Triangles MCQ Quiz - Objective Question with Answer for Triangles - Download Free PDF

Given:

In Δ ABC, AB = 12 cm, BC = 16 cm, AC = 20 cm

Formula used:

Area of the triangle (Δ) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Where s = semi-perimeter = \(\frac{a+b+c}{2}\)

Radius (r) of the inscribed circle = \(\frac{\Delta}{s}\)

Calculations:

a = 12 cm, b = 16 cm, c = 20 cm

s = \(\frac{12+16+20}{2}\) = 24 cm

Area (Δ) = \(\sqrt{24(24-12)(24-16)(24-20)}\)

⇒ Area (Δ) = \(\sqrt{24×12×8×4}\)

⇒ Area (Δ) = \(\sqrt{9216}\)

⇒ Area (Δ) = 96 cm2

Radius (r) = \(\frac{96}{24}\)

⇒ Radius (r) = 4 cm

∴ The correct answer is option (2).

Given:

In triangle ABC, AB = AC = 12 cm.

BC = 5 cm.

D is a point on AC such that DB = BC.

Formula used:

Cosine Rule in a triangle: In any triangle with sides a, b, c and angle C opposite to side c,

c² = a² + b² - 2ab cos(C).

Calculation:

In △ABC, AB = 12 cm, AC = 12 cm, BC = 5 cm.

Since AB = AC, △ABC is an isosceles triangle.

Let's find cos(∠C) in △ABC using the Cosine Rule:

AB² = BC² + AC² - 2 × BC × AC × cos(∠C)

12² = 5² + 12² - 2 × 5 × 12 × cos(∠C)

144 = 25 + 144 - 120 × cos(∠C)

0 = 25 - 120 × cos(∠C)

120 × cos(∠C) = 25

cos(∠C) = 25 / 120

cos(∠C) = 5 / 24

Now consider △DBC.

We are given DB = BC. Since BC = 5 cm, then DB = 5 cm.

We know ∠C is common to both triangles. Let CD = x cm.

Apply the Cosine Rule in △DBC to find CD:

DB² = BC² + CD² - 2 × BC × CD × cos(∠C)

5² = 5² + x² - 2 × 5 × x × (5 / 24)

25 = 25 + x² - (50x / 24)

0 = x² - (25x / 12)

Since x = CD cannot be 0 (as D is a point on AC), we can divide by x:

0 = x - (25 / 12)

x = 25 / 12 cm

∴ The correct answer is option 2.

Given:

In Δ PQR, PS is the median of the triangle.. QR is the base.

the value of PQ is 20 cm and PS is 15 cm and SR is 10 cm.

Formula used:

As per Apollonius theorem,

PQ² + PR² = 2 (Qs² + PS²)

Calculation:

Let PR = x, PS median, so QS = SR

As per the theorem,

20² + x² = 2 (15² + 10²)

⇒ 400 + x² = 2 × 325

⇒ x². = 650 - 400

⇒ x² = 250

⇒ x = 5√10 cm

The value of PR is 5√10 cm.

Given:

∆LMN with medians MX and NY

MX ⊥ NY

MX intersects NY at Z

MX = 20 cm

NY = 30 cm

Formula Used:

The centroid of a triangle divides each median in the ratio 2:1.

Area of a triangle = 1/2 × base × height

Area of ∆LMN = 3 × Area of ∆MNZ (since Z is the centroid)

Calculations:

Since Z is the centroid, it divides the medians in the ratio 2:1.

MZ : ZX = 2 : 1

⇒ MZ = (2/3) × MX = (2/3) × 20 = 40/3 cm

⇒ ZX = (1/3) × MX = (1/3) × 20 = 20/3 cm

NZ : ZY = 2 : 1

⇒ NZ = (2/3) × NY = (2/3) × 30 = 20 cm

⇒ ZY = (1/3) × NY = (1/3) × 30 = 10 cm

Since MX ⊥ NY, ∆MNZ is a right-angled triangle with legs NZ and MZ.

