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

Total number of triangle is:

Hence, the correct answer is "Option 2'.

Given

Ratio of length of each equal side and third side of an isosceles triangle is 3 ∶ 4

Concept used

Area of an isosceles triangle = \(\frac{b}{4}\sqrt{({4{a^2} - {b^2}} )}\)

Where a = same side, b = unequal side

Calculation

Let a = 3x and b = 4x

We know that

Area = \(\frac{b}{4}\sqrt{({4{a^2} - {b^2}})} \)

18√5 = \(\frac{4x}{4}\sqrt{({4 × {(3x)^2} - {(4x)^2}} )}\)

⇒ 18√5 = \(x^2\sqrt{({36 - 16})} \)

⇒ 18√5 = x²√20

⇒ 18√5 = 2x²√5 

⇒ x² = 9

⇒ x = 3 unit

Hence, third side = 4 × 3 = 12

∴ The third side is 12 units.

Given:

In a right-angled triangle:

Ratio of shorter side to longer side = 5:12

Length of hypotenuse = 65 cm

Formula used:

Pythagoras theorem: Hypotenuse² = Shorter side² + Longer side²

Perimeter = Shorter side + Longer side + Hypotenuse

Calculation:

Let shorter side = 5x and longer side = 12x

Hypotenuse = 65 cm

⇒ Hypotenuse² = Shorter side² + Longer side²

⇒ 65² = (5x)² + (12x)²

⇒ 4225 = 25x² + 144x²

⇒ 4225 = 169x²

⇒ x² = 25

⇒ x = 5

Shorter side = 5x = 5 × 5 = 25 cm

Longer side = 12x = 12 × 5 = 60 cm

Perimeter = Shorter side + Longer side + Hypotenuse

⇒ Perimeter = 25 + 60 + 65 = 150 cm

∴ The correct answer is option 1.

Given:

The ratio of the sides of the triangle = 1/2 : 1/3 : 1/4

Perimeter of the triangle = 52 cm

Formula used:

Perimeter = Sum of all sides

Calculation:

Let the sides be x/2, x/3, and x/4 (where x is the common factor)

⇒ Perimeter = x/2 + x/3 + x/4 = 52

To add fractions, find the LCM of 2, 3, and 4, which is 12:

⇒ (6x + 4x + 3x) / 12 = 52

⇒ (13x) / 12 = 52

⇒ 13x = 52 × 12

⇒ 13x = 624

⇒ x = 624 / 13 = 48

The sides are:

Side 1 = x/2 = 48/2 = 24 cm

Side 2 = x/3 = 48/3 = 16 cm

Side 3 = x/4 = 48/4 = 12 cm

∴ The length of the shortest side is 12 cm.

Given:

Side 1 = 0.24 m = 24 cm

Side 2 = 28 cm

Side 3 = 32 cm

Formula Used:

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

Where:

s = Semi-perimeter = (a + b + c) / 2

Calculation:

Convert all sides to the same unit:

Side 1 = 24 cm

Side 2 = 28 cm

Side 3 = 32 cm

Semi-perimeter, s = (24 + 28 + 32) / 2

s = 84 / 2

s = 42

Area = √(s × (s - a) × (s - b) × (s - c))

⇒ Area = √(42 × (42 - 24) × (42 - 28) × (42 - 32))

⇒ Area = √(42 × 18 × 14 × 10)

⇒ Area = √(105840)

⇒ Area = 82 √(15) cm²

The area of the triangle is 82 √(15) cm².

Given:

The sides of an equilateral triangle are increased by 34%.

Formula used:

Effective increment % = Inc.% + Inc.% + (Inc.²/100) 

Calculation:

Effective increment = 34 + 34 + {(34 × 34)/100}

⇒ 68 + 11.56 = 79.56%

∴ The correct answer is 79.56%.

Given:

In an isosceles triangle ABC,

AB = AC = 26 cm and BC = 20 cm.

Calculations:

In this triangle ABC,

∆ADC = 90° ( Angle formed by a line from opposite vertex to unequal side at mid point in isosceles triangle is 90°)

So,

AD² + BD² = AB²   (by pythagoras theorem)

⇒ AD² = 576

⇒ AD = 24

Area of triangle = ½(base × height)

⇒ ½(20 × 24)   (Area of triangle = (1/2) base × height)

⇒  240 cm²

∴ The correct choice is option 2.

Semi-perimeter of the triangle (s) = 28/2 = 14

As we know,

Given:

An equilateral triangle ABC is inscribed in a circle with centre O

∠CBD = 40º

Concept used:

The sum of the opposite angles of a cyclic quadrilateral = 180°

The sum of all three angles of a triangle = 180°

Calculation:

∠ABC = ∠ACB = ∠BAC = 60°  [∵ ΔABC is an equilateral triangle]

Also, ∠BAC + ∠BDC = 180°

⇒ 60° + ∠BDC = 180°

⇒ ∠BDC = 180° - 60° = 120°

Also, ∠CBD + ∠BDC + ∠BCD = 180°

⇒ 40° + 120° + ∠BCD = 180°

⇒ ∠BCD = 180° - 40° - 120° = 20°

∴ The value of ∠BCD is 20°

Shortcut Trick

∠BAC = 75º

∠ABC + ∠ACB + ∠BAC = 180°

∠ABC + ∠ACB + 75° = 180°

∠ABC + ∠ACB = 180° - 75° = 105°

Let, ∠ABC = ∠PBQ = 70° and ∠ACB = ∠RCQ = 35°

So, ∠PQR = 180° - (∠PQB + ∠RQC)

