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

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:

$m_{AD}=\frac{y_D-y_A}{x_D-x_A}=\frac{6-2}{-2-1}=\frac{4}{-3}=-\frac{4}{3}.$

Because AD ⟂ BC, the slope of BC, $m_{BC}$, satisfies

$m_{AD}\cdot m_{BC}=-1⟹(-\frac{4}{3})m_{BC}=-1⟹m_{BC}=\frac{-1}{-\frac{4}{3}}=\frac{3}{4}.$

Equation of the line BC 

$y-6=\frac{3}{4}(x+2).$

$4(y-6)=3(x+2)⟹4y-24=3x+6.$

$4y-24-3x-6=0$

$⟹-3x+4y-30=0⟹3x-4y+30=0.$

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 $P=(1,\sqrt{2}-1)$.

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:

$\tan ({60}^{\circ })=\frac{h}{x}⟹h=\sqrt{3}x$

$\tan ({45}^{\circ })=\frac{h}{QN}⟹h=QN$

$\tan ({30}^{\circ })=\frac{h}{PN}⟹PN=h\sqrt{3}$

from the figure PN = h + a 

$h+a=h\sqrt{3}$

$h=\frac{a}{\sqrt{3}-1}⟹\frac{a\sqrt{3}+1}{2}$

= $(\frac{3+\sqrt{3}}{2})a$

Hence, the correct answer is Option 2.

Concept:

• Median of a triangle: The line segment joining a vertex to the midpoint of the opposite side.
• Centroid (G): The point of intersection of medians; it divides each median in the ratio 2:1.
• Circle touches triangle side: Use geometrical relationships based on circle properties and distances.
• Important property: If AG:AF = AB², and AG = 2×GD, then algebraic equations help to find required lengths.

Calculation:

Let GD = x, DF = y

⇒ AG = 2x, AF = x + y

⇒ 2x(3x + y) = 36

⇒ xy = 4

⇒ 3x² + 4 = 18

⇒ x² = 14/3

⇒ AD = 3x = √42

Now, AC² + AB² = 2(AD² + BD²)

⇒ AC² + 36 = 2(42 + 4)

⇒ AC² + 36 = 92

⇒ AC² = 56

AC² = 56

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 = $\frac{1}{2}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Calculation:

Here, vertices are (3, 13), (5, -8), and (4, -2)

∴ Area of triangle = $\frac{1}{2}\left|\begin{array}{ccc}3& 13& 1\\ 5& -8& 1\\ 4& -2& 1\end{array}\right|$

$=\frac{1}{2}[3(-8+2)-13(5-4)+1(-10+32)]$

$=\frac{1}{2}|-18-13+22|$

$=\frac{1}{2}|-9|$

$=\frac{9}{2}$ 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 = $\frac{1}{3}$

⇒ $\frac{\pi }{3}<\varphi <\frac{\pi }{2}$

So that $\frac{\pi }{3}+\theta <\varphi +\theta <\frac{\pi }{2}+\theta$

⇒ $\frac{7\pi }{12}<\varphi +\theta <\frac{3\pi }{4}$

Concept:

Area of a triangle = $\frac{1}{2}$ × base × altitude

Area of ΔABC = $\frac{1}{2}(\mathrm{a}\cdot \mathrm{b}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C})$

Calculation:

Area of a triangle = $\frac{1}{2}$ × base × altitude

= $\frac{1}{2}$ × c × a sin∠CBA

= $\frac{1}{2}$ × 10 cm × 4cm sin 30° 

= 5 × 4 × $\frac{1}{2}$ (as sin 30° = $\frac{1}{2}$)

= 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:

$\mathrm{A}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$, Where 's' is semi-perimeter of the triangle.

Semi-perimeter of the triangle = s = $\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$

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 = $\frac{9+12+15}{2}=18$

Now, Area of the triangle formed by median as a side = $\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

$$\begin{gathered} =\sqrt{18(18-9)(18-12)(18-15)} \\ =\sqrt{18\times 9\times 6\times 3} \\ =54 \end{gathered}$$

As we know,

Area of triangle = $\frac{4}{3}$ ×(Area of the triangle formed by median as a side) 

= $=\frac{4}{3}\times 54=72\mathrm{c}{\mathrm{m}}^2$

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 $\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0$

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: $\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Here, x₁ = x, y₁ = - 1, x₂ = 2, y₂ = 1, x₃ = 4 and y₃ = 5

So, area of  Δ ABC = $\frac{1}{2}\cdot \left|\begin{array}{ccc}x& -1& 1\\ 2& 1& 1\\ 4& 5& 1\end{array}\right|$

⇒ 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.