Triangles MCQ Quiz - Objective Question with Answer for Triangles - Download Free PDF
Last updated on Jul 8, 2025
Latest Triangles MCQ Objective Questions
Triangles Question 1:
Let the area of the triangle formed by a straight line \(\mathrm{L}: \mathrm{x}+\) by \(+\mathrm{c}=0\) with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of \(45^{\circ}\) with the positive \(x\)-axis, then the value of \(b^{2}+c^{2}\) is
Answer (Detailed Solution Below)
Triangles Question 1 Detailed Solution
Explanation:
\(\frac{c^{2}}{2|b|}=48 \Rightarrow c^{2}=96|b|.\)
\(-\frac{1}{b}=1 \Rightarrow {b} = -1 \Rightarrow {c}^{2}=96 \)
\(b^{2}+c^{2}=97 \)
Hence Option 3 is the correct answer.
Triangles Question 2:
ABC is an equilateral triangle and AD is the altitude on BC. If the coordinates of A are (1,2) and that of D are (−2,6), then what is the equation of BC?
Answer (Detailed Solution Below)
Triangles Question 2 Detailed Solution
Calculation:
Given vertices A(1, 2) and D(−2, 6), where AD is the altitude from A onto BC in an equilateral triangle ABC.
Compute the slope of AD:
\(m_{AD} = \frac{\,y_D - y_A\,}{\,x_D - x_A\,} = \frac{\,6 - 2\,}{\,-2 - 1\,} = \frac{4}{-3} = -\tfrac{4}{3}.\)
Because AD ⟂ BC, the slope of BC, \(m_{BC}\), satisfies
\(m_{AD} \cdot m_{BC} = -1 \;\Longrightarrow\; \Bigl(-\tfrac{4}{3}\Bigr) \,m_{BC} = -1 \;\Longrightarrow\; m_{BC} = \frac{-1}{\, -\tfrac{4}{3}\,} = \tfrac{3}{4}.\)
Equation of the line BC
\(y - 6 = \tfrac{3}{4}\,(x + 2).\)
\(4(y - 6) = 3(x + 2)\;\Longrightarrow\;4y - 24 = 3x + 6.\)
\(4y - 24 - 3x - 6 = 0 \)
\(\;\Longrightarrow\; -3x + 4y - 30 = 0 \;\Longrightarrow\; 3x - 4y + 30 = 0.\)
Hence, the correct answer is Option 4.
Triangles Question 3:
The vertices of a triangle are A(1, 1),B(0, 0) and C(2, 0). The angular bisectors of the triangle meet at P. What are the coordinates of P?
Answer (Detailed Solution Below)
Triangles Question 3 Detailed Solution
Calculation:
Given points A(1,1), B(0,0), and C(2,0). The angle bisectors meet at P (the incenter).
Compute the side lengths:
\(a = BC = \sqrt{(2 - 0)^{2} + (0 - 0)^{2}} = 2\)
\(b = CA = \sqrt{(1 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}\)
\(c = AB = \sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}\)
Therefore, \(a + b + c = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}.\)
By the incenter formula,
\(X_{P} \;=\; \frac{a\,x_{A} + b\,x_{B} + c\,x_{C}}{a + b + c} \;=\; \frac{\,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 2\,}{\,2 + 2\sqrt{2}\,} \;=\; \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} \;=\; 1.\)
\(Y_{P} \;=\; \frac{a\,y_{A} + b\,y_{B} + c\,y_{C}}{a + b + c} \;=\; \frac{\,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 0\,}{\,2 + 2\sqrt{2}\,} \;=\; \frac{2}{2 + 2\sqrt{2}} \;=\; \frac{1}{\,1 + \sqrt{2}\,} \;=\; \sqrt{2} \;-\; 1. \)
∴ The incenter is \(P = \bigl(1,\;\sqrt{2} - 1\bigr) \).
Hence, the correct answer is Option 1.
Triangles Question 4:
Comprehension:
The top (M) of a tower is observed from three points P, Q and R lying in a horizontal straight line which passes directly along the foot (N) of the tower. The angles of elevations of M from P, Q and R are 30°, 45° and 60° respectively. Let PQ = a and QR = b
What is MN equal to?
Answer (Detailed Solution Below)
Triangles Question 4 Detailed Solution
Calculation
\( \tan(60^\circ) = \frac{h}{x} \Rightarrow h = \sqrt{3}x \)
\( \tan(45^\circ) = \frac{h}{b + x} \Rightarrow h = b + x \)
\( h = b + x \Rightarrow \sqrt{3}x = b + x \)
\( x = \frac{b(\sqrt{3} + 1)}{2} \)
\( h = MN = \sqrt{3}x = \left( \frac{3 + \sqrt{3}}{2} \right) b \)
Hence, the correct answer is Option 1.
