Triangles - Aptitude MCQ Questions and Solutions with Explanations

31

Length of the sides of a triangle are a, b and c respectively. If a² + b² + c² = ab + bc + ca then the triangle is:

A) Isosceles B) Equilateral C) Scalene D) Right-angled

Discuss

Answer & Solution

Answer: Option B

Solution:

a² + b² + c² = ab + cb + ca
This is true only when a = b = c
So, triangle will be equilateral

32

PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is:

A) 3 cm B) 6 cm C) 12 cm D) 4.5 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

∵ ΔPQR ∼ ΔPMN
∵ ΔPQR is equilateral
∴ PQ = PR = QR
Triangles mcq solution image

So, ΔPMN must be equilateral
So, MN = PN = 6 cm

33

Incenter of ΔABC is I. ∠ABC = 90° and ∠ACB = 70°. ∠BIC is:

A) 115° B) 100° C) 110° D) 105°

Discuss

Answer & Solution

Answer: Option B

Solution:

∵ Sum of all angles of a triangle = 180°
So, ∠BAC = 180 - (90 + 70)
∠BAC = 20°
Triangles mcq solution image

So, ∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90° + $$\frac{1}{2}$$ × 20°
∠BIC = 100°

34

If in ΔABC, DE || BC, AB = 7.5 cm BD = 6 cm and DE = 2 cm then the length of BC in cm is:

A) 6 cm B) 8 cm C) 10 cm D) 10.5 cm

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \vartriangle ADE \simeq \vartriangle ABC \cr & \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} \cr & \frac{{1.5}}{{7.5}} = \frac{2}{{BC}} \cr & BC = 10\,{\text{cm}} \cr} $$

35

Which of the set of three sides can't form a triangle?

A) 5cm, 6cm, 7cm B) 5cm, 8cm, 15cm C) 8cm, 15cm, 18cm D) 6cm, 7cm, 11cm

Discuss

Answer & Solution

Answer: Option B

Solution:

For a triangle sum of 2 sides is always greater than the third side.
Hence, combination (5, 8, 15) never be possible

36

The orthocenter of a triangle is the point where?

A) The medians meet B) The altitudes meet C) The right bisectors of the sides of D) The bisectors of the angles

Discuss

Answer & Solution

Answer: Option B

Solution:

Orthocenter is a point where the altitudes meet

37

Possible length of the sides of a triangle are:

A) 2cm, 3cm, 6cm B) 3cm, 4cm, 5cm C) 2.5cm, 3.5cm, 6cm D) 4cm, 4cm, 9cm

Discuss

Answer & Solution

Answer: Option B

Solution:

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

38

The centroid of a triangle is G. If area of ΔABC = 72 sq. unit, then the area of ΔBGC is?

A) 16 sq. units B) 24 sq. units C) 36 sq. units D) 48 sq. units

Discuss

Answer & Solution

Answer: Option B

Solution:

G is centroid
Area of ΔBGC = $$\frac{1}{3}$$ area of ΔABC
ΔBGC = $$\frac{1}{3}$$ × 72
ΔBGC = 24 sq. units

39

Three sides of a triangle are 5 cm, 9 cm and x cm. The minimum integral value of x is:

A) 2 B) 3 C) 4 D) 6

Discuss

Answer & Solution

Answer: Option D

Solution:

For triangle's side must be
5 + x > 9
or
9 - 5 < x
Only option (d) satisfy
So, x = 6

40

ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12$$\sqrt 5 $$ and BC = 24 cm then radius of circle is:

A) 10 cm B) 15 cm C) 12 cm D) 14 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {R_2} = \frac{{abc}}{{4\vartriangle }} \cr & \vartriangle = \sqrt {S\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr} $$
$$ = \sqrt {12\left( {\sqrt 5 + 1} \right)\left( {12} \right) \times 12 \times 12\left( {\sqrt 5 - 1} \right)} $$
where a = 12$$\sqrt 5 $$ , b = 12$$\sqrt 5 $$ & c = 24
$$\eqalign{ & S = \frac{{a + b + c}}{2} \cr & S = \frac{{24\sqrt 5 + 24}}{2} \cr & S = 12\left( {\sqrt 5 + 1} \right) \cr & {R_2} = \frac{{12\sqrt 5 \times 12\sqrt 5 \times 24}}{{4 \times 12 \times 12 \times 2}} \cr & {R_2} = \frac{{30}}{2} \cr & {R_2} = 15\,{\text{cm}} \cr} $$