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1
Given that the ratio of altitudes of two triangles is 4 : 5, ratio of their areas is 3 : 2, the ratio of their corresponding bases is :
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \frac{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,1}}{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,2}} = \frac{3}{2} \cr & \Rightarrow \frac{{\frac{1}{2} \times {B_1} \times 4}}{{\frac{1}{2} \times {B_2} \times 5}} = \frac{3}{2} \cr & \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{3}{2} \times \frac{5}{4} \cr & \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{{15}}{8} \cr & \therefore {B_1}:{B_2} = 15:8 \cr} $$
2
In ΔABC, the external bisectors of the angles ∠B and ∠C meet at the point O. If ∠A = 70°, then the measure of ∠BOC is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Triangles mcq solution image
As we know
⇒ The external bisectors of the angle ∠B and ∠C meet at the point O
∠BOC = 90° - $$\frac{{\angle A}}{2}$$
∠BOC = 90° - $$\frac{{70}}{2}$$
∠BOC = 90° - 35°
∠BOC = 55°
3
If I be the incentre of ΔABC and ∠B = 70° and ∠C = 50°, then the magnitude of ∠BIC is
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
As we know that
   ∠A + ∠B + ∠C = 180°
∴ ∠A = 180° -70° - 50°
   ∠A = 60°
∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 60°
   ∠BIC = 120°
4
In ΔABC, if AD ⊥ BC, then AB2 + CD2 is equal to
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Triangles mcq solution image
Given :
AD ⊥ BC
∴ In ΔADB
AB2 = BD2 + AD2
AD2 = AB2 - BD2 . . . . . . . . (i)
In ΔADC
AC2 = AD2 + CD2
AD2 = AC2 - CD2 . . . . . . . (ii)
Compare equation (i) and (ii)
AB2 - BD2 = AC2 - CD2
AB2 + CD2 = BD2 + AC2
5
If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB = 19 cm and AC = 22 cm then the length of BC is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Triangles mcq solution image
Given :
AB = 19 cm, AC = 22 cm
∵ BE ⊥ CF (given), [Medians CF & BE are perpendicular to each other]
⇒ In this case
⇒ We know,
   AB2 + AC2 = 5(BC)2
⇒ 192 + 222 = 5(BC)2
⇒ 361 + 484 = 5(BC)2
⇒ 845 = 5(BC)2
⇒ (BC)2 = 169
⇒ BC = 13 cm
6
ABC is a triangle in which ∠A = 90°. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
BC = 10 cm, AC = 8 cm, BP = 9 cm
In ΔCAB,
BC2 = AB2 + AC2
102 = AB2 + 82
Triangles mcq solution image
AB2 = 100 - 64
AB2 = 36
AB = 6 cm
In ΔPAB
BP2 = AB2 + AP2
92 = 62 + AP2
AP2 = 81 - 36
AP2 = 45
AP = $$3\sqrt 5 $$  cm
7
∠A of ΔABC is a right angle. AD is perpendicular on BC. If BC = 14 and BD = 5 cm, then measure of AD is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Triangles mcq solution image
AD2 = BD × DC
AD2 = 5 × 9
AD = $$\sqrt {45} $$
AD = $$3\sqrt 5 $$  cm
8
For a triangle ABC, D, E, F are the mid - point of its sides. If ΔABC = 24 sq. units then ΔDEF is :
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Triangles mcq solution image
As we know that
Given: area of ΔABC = 24 square units
As we know that
D, E and F are the midpoint of AB, AC and BC
∴ Area of ΔADE = area of ΔDBF
= area of ΔDEF = area of ΔEFC
∴ Area of ΔDEF = $$\frac{1}{4}$$ area of ΔABC
Area of ΔDEF = $$\frac{1}{4}$$ × 24
Area of ΔDEF = 6 sq. units
9
∠A + $$\frac{1}{2}$$ ∠B + ∠C = 140°, then ∠B is :
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
∠A + $$\frac{1}{2}$$ ∠B + ∠C = 140° . . . . . . . . (i)
As we know that
∠A + ∠B + ∠C = 180° . . . . . . . . . (ii)
Compare equation (i) and (ii)
$$\frac{1}{2}$$ ∠B = 40°
∴ ∠B = 80°
10
If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60°, BC = 8 cm, then area of the triangle ABC is:
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°
Triangles mcq solution image
⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{{\sqrt 3 }}{4}$$ a2
= $$\frac{{\sqrt 3 }}{4}$$ 82
= 16$${\sqrt 3 }$$ cm2
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$${\sqrt 3 }$$
⇒ Area of ΔABC = 48$${\sqrt 3 }$$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}