Properties of Triangle MCQ Quiz - Objective Question with Answer for Properties of Triangle - 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 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:

In ΔABC, O is the point of intersection of the bisectors of ∠B and ∠A

∠BOC = 108° 

Concept used:

If the bisector of ∠ABC and ∠ACB of a triangle ABC meet at a point O then, ∠BOC = 90° + \(\frac{1}{{2}} \)∠A

Angle bisector theorem - An angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

Calculation:

If the bisector of ∠ABC and ∠ACB of a triangle ABC meet at a point O then, ∠BOC = 90° + \(\frac{1}{{2}} \)∠A

⇒ 108° = 90° + \(\frac{1}{{2}} \)∠A

⇒ \(\frac{1}{{2}} \)∠A = (108° – 90°)

⇒ \(\frac{1}{{2}}\)∠A = 18° 

⇒ ∠A = 36° 

Now,

We know that, an angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

So,

⇒ ∠BAO = \(\frac{∠ A}{{2}}\)

⇒ ∠BAO = \(\frac{36}{{2}}\)

⇒ ∠BAO = 18° 

∴ The value of ∠BAO = 18° 

Confusion Points Bisector of ∠A, ∠B and ∠C will intersect at a single point i.e O. 

Given:

One side of a right-angled triangle is thrice the other.

The hypotenuse is 8 cm

Concept used:

Area of a right angle triangle = (1/2) × p × b

Where, p = perpendicular, b = base

Calculation:

Let the sides be 3x cm and x cm

⇒ 3x² + x² = 8²

⇒ 9x2 + x2 = 64

⇒ 10x2 = 64

⇒ x2 = 64/10

Area of the triangle = (1/2) × 3x × x

⇒ (1/2) × 3x²

⇒ (1/2) × 3 × 64/10

⇒ 96/10 = 9.6 cm²

∴ The area of the triangle is 9.6 cm2.

Given:

Two acute angles of a right triangle are in the ratio 4:5.

Formula Used:

Sum of acute angles in a right triangle = 90º

Calculation:

Let the acute angles be 4x and 5x.

Sum of acute angles = 90º

4x + 5x = 90º

⇒ 9x = 90º

⇒ x = 90º / 9

⇒ x = 10º

Acute angles = 4x and 5x

⇒ 4 × 10º and 5 × 10º

⇒ 40º and 50º

The acute angles are 40º and 50º.

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:

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:

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:

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

Concept used:

An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Calculation:

​According to the concept,

Considering ΔACD, y + 110° = 120°

⇒ y = 10°

Considering ΔABC, the sum of two interior angles of a triangle is equal to the exterior angle of the third angle.

hence, x + z = 110°

Now, x + y + z

⇒ 110° + 10° = 120°

∴ The measure of x + y + z is 120°.

Given:

The sides of a triangle are in the ratio 4 ∶ 6 ∶ 8. 

Concept used:

The triangle is an obtuse triangle if the sum of the squares of the smaller sides is less than the square of the largest side.

Calculation:

Let the sides of the triangle be 4x, 6x, and 8x respectively.

Now,

(4x)² + (6x)² < (8x)²

⇒ 52x² < 64x²

So, the triangle is an obtuse-angled triangle.

∴ The triangle is obtuse-angled.

Calculation : 

∠BAC = 72° 

By angle sum property