= 180° - [(180° - 2∠PBQ) + (180° - 2∠RCQ)  [∵ BQ = PQ; QC = QR]

= 180° - [(180° - 2 × 70°) + (180° - 2 × 35°)]

= 180° - (40° + 110°)

= 180° - 150°

= 30°

Alternate Method

Given:

In a ΔABC, ∠BAC = 75º

BQ = PQ and QC = QR

Concept used:

The sum of all three angles of a triangle = 180°

The sum of all angles on a straight line = 180°

Calculation:

Let, ∠ABC = x and ∠ACB = y

So, ∠ABC = ∠PBQ = ∠QPB = x  [∵ BQ = PQ]

∠ACB = ∠RCQ = ∠QRC = y  [QC = QR]

In ΔABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ x + y + 75° = 180°

⇒ x + y = 180° - 75° = 105°   .....(1)

For ΔBPQ and ΔCRQ,

(∠PBQ + ∠QPB + ∠PQB) + (∠RCQ + ∠QRC + ∠RQC) = 180° + 180° = 360°

⇒ (x + x + ∠PQB) + (y + y + ∠RQC) = 360°

⇒ 2x + 2y + ∠PQB + ∠RQC = 360°

⇒ 2 (x + y) + ∠PQB + ∠RQC = 360°

⇒ (2 × 105°) + ∠PQB + ∠RQC = 360°  [∵ x + y = 105°]

⇒ ∠PQB + ∠RQC = 360° - 210° = 150°   .....(2)

Also, ∠PQB + ∠RQC + ∠PQR = 180°

⇒ 150° + ∠PQR = 180°  [∵ ∠PQB + ∠RQC = 150°]

⇒ ∠PQR = 180° - 150° = 30°

∴ The measure of ∠PQR (in degrees) is 30°

Given,

One of two equal sides of an isosceles triangle, a = 20 cm

Perimeter of the triangle = 72 cm

Formula:

Perimeter of an isosceles triangle = 2a + b

Area of an isosceles triangle = (b/4) × √(4a² – b²)

Calculation:

Let a = 20 cm

2a + b = 72

⇒ 2 × 20 + b = 72

⇒ 40 + b = 72

⇒ b = 72 – 40

⇒ b = 32

Area of isosceles triangle = (32/4) × √(4 × 20² – 32²)

⇒ 8 × √(4 × 400 – 1024)

⇒ 8 × √(1600 – 1024)

⇒ 8 × √576

⇒ 8 × 24

⇒ 192 cm²

∴ Area of the triangle is 192 cm².

Alternate solution

Third side = 72 – 2 × 20 = 72 – 40 = 32

Semi perimeter, s = 72/2 = 36

Now,

Area = √[s (s – a) (s – b) (s – c)] = √[36(36 – 32)(36 - 20)(36 - 20)] = √(36 × 4 × 16 × 16) = 16 × 4 × 3 = 192 cm²

Given 

Area of Quadrilateral = 12√3 cm²

Concept Used 

 We know that The median of a triangle is cut the triangle in equal areas

Area of triangle = (√3/4) (side)² 

Calculation 

⇒ Area of ΔABC = Area of Quadrilateral BDGF × 3 

⇒ Area of ΔABC = 12√3 × 3 

⇒Area of ΔABC = 36√3 cm²

⇒ 36√3 = (√3/4) × (side)²

⇒ side = 12 cm

∴ the side of ΔABC is 12 cm   

Given:

The side of an equilateral triangle is 12 cm.

Concept used:

The radius of a circle circumscribing an equilateral triangle = side/√3

Calculation:

According to the concept,

Radius of the circle = 12/√3

⇒ (4 × 3)/√3

∵ 3 = √3 × √3 = (√3)2

⇒ 4 ×(√3)²/√3

⇒ 4√3

∴ The radius of the circle circumscribing this equilateral triangle is 4√3 cm.

Given data:

CIrcumradius = 3.2 cm

Calculations:

From the equilateral triangle property, O is the circumcenter as well as the centroid.

∴ OD = \(1\over 3\) × AD

⇒ AD = 3 × OD

⇒ AD = 3 × 3.2 = 9.6 cm

∴ The length of the altitudes of the equilateral triangle is 9.6 cm.  

Shortcut TrickWe know that the height of an equilateral triangle = a√3/2

Here, Height = 9 = 6√3 × √3/2 

a = 6√3, so Height = a√3/2.

∴ Given triangle is an equilateral triangle.

So, ∠A = 60°.

Traditional method:

Concept:

Pythagoras theorem:

(AB)² = (BD)² + (AD)²

Equilateral Triangle:

AB = BC = AC

∠A = ∠B = ∠C = 60°

Calculation:

(6√3)² = (BD)² + 9²

⇒ BD = 3√3 cm 

∵ DC = BD = 3√3 cm

∴ BC = AC = AB =  6√3 cm &

∠A = ∠B = ∠C = 60°

∴ ABC is an equilateral triangle.