 8 Alternate : From option approach $$\eqalign{ & \left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 2} \right) \cr & \,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Triplet}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} = {\text{8}} \cr} $$

According to question, ABC is a right angle triangle, ∴ AB = 8 cm BC = 6 cm ∴ AC 2 = AB 2 + BC 2 AC = 64 + 36 AC = $$\sqrt {100} $$ AC = 10 cm In right triangle Circum Radius I R = $$\frac{{AC}}{2}$$ = $$\frac{{10}}{2}$$ = 5 cm