Triangles Question 5:
Comprehension:
The top (M) of a tower is observed from three points P, Q and R lying in a horizontal straight line which passes directly along the foot (N) of the tower. The angles of elevations of M from P, Q and R are 30°, 45° and 60° respectively. Let PQ = a and QR = b
What is PN equal to?
Answer (Detailed Solution Below)
Triangles Question 5 Detailed Solution
Calculation:
The angles of elevation are:
From point P,\(\theta = 30^\circ \); from point Q, \(\theta = 45^\circ \); and from point R, \(\theta = 60^\circ \)
Using the tangent formula for each angle:
\( \tan(60^\circ) = \frac{h}{x} \implies h = \sqrt{3}x \)
\( \tan(45^\circ) = \frac{h}{QN} \implies h = QN\)
\( \tan(30^\circ) = \frac{h}{PN} \implies PN = h{\sqrt{3}} \)
from the figure PN = h + a
\(h+a = h\sqrt3\)
\(h = \frac{a}{\sqrt3 -1} \implies \frac{a \sqrt3 +1}{2}\)
= \((\frac{3 + \sqrt3}{2}) a\)
Hence, the correct answer is Option 2.
Top Triangles MCQ Objective Questions
In quadrilateral ABCD, ∠C = 72° and ∠D = 80°. The bisectors of ∠A and ∠B meet at a point O. What is the measure of ∠AOB ?
Answer (Detailed Solution Below)
Triangles Question 6 Detailed Solution
Download Solution PDFDetailed Solution:
As we know,
∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + 72° + 80° = 360°
⇒ ∠A + ∠B = 360° – 152° = 208°
In ΔAOB
∠A/2 + ∠B/2 + ∠AOB = 180°
⇒ ∠A/2 + ∠B/2 + ∠AOB = 180°
⇒ ∠AOB = 180° - (∠A + ∠B)/2
⇒ ∠AOB = 180° – 208°/2
∴ ∠AOB = 180° – 104° = 76°
Additional Information Short Trick:
As we know,
2∠AOB = ∠C + ∠D
⇒ 2∠AOB = 72° + 80°
⇒ ∠AOB = 152°/2 = 76°
Find the area of triangle whose vertices are (3, 13), (5, -8), and (4, -2)
Answer (Detailed Solution Below)
Triangles Question 7 Detailed Solution
Download Solution PDFConcept:
Area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression
Area = \(\frac{1}{2} \left |\begin{array}{ccc} \rm x_1&\rm y_1&1\\\rm x_2&\rm y_2& 1\\\rm x_3&\rm y_3&1\end{array}\right|\)
Calculation:
Here, vertices are (3, 13), (5, -8), and (4, -2)
∴ Area of triangle = \(\frac{1}{2} \left |\begin{array}{ccc} \rm 3&\rm 13&1\\\rm 5&\rm -8& 1\\\rm 4&\rm -2&1\end{array}\right|\)
\(=\frac12[3(-8+2)-13(5-4)+1(-10+32)]\)
\(=\frac12|-18-13+22|\)
\(=\frac12|-9|\)
\(=\frac92\) sq. units
Hence, option (4) is correct.
If perpendicular of a right angled triangle is 8 cm and its area is 20 cm2, the length of base is?
Answer (Detailed Solution Below)
Triangles Question 8 Detailed Solution
Download Solution PDFGiven:
Perpendicular of right angled triangle = 8 cm
Area = 20 cm2
Formula used:
Area of right angled triangle = (1/2) × perpendicular × base
Calculation:
⇒ 20 cm2 = (1/2) × 8 × base
⇒ base = 20/4
⇒ 5 cm
∴ The length of base is 5 cm
Let θ and ϕ be an acute angle such that sin θ = \(\frac{1}{\sqrt2}\) and cos ϕ = \(\frac{1}{3}\), the value of θ + ϕ is:
Answer (Detailed Solution Below)
Triangles Question 9 Detailed Solution
Download Solution PDFSolution:
Given, sin θ = \(\frac{1}{\sqrt{2}}\)
⇒ \(\theta=\frac{\pi}{4} \)
and cos ϕ = \(\frac{1}{3}\)
⇒ \(\frac{\pi}{3}<\phi <\frac{\pi}{2}\)
So that \(\frac{\pi}{3}+\theta <\phi+\theta <\frac{\pi}{2}+\theta \)
⇒ \(\frac{7\pi}{12} <\phi+\theta <\frac{3\pi}{4} \)
What is the area of the triangle ABC with sides a = 10cm and c = 4cm angle B = 30°?
