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 = $\frac{\mathrm{\Delta }}{s}$

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) = $\sqrt{24\times 12\times 8\times 4}$

⇒ Area (Δ) = $\sqrt{9216}$

⇒ Area (Δ) = 96 cm2

Radius (r) = $\frac{96}{24}$

⇒ Radius (r) = 4 cm

∴ The correct answer is option (2).

Given:

The sides of a triangle are k, 1.5k, and 2.25k.

Formula used:

The sum of the squares of medians of a triangle is given by:

$\text{Sum of squares of medians}=\frac{3}{4}(a^2+b^2+c^2)$

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

Calculation:

Let a = k, b = 1.5k = 3k/2, c = 2.25k = 9k/4.

$a^2+b^2+c^2=k^2+\frac{9}{4}{k}^2+\frac{81}{16}{k}^2$

$a^2+b^2+c^2=\frac{16}{16}{k}^2+\frac{36}{16}{k}^2+\frac{81}{16}{k}^2=\frac{133}{16}{k}^2$

Sum of squares of medians = $\frac{3}{4}\times \frac{133}{16}{k}^2$

Sum of squares of medians = $\frac{3\times 133}{4\times 16}{k}^2=\frac{399}{64}{k}^2$

Therefore, the sum of the squares of the medians is: $\frac{399}{64}{k}^2$

Given:

Perimeter of triangle ABC = 105 cm

Ratio of sides AB : BC : CA = 5 : 10 : 6

Formula used:

Heron's Formula for the area of a triangle:

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Where a, b, c are the lengths of the sides of the triangle, and

s is the semi-perimeter s = $({\displaystyle \frac{\text{a + b + c}}{2}})$

Calculations:

From the previous problem, we established

The ratio of the sides AB : BC : CA = 5 : 10 : 6.

Let the sides be AB = 5x, BC = 10x, and CA = 6x.

The perimeter is the sum of the sides:

Perimeter = AB + BC + CA = 5x + 10x + 6x = 21x

Given Perimeter = 105 cm.

⇒ 21x = 105

⇒ x = $({\displaystyle \frac{105}{21}})$

⇒ x = 5

a (BC) = 10x = 10 × 5 = 50 cm

b (CA) = 6x = 6 × 5 = 30 cm

c (AB) = 5x = 5 × 5 = 25 cm

The semi-perimeter (s): s = $({\displaystyle \frac{\text{Perimeter}}{2}})$ = $({\displaystyle \frac{105}{2}})$ = 52.5 cm

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Area = $(\sqrt{52.5\times (52.5-50)\times (52.5-30)\times (52.5-25)})$

Area = $(\sqrt{52.5\times 2.5\times 22.5\times 27.5})$

Area = $(\sqrt{99738.28125})$

Area ≈ 284.975 cm²

Area ≈ 285 cm²

∴ The correct answer is option 4.

Given:

Perimeter of triangle ABC = 105 cm

Ratio of altitudes AD : BE : CF = 3 : 5 : 6

Formula used:

The area of a triangle (Area) can be expressed as: Area = $({\displaystyle \frac{1}{2}})\times base\times height.$

For a triangle with sides a, b, c and corresponding altitudes ha, hb, hc, we have:

2 × Area = a × ha = b × hb = c × hc

This implies that the sides of a triangle are inversely proportional to their corresponding altitudes. That is, $a:b:c=({\displaystyle \frac{1}{h_a}}):({\displaystyle \frac{1}{h_b}}):({\displaystyle \frac{1}{h_c}})$

Calculations:

Let the sides of the triangle be a, b, c, where:

a = BC (side opposite to vertex A)

b = CA (side opposite to vertex B)

c = AB (side opposite to vertex C)

Let the altitudes be ha, hb, hc, where:

ha = AD

hb = BE

hc = CF

Given the ratio of altitudes: ha : hb : hc = 3 : 5 : 6.

Using the property that sides are inversely proportional to altitudes:

$a:b:c=({\displaystyle \frac{1}{3}}):({\displaystyle \frac{1}{5}}):({\displaystyle \frac{1}{6}})$

⇒ $a:b:c=({\displaystyle \frac{1}{3}})\times 30:({\displaystyle \frac{1}{5}})\times 30:({\displaystyle \frac{1}{6}})\times 30$

⇒ a : b : c = 10 : 6 : 5

This means the ratio of the sides BC : CA : AB is 10 : 6 : 5.

Therefore, AB : BC : CA = c : a : b = 5 : 10 : 6.

∴ The correct answer is option 2.

Given:

∠P = 54º

Sides AB and AC of ΔABC are produced to D and E, respectively.

The bisectors of ∠CBD and ∠BCE meet at P.

Formula Used:

Exterior Angle Property: ∠P = ∠CBD + ∠BCE

Sum of angles in a triangle: ∠A + ∠B + ∠C = 180º

Calculation:

Using the exterior angle property:

∠P = ∠CBD + ∠BCE

⇒ 54º = ∠B + ∠C

Using the sum of angles in a triangle:

∠A + ∠B + ∠C = 180º

⇒ ∠A + 54º = 180º

⇒ ∠A = 180º - 54º

⇒ ∠A = 72º

The measure of ∠A is 72º.

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.

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

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:

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.

⇒ $\frac{BC}{AC}=\frac{AC}{DC}$

⇒ 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:

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:

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 = 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 = $\frac{AP}{2}$ = $\frac{5}{2}$ = 2.5 cm