Triangles MCQ Quiz - Objective Question with Answer for Triangles - Download Free PDF
Last updated on Jun 3, 2025
Latest Triangles MCQ Objective Questions
Triangles Question 1:
In Δ ABC, AB = 12 cm, BC = 16 cm and AC = 20 cm. A circle is inscribed inside the triangle. What is the radius (in cm) of the circle?
Answer (Detailed Solution Below)
Triangles Question 1 Detailed Solution
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).
Triangles Question 2:
In ΔABC, AB = AC = 12 cm, BC = 5 cm and D is a point on AC such that DB = BC. What is the measure of CD?
Answer (Detailed Solution Below)
Triangles Question 2 Detailed Solution
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.
Triangles Question 3:
In Δ PQR, PS is the median of the triangle. QR is the base. If the value of PQ is 20 cm and PS is 15 cm and SR is 10 cm. Then find the value of PR.
Answer (Detailed Solution Below)
Triangles Question 3 Detailed Solution
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.
Triangles Question 4:
In ∆LMN, medians MX and NY are perpendicular to each other and intersect at Z. If MX = 20 cm and NY = 30 cm, what is the area of ∆LMN (in cm2)?
Answer (Detailed Solution Below)
Triangles Question 4 Detailed Solution
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.
Triangles Question 5:
In Δ ABC = ∠A = 2∠B. The bisector of ∠A meets BC at D such that CD = 4 cm and BD = 6 cm. The length of side AC is:
Answer (Detailed Solution Below)
Triangles Question 5 Detailed Solution
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.
Top Triangles MCQ Objective Questions
In the triangle ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°. What is the value of the length of the side BC?
Answer (Detailed Solution Below)
Triangles Question 6 Detailed Solution
Download Solution PDFGiven:
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.
The perimeter of a triangle with sides of integer values is equal to 13. How many such triangles are possible?
Answer (Detailed Solution Below)
Triangles Question 7 Detailed Solution
Download Solution PDFConcept 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.
What is the radius of circle which circumscribes the triangle ABC whose sides are 16, 30, 34 units, respectively?
Answer (Detailed Solution Below)
Triangles Question 8 Detailed Solution
Download Solution PDFGiven:
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
The lengths of the three sides of a triangle are 30 cm, 42 cm and x cm. Which of the following is correct?
Answer (Detailed Solution Below)
Triangles Question 9 Detailed Solution
Download Solution PDFGiven:
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.
ABC is a triangle and D is a point on the side BC. If BC = 16 cm, BD = 11 cm and ∠ADC = ∠BAC, then the length of AC is equal to:
Answer (Detailed Solution Below)
Triangles Question 10 Detailed Solution
Download Solution PDFGiven:
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.
In a triangle ABC, angle B = 90° and p is the length of the perpendicular from B to AC. If BC = 10 cm and AC = 12 cm, then what is the value of p?
Answer (Detailed Solution Below)
Triangles Question 11 Detailed Solution
Download Solution PDFGiven:
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
'O' is a point in the interior of an equilateral triangle. The perpendicular distance from 'O' to the sides are \(\sqrt3\) cm, 2\(\sqrt3\) cm, 5\(\sqrt3\) cm. The perimeter of the triangle is :
Answer (Detailed Solution Below)
Triangles Question 12 Detailed Solution
Download Solution PDFGiven:
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.
In triangle ABC, AD is the angle bisector of angle A. If AB = 8.4 cm and AC = 5.6 cm and DC = 2.8 cm, then the length of side BC will be:
Answer (Detailed Solution Below)
Triangles Question 13 Detailed Solution
Download Solution PDFGiven:
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.
In the given figure, AB = DB and AC = DC. If ∠ABD = 58° and ∠DBC = (2x - 4)°, ∠ACB = (y + 15)° and ∠DCB = 63°, then the value of 2x + 5y is :
Answer (Detailed Solution Below)
Triangles Question 14 Detailed Solution
Download Solution PDFGiven:
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°.
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. What is the length (in cm) of MN, if BC = 10 cm and AC = 15 cm?
Answer (Detailed Solution Below)
Triangles Question 15 Detailed Solution
Download Solution PDFGiven:
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