Answer (Detailed Solution Below)
Triangles Question 10 Detailed Solution
Download Solution PDFConcept:
Area of a triangle = \(1\over 2\) × base × altitude
Area of ΔABC = \(\rm {1\over2} (a \cdot b\cdot sin \ C)\)
Calculation:
Area of a triangle = \(1\over 2\) × base × altitude
= \(1\over 2\) × c × a sin∠CBA
= \(1\over 2\) × 10 cm × 4cm sin 30°
= 5 × 4 × \(1\over 2\) (as sin 30° = \(1\over 2\))
= 10 cm2
The length of three medians of a triangle are 9 cm, 12 cm and 15 cm. Then the area of triangle is:
Answer (Detailed Solution Below)
Triangles Question 11 Detailed Solution
Download Solution PDFConcept:
Area of triangle = \(\frac{4}{3}\) ×(Area of the triangle formed by median as a side)
The area of a triangle whose side lengths are a, b and c is given by:
\(\rm A = \sqrt{s(s-a)(s-b)(s-c)}\), Where 's' is semi-perimeter of the triangle.
Semi-perimeter of the triangle = s = \(\rm \frac{a+b+c}{2}\)
Calculation:
Given: length of three medians of a triangle are 9 cm, 12 cm and 15 cm
Let s be semi-perimeter of the triangle formed by median as a side
∴ s = \(\rm \frac{9+12+15}{2} = 18\)
Now, Area of the triangle formed by median as a side = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{18(18-9)(18-12)(18-15)} \\= \sqrt{18 \times 9\times 6\times 3} \\= 54\)
As we know,
Area of triangle = \(\frac{4}{3}\) ×(Area of the triangle formed by median as a side)
= \(\rm = \frac{4}{3}\times 54 = 72 cm^2\)
On the basis of angle, how many types of triangle are there?
Answer (Detailed Solution Below)
Triangles Question 12 Detailed Solution
Download Solution PDFCalculation:
On the basis of angle, there are 3 types of triangle
(i) Obtuse angled traingle
(ii) Acute angled triangle
(iii) Right angled triangle
∴ There are 3 types of triangle on the basis of angle
A circle is circumscribing a triangle whose sides are 30 cm, 40 cm and 50 cm. Find the circumference of the circle.
Answer (Detailed Solution Below)
Triangles Question 13 Detailed Solution
Download Solution PDFGiven:
A circle is circumscribing a triangle whose sides are 30 cm, 40 cm and 50 cm
Concept Used:
Circumference of circle = 2πr
Calculation:
The triangle described is a right triangle (since \(30^2 + 40^2 = 50^2\)), also known as a Pythagorean triplet.
In a right triangle, the circumradius (radius of the circle circumscribing the triangle) is half the length of the hypotenuse.
Since the hypotenuse is 50 cm, the circumradius is 50/2 = 25 cm.
The circumference of the circumscribing circle is 2 × π × 25 cm = 50π cm.
∴ Option 4 is the correct answer.
If O is the orthocentre of a ΔABC, ∠BOC = 100° and ∠AOB = 90°, the measure of ∠ABC is
Answer (Detailed Solution Below)
Triangles Question 14 Detailed Solution
Download Solution PDFGiven:
If O is the orthocentre of a ΔABC, ∠BOC = 100° and ∠AOB = 90°
Definitions:
Orthocentre is the intersection point of all the altitudes from all the three vertices of a triangle
if In ΔABC, CE and BD are altitudes and they intersect at O then,
∠BOC + ∠BAC = 180∘
The sum of all the angles around a point is 360°.
Calculations:
According to the question,
∠AOB + ∠BOC + ∠AOC = 360° (complete angle)
⇒ 90° + 100° + ∠AOC = 360°
⇒ ∠AOC = 360° - 190° = 170°
Now, ∠AOC = 180° – ∠ABC
⇒ ∠ABC = 180° – ∠AOC
⇒ ∠ABC = 180° – 170° = 10°
∴ The measure of ∠ABC is 10°.
The value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear is
Answer (Detailed Solution Below)
Triangles Question 15 Detailed Solution
Download Solution PDFCONCEPT:
If three points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of Δ ABC is zero i.e \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)
CALCULATION:
Given: The points (x, -1), (2, 1) and (4, 5) are collinear
Let A = (x, - 1), B = (2, 1) and C = (4, 5)
Let's find the area of the Δ ABC
As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the Δ ABC then area of Δ ABC is given by: \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)
Here, x1 = x, y1 = - 1, x2 = 2, y2 = 1, x3 = 4 and y3 = 5
So, area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x}}&{{-1}}&1\\ {{2}}&{{1}}&1\\ {{4}}&{{5}}&1 \end{array}} \right|\)
⇒ Area of Δ ABC = 2 - 2x
∵ The points A, B and C are collinear ⇒ Area of ΔABC = 0
⇒ 2 - 2x = 0
⇒ x = 1
Hence, option C is the correct answer.