Triangles - Study Mode

[#76] The orthocentre of a right angled triangle lies
Correct Answer

(B) At the right angular vertex

Explanation

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

[#77] O is the incentre of ΔABC and ∠A = 30°, then ∠BOC is
Correct Answer

(B) 105°

Explanation

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

[#78] If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5 cm then AB is:
Correct Answer

(C) 5$$sqrt 2 $$ cm

Explanation

Solution: 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

[#79] In a triangle ABC, ∠BAC = 90° and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm then the length of BC is:
Correct Answer

(D) 13 cm

Explanation

Solution: 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 $$eqalign{
& herefore frac{{BC}}{{AB}} = frac{{AB}}{{BD}} cr
& herefore frac{{BC}}{{sqrt {52} }} = frac{{sqrt {52} }}{4} cr
& BC = frac{{52}}{4} cr
& BC = 13,cm cr} $$ Alternate : $$eqalign{
& A{B^2} = BD.BC cr
& {left( {sqrt {B{D^2} + A{D^2}} }
ight)^2} = BD.BC cr
& {left( {sqrt {{4^2} + {6^2}} }
ight)^2} = 4.BC cr
& frac{{52}}{4} = BC, cr
& herefore BC = 13,cm cr} $$

[#80] Let O be the in-centre of a triangle ABC and D be a point on the side BC of ΔABC, such that OD ⊥ BC. If ∠BOD = 15°, then ∠ABC = ?
Correct Answer

(C) 150°

Explanation

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