Quantitative Aptitude
TRIANGLES MCQs
Total Questions : 83
| Page 5 of 9 pages
Answer: Option B. -> 24 sq. units
G is centroid
Area of ΔBGC = $$\frac{1}{3}$$ area of ΔABC
ΔBGC = $$\frac{1}{3}$$ × 72
ΔBGC = 24 sq. units
G is centroid
Area of ΔBGC = $$\frac{1}{3}$$ area of ΔABC
ΔBGC = $$\frac{1}{3}$$ × 72
ΔBGC = 24 sq. units
Answer: Option B. -> 3cm, 4cm, 5cm
If triangle's side are a, b, c then must be:-
a + b > c
or a - b < c
only option (B) satisfy
3 + 4 > 5
7 > 5
If triangle's side are a, b, c then must be:-
a + b > c
or a - b < c
only option (B) satisfy
3 + 4 > 5
7 > 5
Answer: Option B. -> 15 cm
$$\eqalign{
& {R_2} = \frac{{abc}}{{4\vartriangle }} \cr
& \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$ & c = 24
$$\eqalign{
& S = \frac{{a + b + c}}{2} \cr
& S = \frac{{24\sqrt 5 + 24}}{2} \cr
& S = 12\left( {\sqrt 5 + 1} \right) \cr
& {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr
& {R_2} = \frac{{30}}{2} \cr
& {R_2} = 15\,{\text{cm}} \cr} $$
$$\eqalign{
& {R_2} = \frac{{abc}}{{4\vartriangle }} \cr
& \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$ & c = 24
$$\eqalign{
& S = \frac{{a + b + c}}{2} \cr
& S = \frac{{24\sqrt 5 + 24}}{2} \cr
& S = 12\left( {\sqrt 5 + 1} \right) \cr
& {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr
& {R_2} = \frac{{30}}{2} \cr
& {R_2} = 15\,{\text{cm}} \cr} $$
Answer: Option D. -> 15°
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,
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°
Answer: Option D. -> 24°
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,
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°
Answer: Option D. -> ∠ABC = ∠DEF
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,
∠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).
Answer: Option B. -> Acute or equilateral
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,
In equilateral triangle
∠A + ∠B > ∠C
60° + 60° > 60°
120° > 60°
In acute angle triangle
∠P + ∠Q > ∠R
60° + 40° > 80°
100° > 80°
Answer: Option A. -> 30°
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°
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°
Answer: Option C. -> Isosceles and equilateral
According to question,
AB = AC
BD = DC
The triangle will be isosceles and equilateral triangle
According to question,
AB = AC
BD = DC
The triangle will be isosceles and equilateral triangle
Answer: Option A. -> 3 units
According to question,
ABC is a right angle triangle
By using Pythagoras theorem
AC2 = BC2 + AB2
52 = 32 + 42
25 = 9 + 16
25 = 25 (satisfied)
∴ Smallest length of right angle triangle is 3 units
According to question,
ABC is a right angle triangle
By using Pythagoras theorem
AC2 = BC2 + AB2
52 = 32 + 42
25 = 9 + 16
25 = 25 (satisfied)
∴ Smallest length of right angle triangle is 3 units