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

Given:

ΔABC is inscribed in a circle with Centre O.

AB = 21 cm, BC = 20 cm, AC = 29 cm

Formula used:

Circumradius (R) of a triangle = \(abc\over4Δ\)

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

Calculation:

Sides of the triangle are 21 cm, 20 cm, and 29 cm

Since (21, 20, 29) is the triplet. so the given triangle is a right-angle triangle.

So, area of the triangle = 1/2 × base × perpendicular

⇒ Area = 1/2 × 20 × 21 = 210 cm²

Now,

Circumradius = (21 × 20 × 29) / (4 × 210)

⇒ 29/2 = 14.5 cm

∴ The correct answer is option (4).

Given:

Perimeter of equilateral ΔABC = 72 cm

O is the circumcenter, D is the midpoint of BC

Formula used:

Side length of an equilateral triangle: \( a = \dfrac{\text{Perimeter}}{3} \)

Radius of circumcircle: \( R = \dfrac{a}{\sqrt{3}} \)

Median length: \( AD = \dfrac{\sqrt{3}}{2} \times a \)

Centroid divides median in 2:1 ratio: \( OD = \dfrac{1}{3} \times AD \)

Calculation:

\( a = \dfrac{72}{3} \) = 24 cm

\( R = \dfrac{24}{\sqrt{3}} \)

⇒ \( R = \dfrac{24\sqrt{3}}{3} \) = 8√3 cm

\( AD = \dfrac{\sqrt{3}}{2} \times 24 \)

⇒ \( AD = 12\sqrt{3} \)

\( OD = \dfrac{1}{3} \times 12\sqrt{3} \)

⇒ \( OD = 4\sqrt{3} \)

∴ The length of OD is 4√3 cm.

Given:

O is the incenter of triangle PQR.

S is a point on QR such that OS ⊥ QR.

∠QOS = 15°.

Formula used:

The incenter is the intersection of the angle bisectors of a triangle.

The sum of angles in a triangle is 180°.

In a right-angled triangle, the sum of the two acute angles is 90°.

Calculation:

In triangle OQS, ∠OSQ = 90° (given OS ⊥ QR).

∠QOS = 15° (given).

Therefore, ∠OQS = 180° - (90° + 15°) = 180° - 105° = 75°.

Since O is the incenter, OQ bisects ∠PQR.

Therefore, ∠PQR = 2 × ∠OQS = 2 × 75° = 150°.

∴ ∠PQR = 150°.

Given:

In triangle ΔPQR, PT is the median. We are asked to determine the correct relationship among the given options.

Concept used:

In any triangle, the following relationship holds when a median is drawn from a vertex to the midpoint of the opposite side:

The correct formula is:

PQ² + PR² = 2(PT² + QT²)

This formula is derived from the median theorem, which states that the sum of the squares of the two sides of a triangle (PQ and PR) is equal to twice the sum of the square of the median (PT) and the square of half the base (QT), i.e., the length of the segment from P to the midpoint of QR.

Conclusion:

The correct answer is: PQ² + PR² = 2(PT² + QT²)

Given:

ABC is a right-angled triangle. A circle is inscribed in it.

The length of the two sides containing the right angle are 10 cm and 24 cm

Calculations:

Hypotenuse² = 10² + 24²    (Pythagoras theorem)

Hypotenuse = √676 = 26

Radius of the circle (incircle) inside a triangle = ( Sum of sides containing right angle – Hypotenuse)/2

⇒ (10 + 24 - 26)/2

⇒ 8/2

⇒ 4

∴ The correct choice is option 4.

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:

In ΔABC, O is the orthocenter and I is the incenter for the given triangle,  

If ∠BIC - ∠BOC = 90∘.

Formula used:

(1) In ΔABC, I is the incenter for the given triangle,

(1.1) ∠BIC = 90^∘ + \(\frac{1}{2}\)∠A

(1.2) ∠AIC = 90∘ + \(\frac{1}{2}\)∠B

(1.3) ∠AIB = 90∘ + \(\frac{1}{2}\)∠C

(2) In ΔABC, O is the orthocenter for the given triangle,

(2.1) ∠BOC = 180^∘ - ∠A

(2.2) ∠AOB = 180∘ - ∠C

(3.3) ∠AOC = 180∘ - ∠B

Calculation:

According to the question, the required image is:

As we know,

∠BOC = 180∘ - ∠A     ----(1)

∠BIC = 90∘ + \(\frac{1}{2}\)∠A    ----(2)

Now, subtract equation (1) from (2).

⇒ ∠BIC - ∠BOC = 90∘ + \(\frac{1}{2}\)∠A  - (180∘ - ∠A )

⇒ 90^∘ = 90∘ + \(\frac{1}{2}\)∠A  - 180∘ + ∠A

⇒ 90∘ = \(\frac{3}{2}\)∠A  - 90∘

⇒ 180∘ = \(\frac{3}{2}\)∠A

⇒ ∠A = 120^∘

∴ The required answer is 120∘.

Additional Information

(1) Incenter - It is the intersection point of all three angle bisectors of a triangle.

(1.1) Angle bisector cuts the angle into two equal half.

(2) Orthocenter - It is the intersection point of all three altitudes drawn from the vertex to the opposite side of the triangle.