Area of ∆MNZ = 1/2 × base × height = 1/2 × NZ × MZ

⇒ Area of ∆MNZ = 1/2 × 20 × (40/3)

⇒ Area of ∆MNZ = 10 × (40/3) = 400/3 cm²

Area of ∆LMN = 3 × Area of ∆MNZ

⇒ Area of ∆LMN = 3 × (400/3)

⇒ Area of ∆LMN = 400 cm²

∴ The area of ∆LMN is 400 cm².

Given:

ΔABC, ∠A = 2∠B

AD bisects ∠A

CD = 4 cm

BD = 6 cm

Image of triangle ABC

Formula used:

Congruent triangles: Corresponding sides and angles are equal.

Calculation:

DC = 4 cm 

BC = 6 + 4 = 10 CM

Now in Δ ACB and Δ DCA

ΔACB, ∠A = 2∠B = 2X

∠BAC = 2X, ∠ABC = X

And, Now in Δ DCB

∠ADC = 2X (Sum of interior angle is equal to opposite exterior angle.)

∠DAC = X (As AD is bisector of ∠A)

And AC is common in both the triangle.

Then by ASA(angle side angle) Δ ACB and Δ DCA is congruent. 

Then, 

AC/DC = CB/CA

AC² = DC × CB

AC2 = 4 × 10

AC = 2√10

∴ The length of side AC is 2√10 cm.

Given:

In the triangle, ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°.

Concept used:

According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a² = b² + c² - 2bc × cos∠A

Calculation:

​According to the concept,

BC² = AB² + AC² - 2 × AB × AC × cos60°

⇒ BC² = 12² + 10² - 2 × 12 × 10 × 1/2

⇒ BC² = 124

⇒ BC ≈ 11.13

∴ The measure of BC is 11.13 cm.

Concept used:

If the perimeter of the triangle is "p"

Let Total possible triangles "t"

If p = even, then

t = p²/48

If p = odd, then

t = (p + 3)²/48

Calculation:

According to the question,

Total possible triangles = (13 + 3)²/48

⇒ 5.33 ≈ 5

∴ Total possible triangles are 5.

Given:

Triangle first side (a) = 16 units

Triangle second side (b) = 30 units

Triangle third side (c) = 34 units

Formula used:

Heron's formula:

Area of triangle = √{s × (s - a) × (s - b) × (s - c)}

Where, semi - perimeter (s) = (a + b + c)/2

and a, b and c are the sides of a triangle.

Circum-radius of a triangle = (a × b × c)/(4 × area of triangle)

Calculation:

Semi - perimeter = (16 + 30 + 34)/2 = 80/2 = 40 units

Area of triangle = √{s × (s - a) × (s - b) × (s - c)}

⇒ √{40 × (40 - 16) × (40 - 30) × (40 - 34)}

⇒ √{40 × 24 × 10 × 6} = √57600 = 240 unit²

Circum-radius of a triangle = (a × b × c)/(4 × area of triangle)

⇒ (16 × 30 × 34)/(4 × 240) = 17 units

∴ The correct answer is 17 units.

Shortcut TrickCalculation:

The sides of a triangle given are Pythagorean triples.

So the hypotenuse = 34 units

and the circumradius of right angled triangle = 34/2 = 17 units

Given:

First side of triangle = 30 cm

Second side of triangle = x cm

Third side of triangle = 42 cm

Concept used:

(3rd side - 1st side) < second side < (3rd side + 1st side)

Calculation:

Range of second side = (42 - 30) < x < (42 + 30)

⇒ 12 < x < 72

∴ The correct option is 3.

Given:

BC = 16 cm, BD = 11 cm and ∠ADC = ∠BAC

Concept:

If two angles and a side of the two triangles are equal, then both triangles will be similar by AA property.

Calculation:

In ΔABC and ΔDAC

⇒ ∠ADC = ∠BAC

⇒ ∠C = Common angle in both triangle

So, ΔABC and ΔDAC are similar triangle.

⇒ \({BC\over AC}={AC\over DC}\)

⇒ AC² = BC × DC

⇒ AC² = 16 × 5 = 80

⇒ AC = 4√5

∴ The required result will be 4√5.

