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

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. 

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.

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.

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.

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.

Detailed 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°

Concept:

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. 

Given:

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

Solution:

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 \(\frac{\pi}{3}+\theta <\phi+\theta <\frac{\pi}{2}+\theta \)

⇒ \(\frac{7\pi}{12} <\phi+\theta <\frac{3\pi}{4} \)

Concept:

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

Concept:

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) 

= \(\rm = \frac{4}{3}\times 54 = 72 cm^2\)

Calculation:

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

Given:

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.

Given:

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°.

CONCEPT:

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.