(2.1) The altitude of a triangle is perpendicular to the opposite side.

Given:

In ΔPQR,

PQ = 12 cm, QR = 16 cm and PR = 20 cm

Concept used:

a, b and c denote the sides of the triangle and A denotes the area of the triangle,

Then the measure of the circumradius(r) is.

r = [abc/4A]

If the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the biggest side is a right angle.

Hypotenuse² = Perpendicular² + Base² 

Calculation:

Here, we can see that 

(20)2 = (16)2 + (12)2  = 400 

⇒ PR² = QR² + PQ²

So, ΔPQR is a right angle triangle.

Area of ΔPQR = (½ ) × base × Perpendicular

⇒ Area of ΔPQR = (½ ) × 16 × 12

⇒ Area of ΔPQR = 96 cm²

Circumradius (r) = [abc/4 × Area] 

Circumradius (r) = [(12 × 16 × 20)/4 × 96] = 10 cm

∴ The circumradius of a triangle is 10 cm.

For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse. All the vertices of a triangle are at an equal distance from the circumcenter.

PO = QO = OR = r

⇒ PO = PR/2

⇒ PO = 20/2 = 10 cm

∴ The circumradius of a triangle is 10 cm.

Concept used:

Length of circumradius = (A × B × C)/(4 × Δ)    

where A, B, C are the lengths of the 3 sides of the triangle

Δ → Area of the triangle.

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

where a, b, c are the lengths of the 3 sides of the triangle

s → Semiperimeter

Calculation: 

s = (a + b + c)/2

⇒ s = (10 + 9 + 17)/2

⇒ s = 18

Area of ΔABC = √{18(18 - 10)(18 - 9)(18 - 17)

⇒ √(18 × 8 × 9 × 1)

⇒ 36 cm²

Circumradius of the triangle = (a × b × c)/4Δ

⇒ (10 × 17 × 9)/(4 × 36)

⇒ 10.6 cm

∴ Circumradius of the triangle is 10.6 cm

Given:

In-centre O and ∠QOR = 110° 

Formula used:

In triangle PQR, If O is the in-centre then

∠QOR = 90° + ∠QPR/2

Calculation:

According to the question

∠QOR = 90° + ∠QPR/2

110° = 90° + ∠QPR/2

⇒ ∠QPR/2 = 20° 

⇒ ∠QPR = 40° 

∴ ∠QPR is 40°.

Additional Information 

Incenter, ∠BIC = 90° + ∠P/2

Circumcenter, ∠BcC = 2∠A

Given:

XY = 20 cm

XZ = 25 cm

YP = a cm

PZ = (a + 3) cm

Concept used:

In a given triangle, if AD is angle bisector of ∠BAC, and I is incenter then

BD ∶ DC = c ∶ b

AI ∶ ID = (c + b) ∶ a

Calculation:

As, XP is angle bisector and I is incenter, then

YP ∶ PZ = XY ∶ XZ

⇒ a ∶ (a + 3) = 20 ∶ 25

⇒ a/(a + 3) = 4/5

⇒ 5a = 4a + 12

⇒ a = 12 cm

YZ = a + a + 3 = 2a + 3

⇒ YZ = 27 cm

XI ∶ IP = (XY + XZ) ∶ YZ

⇒ XI ∶ IP = (20 + 25) ∶ 27 = 45 ∶ 27

∴ XI ∶ IP is 5 ∶ 3

Given:

POR = 140 degree

Concept used:

The incenter of a triangle is equally inclined to all the sides of the triangle.

Angle at incentre = 90° + vertex angle/2

Calculation:

According to the concept,

90° + ∠PQR/2 = 140°

⇒ ∠PQR/2 = 140° - 90°

⇒ ∠PQR/2 = 50°

⇒ ∠PQR = 100°

∴ The angle PQR is 100°.

Given:

In ΔABC, perpendicular bisectors AG, BH and CI meet at O.

∠B = 44°, ∠C = 66°

Concepts used:

Sum of all three angles of triangle is 180°.

The points where perpendiculars meet is orthocentre.

Angle formed at orthocentre by given side = [180° - (angle opposite to given side)]

Calculation:

According to sum angle property of triangle,

∠A + ∠B + ∠C = 180°

⇒ ∠A + 44° + 66° = 180°

⇒ ∠A = 180° - 110°

⇒ ∠A = 70°

Angle formed at orthocentre by given side = [180° - (angle opposite to given side)]

⇒ ∠BOC = 180° - 70° = 110°

∴ ∠BOC is equal to 110°.

Confusion Points The given point is orthocentre, according to the property of orthocentre,

The angle formed on orthocentre = [180° - (angle opposite to given side)]

Given:

ZL and YM intersect each other at G.

Concept:

Median divides the triangle into two equal parts.

Calculation:

If mid points of sides of a triangle joined, then the triangle is divided into four equal parts, [see figure (1)]

Let area of each triangle ΔXLM = ΔYLP = ΔPMZ = ΔLMP = 6 units

Total area of ΔXYZ = 24 units

Median divides the triangle into two equal parts, [see figure (2)]

Area of each triangle ΔLGM = ΔMGP = ΔPLG = 2 units

∴ Area of ΔGLM : Area of ΔXYZ = 2 : 24 = 1 : 12.