Given:

ABC is right angle triangle at angle B, BC = 10 cm

 AC = 12 cm, p is length of the perpendicular from B to AC

Formula used:

ArΔ = 1/2 × base × height

Calculation:

In an Δ ABC, by using the Pythagoras theorem

AC² = AB² + BC²

144 = AB² + 100

AB² = 44

AB = √44

Here, We can find the area in two ways,

1) By taking AC as the base & length p as the perpendicular.

2) By taking BC as base & AB as the perpendicular

As, Area (ΔABC) = Area (ΔABC)

⇒ 1/2 × 10 × √44 = 1/2 × 12 × p

⇒ 5 × 2√11 = 6p

⇒ p = (5√11)/3 cm

∴ The correct answer is (5√11)/3 cm

Given:

The perpendicular distance:

P₁ = √3; P₂ = 2√3; P₃ = 5√3

Concept used:

Height of an equilateral triangle = (√3 × side)/2 

Height of equilateral triangle = sum of perpendicular distance with point

Perimeter of an equilateral triangle = 3 × side

Calculation:

Height of equilateral triangle = sum of perpendicular distance

⇒ (√3 × side)/2 = P1 + P2 + P3 

⇒ (√3 × side)/2 = √3 + 2√3 + 5√3

⇒ side = 8 × 2 = 16 cm

Perimeter of an equilateral triangle = 3 × side

⇒ 3 × 16 = 48 cm

∴ The correct answer is 48 cm.

Given:

AB = 8.4 cm, and AC = 5.6 cm, DC = 2.8 cm

Concept used:

The angle bisector of a triangle divides the opposite side into two parts proportional to the other two sides of the triangle.

Calculation:

According to the concept,

AB/AC = BD/DC

⇒ 8.4/5.6 = BD/2.8

⇒ 8.4/2 = BD

⇒ 4.2 = BD

So, BD + DC = BC

BC = 4.2 + 2.8

⇒ 7 cm

∴ The length of side BC will be 7 cm.

Given:

AB = DB and AC = DC.

∠ABD = 58° and ∠DBC = (2x - 4)°, 

∠ACB = (y + 15)° and ∠DCB = 63°

Concept used:

If all three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS (Side - Side - Side) rule.

Calculation:

As AB = DB, AC = DC, and BC is common for two triangle 

So, ΔABC ≅ ΔDBC

So, ∠ABC = ∠DBC = ∠ABD/2

⇒ 58°/2 = 29°

So,

(2x - 4)° = 29°

⇒ 2x = 33°

Again,

∠ACB = ∠DCB = 63°

So,

(y + 15)° = 63°

⇒ y = 48°

So,

2x + 5y = 33° + 5 × 48°

⇒ 33° + 240°

⇒ 273°

∴ The required answer is 273°.

Given:

In ΔABC, M is the midpoint of the side AB

N is a point in the interior of ΔABC such that CN is the bisector of ∠C and CN ⊥ NB

BC = 10 cm

AC = 15 cm

Concept used:

Midpoint theorem - The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side

Calculation:

Construction: Produce BN to P which meets AC at P.

And Join MN

According to the question

In ΔNPC and ΔNBC

∠N = ∠N  [90°]

BC = PC [corresponding side]

BN = NP [corresponding angle]

⇒ ΔNPC ≅ ΔNBC 

Hence, NB = NP (It means Point N is the midpoint of side BP)

And BC = PC = 10 cm

So, AP = AC – PC

⇒ AP = (15 – 10) cm

⇒ AP = 5 cm

Now, In ΔABP

M and N are the midpoints of AB and BP

So, According to the midpoint theorem

⇒ MN = \(\frac{AP}{{2}}\)

⇒ \(\frac{5}{{2}}\) cm

⇒ 2.5 cm

∴ The length of MN is 2.5 cm

Shortcut Trick  

The using mid-point theorem,

In ΔBAP

MN = \(AP\over2\) = \(\frac{5}{{2}}\) = 2.5 cm